Logarithmic Single X Axis Graph Paper: Fill & Download for Free

“The problem of distinguishing prime numbers from composite numbers and of resolving the latter into their prime factors is known to be one of the most important and useful in arithmetic. It has engaged the industry and wisdom of ancient and modern geometers to such an extent that it would be superfluous to discuss the problem at length…”-- Karl Friedrich Gauss, Disquisitiones Arithmeticae (translation: A. A. Clarke)Number theory, or higher arithmetic, is the study of integers. It's a collection of "magic tricks" that enable mathematicians to manipulate and understand big numbers more easily. However, there appears to be one thing that's missing from that box of tricks:A means to relate the quadratic growth of square numbers with the asymptotic growth of prime numbers.Gauss was not interested in the next prime number, per se, but was interested in how they were distributed. How does the composition sound, as a whole? On the back page of his book of logarithms, he wrote:The prime counting function, denoted by π(x), is the number of primes less than or equal to a real number x. Gauss’ conjecture that π(x) is very well approximated by the logarithmic integral Li(x):The first table of Gauss, covers the primes from 1 to 50,000, where each entry in the table represents an interval of length 1000. Thus, for example, there are 168 primes from 1 to 1000; 135 primes from 1001 to 2000; 127 primes from 3001 to 4000; and so forth. Gauss suspected that the density with which primes occurred in the neighborhood of the integer n was 1/ ln n, so that the number of primes in the interval [a, b] should be approximately equal to:The prime number theorem, in the form π(x) ∼ Li(x), was conjectured by Gauss in 1849, in a letter to Enke. Gauss made his conjecture on the basis of his calculations (done by hand) of π(x) and contemporary tables of values of Li(x). A short table of his findings appears in his letter to Enke:Gauss investigates the distribution of primes up to 3,000,000 and compares the number of primes found with the above integral, the agreement turning out to be striking. The only information beyond Gauss' tables concerning Gauss' work on the distribution of primes is contained, in a 1849 letter to Encke.In his letter, Gauss describes his numerical experiments and his conjecture concerning π(x), while, on the second page of the letter, he compares his approximation to π(x), namely li(x), with Legendre’s formula. The results are tabulated at the top of the second page and Gauss' formula yields a much larger numerical error. In a very prescient statement, Gauss defends his formula by noting that although Legendre's formula yields a smaller error, the rate of increase of Legendre's error term is much greater than his own. Anyway, as we shall see later, in a certain sense Gauss anticipated what is today known as the Riemann hypothesis .Gauss demonstrated that the rate of accumulation of primes slows down. We can make a diagram of this, where we count out the numbers along a horizontal line, and place a mark above each number which represents the number of primes beneath that number. You will find that chart below, compared with the rate of growth of Gauss' natural logarithm function:So, they're not exactly equal. As will be seen with more study of Gauss's work, he never relied on a formula, though his papers are absolutely full of them. He performed experiments, developed hypotheses, tested them with more experiments, and constantly changed his hypotheses until his death.What does this mean? Is that higher principle which orders the prime numbers, the same principle which generates the arithmetic and geometric progressions?Gauss had recognized that, no matter how the primes were grouped, their distribution would always be inversely proportional to the natural logarithm. Here is a graph showing how many primes are between each successive group of a thousand numbers. The red curve is [1/(Log n)].Less than a decade later, a young student of Gauss, Bernhard Riemann, gave his first lecture at Göttingen University. It was a lecture on the distribution of prime numbers, and he asserted that the distribution hypothesized by Gauss would always get closer and closer to the actual distribution of the prime numbers. This has been known since then as the "Riemann Hypothesis," and scientists have been fighting to verify it ever since. Some still argue that, since Riemann hasn't demonstrated the hypothesis for the last prime number, then it can't be asserted as true!The infinitude of primeThe prime numbers have fascinated mathematicians for many centuries. As far back as 300 BCE, Euclid proved that there are infinitely many primes but like all others was not able to show how this works. There appears to be nothing that can be used to uniquely identify prime number positions and we are left struggling to explain why these positions continue to appear as numerical ranges increase. There is no need to reinvent the wheel, in this matter, but there is a definite need to find a pathway to supersede it. Therefore, I am going to offer a short description of why I think there are an infinite number of prime numbers.Firstly, there is a very tough problem called Goldbach's Conjecture that states that all even integers are formed by the addition of two prime numbers. It is clear that the most common difference between prime numbers is a multiple of 2. The next common number is 30. Anyone with a quick mind will realize that this is merely 2 x 3 x 5. Multiples of 3 and 5 repeat in blocks of 30. The simple reason behind this is if you take any other even number less than 30, and add multiples of it to any prime number, it will land on many prime number positions , but it will also land on multiples of 3 and 5, which cannot be prime numbers. The numbers 6, 12, 18, and 24 do land less on multiples of 3 and 5, whereas 30, when successfully added to prime numbers other then 3 and 5, never lands on multiples of these numbers.In order to get some perspective, you need to put into a table, odd numbers arranged in blocks of 30. Any number that is not a multiple of 3 or 5 falls into one of the Prime number channels ( 1. 7. 11. 13. 17. 19. 23. 29.) we can create a multiplication table based on the interaction between numbers that are in these channels to see what distribution patterns exist. This table is known as a Cayley table and it is Abelian which means it is symmetrical.I have called the channels, the Prime Number Channels.Because all these channels represent a number of 30s + either 1 (or 30Mod1), 7 (or 30Mod7), 11 or (30Mod11), 13 (or 30Mod13), 17 (or 30Mod17), 19 (or 30Mod19), 23 (or 30Mod23) or 29 (or (30Mod29), we can derive a series of simple parametric equations that describe all the non-prime numbers that fall into these spaces.Equations for Channel 1.A) (900AB + 30A + 30B) + 1B) (900AB - 30A- 30B) + 1C) (900AB + 330A + 330B) + 121D) (900AB - 330A - 330B) + 121E) (900AB + 390A + 210B) + 91F) (900AB - 390A - 210B) + 91Equations for Channel 19A) (900AB + 210A + 210B) + 49B) (900AB – 210A – 210B) + 49C) (900AB +390A + 390B) + 169D) (900AB – 390A – 390B) + 169E) (900AB + 570A + 30B) + 19F) (900AB – 570A – 30B) + 19Equations for Channel 11A) (900AB + 330A + 30B) + 11B) (900AB – 330A – 30B) + 11C) (900AB + 690A + 210B) + 161D) (900AB – 690A – 210B) + 161Equations for Channel 13A) (900AB + 390A + 30B) + 13B) (900AB – 390A – 30B) + 13C) (900AB + 570A + 210B) + 133D) (900AB – 570A – 210B) + 133Equations for Channel 17A) (900AB + 330A + 210B) + 77B) (900AB – 330A – 210B) + 77C) (900AB + 510A + 30B) + 17D) (900AB - 510A - 30B) + 17Equations for Channel 23A) (900AB +390A + 330B) +143B) (900AB – 390A -330B) + 143C) (900AB + 570A + 510B) + 323D) (900AB - 570A – 510B) +323Equations for Channel 29A) (900AB + 510A + 210B) + 119B) (900AB – 510A – 210B) + 119C) (900AB + 570A + 330B) + 209D) (900AB – 570A – 330B) + 209There are four basic effects to consider;1) for every 30 consecutive numbers that a number is multiplied by, 15 of these products fall on even numbers, 7 fall on multiples of 3 and 5, and the remaining 8 fall into the prime number channels.2) For each new prime number, the expanse over which it's multiplication products span for a consecutive series of 30 numbers is larger than the previous ones.3) Larger numbers absorb more numbers as factors4) Many numbers fail to fall into a zone through multiplication.All these effects combine to form an overall balancing effect that ensures continuing prime numbers being found.Euclid's proof for an infinite number of prime numbers can be modified in the light of knowledge about the Prime Number Channels. For example,Channel 7 Equations (30Mod7)From the list of factors for channel 7, we have 1 and 7, 11 and 17, 13 and 19, and 23 and 29.A) 1 and 7 => (30A+1) (30B+7) = (900AB + 210A + 30B) + 7B) 11 and 17 => (30A+11)(30B+17) = (900AB + 510A + 330B) + 187Look at each set of factors for each channel and you can see that there are pairs of numbers that are equidistant from each 30 boundary and in the next two equations this can be used to demonstrate the behaviour of all non-prime numbers above a certain number of being equidistant from other numbers.C) 13 and 19 are treated as -17 and -11=> (30A-17)(30B-11) = (900AB - 510A - 330B) + 187 D) Similarly 23 and 29 are -7 and -1=> (30A-7)(30B-10 = (900AB - 210A - 30B) + 7If pairs of numbers from each value of A and B are used as products such that for equation A)(7 x 31) x (37 x 61) x (67 x 91) x ... and then add 30, the answer is either a prime number or has factors that include a number higher than any of the numbers used above, or has factors from the other equations.If this exercise is repeated with pairs of numbers from each of the individual equations, then the answer will be a prime or the product of higher prime numbers.Goldbach’s ConjectureGoldbach posed in 1742 in a letter to Leonhard Euler, that every even integer > 2 can be written as the sum of two primes.The relevant part of the letter to Euler reads:The Prime Number Channels give one way of gaining a deep understanding of Goldbach's conjecture, using possible alignments that give the sum of two different primes as shown in the table below,The following graph of prime number pairs for even integers up to 1 million has some obvious patterns but no previous simple explanation exists for these to explain them.However, it can be shown that the prime number channel alignments per block of 30 make a perfect fit for these bands. That is, plotting the maximum number of alignments against the number of prime pairs.The way I built the datasets is relatively easily done. First create a database or array of the first 500,000 prime numbers. Then create a second database or array representing the even numbers up to 1,000,000, each record or element having a field to store the number of prime number pairs and a field to store what alignment type it is.Then create two loops where one starts at the first prime, while the second loop goes through the remainder of the prime numbers, forming different sums which then become an index into the second database or array which has one added to its total of prime pairs seen.Next set the alignment type as either 3, 4, 6 or 8 for each record.You now have a complete data set from which to pull different combinations of records by alignment type and other conditions.It is clear there are very specific dynamics going on which may help explain some very deep details about prime numbers. Prime number pairs according to the above graphs are concentrated largely around very specific regions with lesser regions attracting lesser quantities.When looking at the raw unplotted data, some of the points that represent high numbers of prime number pairs are divisible by 11 or 13, but most are divisible by 7. Those that are divisible by 11 and have higher numbers of prime number pairs tend to also be divisible by other low prime numbers such as 7, 13, 17 and 19.Why does this make such a difference? Take 98 which is divisible by 7. Each pair of odd numbers that are divisible by 7 are also divisible by other factors and its likely that this prevents the same factors restricting the number of pairs of primes. When a member of one pair of odd numbers is divisible by several factors, the next and previous multiples of those factors are even numbers and therefore only able to reduce the numbers of prime number pairs when they are at least double the factor in distance away. Here are the pairs of odd numbers that are divisible by 7;Therefore when N is divisible it itself by many factors, then multiples of those factors take out many other factors and increase the likelihood of more prime number pairs.Conversely if N has only 2 as a factor, does that mean that it has fewer pairs of primes? Powers of 2 cycle through the same four positions within blocks of multiples of 30. Multiples of 3 and 5 are ignored as they play an increasingly insignificant part in larger value integers.For different powers of 2, something completely different is happening. Multiples of 3 and 5 cannot pair up to their own multiples and instead pair up with all other available prime numbers. In fact all prime numbers cannot align with multiples of themselves when N = different powers of 2.Powers of 2 therefore largely fall into the bottom zone of alignments of 3.There is a deeper level that operates on a more independent basis and this relies on the attribute that prime numbers and squares of primes do not exist as the difference between the squares of non-consecutive numbers.Legendre’s ConjectureThere is a prime number between n^2 and (n+1)^2 for every positive integer n.As part of this view is the concept that groups of prime numbers are generated in-between boundaries that are consecutive squares themselves. Substantiating that concept is the observation of a relationship between the number of prime numbers that exist between successive prime number boundaries and the square of the upper or lower boundary. Gauss's equation was refined several times to show an exponential relationship between the number of primes and the size of a number. This new relationship is different in that it looks at the number of prime numbers between explicit boundaries and is based on an additional attribute of prime numbers. When this relationship is plotted out over large numbers, it starts as a curve and then decreases very slowly to form an almost linear relationship. New prime numbers occur between the squares of consecutive numbers, again something simple that has never been shown before.The graph appears below. The number of prime numbers between consecutive squares are plotted against the square root of the lower of the two consecutive squares. For example, the range of 9^2 to 10^2 yields primes of 83, 89, and 97 against 9.The trend downward starts to tail off, and this type of curve is asymptotic. It lends great weight to the concept that prime numbers are naturally created between consecutive squares. As the numbers increase, the noise, or amplitude, lessens and tends to even out.The result is asymptotic. The following graph shows the first 5000 boundaries. The first 1000 boundaries (integers) show a lot of noise, but this starts to settle down and smooth out in the next 3000 boundaries. The amplitude also grows smaller as the square of the integer or boundary increases.The graph below shows a comparison between a square interval count vs log count.The Y axis is the number of primes; the X axis is the root of the perfect square. For example, in this graph, 324 (on the X axis) means the 324th perfect square root, so the perfect square is 324 * 324 = 104976. The previous perfect square is 323 * 323 = 104329. The difference (104976 - 104329 = 647) is the quadratic interval. The natural log of both perfect squares is approximately 11.56, so the expected count is (104976 / 11.56) - (104329 / 11.56) = 51. The actual prime count for this interval is 58.This graph does NOT show the cumulative prime count: real or expected. It shows the prime count from one perfect square to the next. This is something quite different than you will most likely see at other math sites where the prime-counting function is explained and graphed as a cumulative total, with the line showing a step-like pattern. This graphs show the total number of primes for each successive quadratic interval - and it demonstrates that the number of primes is increasing absolutely for successive intervals.Gauss's equation operates in a large space without identifiable boundaries other than a log scale. There is no obvious relationship between the log scale other than the one demonstrated by a graph. But this new graph is a representation of the boundaries between groups of prime numbers and thus is much much closer to the nature of prime numbers.I could go further and say the ratio of the real prime count to the square interval is nearly linear!It would also be possible to plot another graph where the number of prime numbers is used that occurs in-between the squares of positive numbers and this will give a similar asymptote. It is also possible to plot the distance between squares against the number of prime numbers to give a similar result to the above graph.All these graphs show an unmistakable relationship between prime number boundaries and prime number density.Secondly, the idea of the general number field sieve is based on a very old factoring strategy going back to Fermat and Legendre which is to write a number as a difference of two squares. Many factoring algorithms exploit this idea include the continued fraction method, the random squares method, the quadratic sieve, and, of course, the special number field sieve.Consider the following example – the number 21 has the following pairs of squares and numbers that produce this sum.2^2 + 173^2 + 124^2 + 5The middle one has both the square and number being divisible by 3 and so 21 is not a prime number. Therefore a prime number can be defined as a number where all the pairs of squares and numbers that sum together to produce it have no factors in common between each pair.This definition has a utility by allowing a variation in the way that a number is tested to see if it is a prime. The first test is to see if the test number is a square of another number. If it is not then subtract 4 from it and test the remainder to see if it is divisible by 2 which it can never be.Take the power series of squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 101, 121, etc., which represents the entire series of rational whole numbers. We shall call them Perfect squares or PS for argument sake. The difference between consecutive perfect squares (2*2, 3*3, 4*4) is an odd number that grows by 2 at each step.Now 1 is 1, but it's also the first perfect square: 1 * 1 = 1, so is it the oneness of 1 that ensures that a PS -1 is always a composite number - or is it because of its property of squareness...?It seemed possible that if this rule held for PS-1, it might hold for PS-N, where N is another perfect square - just maybe....Fermat’s Last TheoremThe initial idea comes from Fermats Last Theorem, consequently, I restate, here, literally, what Fermat had written, in Latin, as a marginal note, in his copy of Diophante’s {Arithmetica}:“It is impossible for a cube to be the sum of two cubes, or for a fourth power to be the sum of two other fourth powers, or, in general, for any power other than two to be the sum of two similar powers.”Fermat’s idea was a direct response to the earlier challenge issued by Pythagoras, in confirming that the ancient Egyptians and Greeks had already understood that within the domain of {Sphaerics}, X^2+ Y^2 = A^2 represented the only form of summation of powers.In the same note, Fermat added:“I have discovered a truly marvelous demonstration of this proposition, but this margin is too narrow to contain it.”The crucial point that I wish to make, here, is that Fermat’s demonstration was never found because empiricists refused to look for it in Diophante. In his {Arithmetica}, Diophante proved the crucial point, in Book II, sections VIII, and IX, where Fermat had appended his note, and where Diophante showed his method of how to divide a square into two other squares. Diophante wrote:“Divide a given square into two other squares. {Example} Let 16 be the given square, and I will call N^2 and 16 - N^2 the sought for squares. N remains to be found, in such a way that 16 - N^2 be a square. I pose that 16- N^2 = (2N – 4)^2, thus N = 16/5.” ({Précis des oeuvres mathématiques de P. Fermat et de l’arithmétique de Diophante}, par E. Brassinne, Imprimerie de Jean-Mathieu Douladoure, Toulouse, 1853, p. 53.) The interesting question, here is, how did Diophante derive 16/5?Consider for example: Diophantus’ treatment of problem VIII, and IX where it is required that the square A^2 be divided into a sum of two squares:X^2 + Y^2 = A^2…………….A^2 = 16……………..(3)Rational solutions of this equation are e.g. (0, a) and (0, -a). In order to determine other solutions, Diophante substitutes:Y = kX - A…………….(4)Indeed he says: “I form the square from any number of X minus as many units as there are in the side of 16, and let it be 2X - 4.” Diophantus chooses 2 as one of the possible numbers while 4 is held fixed so that Eq. (4), with a fixed value of A, is an adequate letter formula of Diophantus’ substitution.The solution of Eq. (3) now leads to(kX - A)^2 = A^2 - X^2 ,so thatX = (2Ak)/k^2+1 and Y = A (k^2 - 1)/(k^2 + 1)Thus X and Y are represented by the rational parameter k. Assuming A^2 = 16 and k = 2, Diophantus arrives at X = 16/5 and Y = 12/5, saying nothing, however about the number of solutions, although this method makes it possible to calculate an infinity of them. He himself mentions this fact elsewhere, (Book III problem 19): “We saw how to divide a square into an infinite number of ways”Essentially, any theoretical FLT equation has a mirror equation with similar attributes. It’s relatively easy to show that for any Z^N - Y^N, the difference is divisible by (Z-Y). And also that Z^N - X^N is divisible by (Z-X). You can do this by treating the difference between Z and Y as W and rewriting the equation as Z^N - (Z-W)^N = X^N. No matter what N is (2 or more), X^N, if it exists, must have W as a factor.Wiles’ proof must be an absolute overkill, because it assumes that the problem is the difference between two numbers raised to the same power overlapping with a third number to the same power. From what I understand, it doesn’t take into account that there is a tight relationship between the numbers, which means that the correct solution for this problem would show why this is important.Two variables I’ll call Z2 and Y2, which are also (Z-Y) apart from each other, can be shown to definitely exist, which must have the same answer as X^N. In other words, if a number has a factor of (Z2-Y2) then there are there must be a Z2 and Y2 that form (Z2-Y2)(Z2+Y2). Note, therefore, that all odd numbers that are not prime or the squares of primes are the difference between two non-consecutive squares. The reason is because (Z2-Y2) cannot equal (Z2+Y2) and therefore can only be a non-prime number that has either 2 different factors or multiple factors or the same factor raised to a power higher than 2. And all even numbers are also the difference between the squares, although they are not always integers, for example, 2 x 7 = 14 is 4.52 - 2.52The difference between non-consecutive squares can be understood with what happens when numbers either side of 8 are multiplied by each other.Both the lower and upper factors are equidistant from 8, and the product equals the difference between two non-consecutive squares, where the difference between those square roots just happens to be the same as the lower factor. In FLT equations, the lower factor must also be the same difference between a set of non-consecutive squares.What is intriguing about this problem is the fact that the difference between Z2 and Y2 has to be (Z-Y). If X = (Z-Y) then it is clear that no matches can be found, and somehow the difference being divisible by (Z-Y) still has a part in ensuring that no FLT equations exist.From above, every potential FLT equation also equals a difference between squares (Z2^2 - Y2^2) which also equals X^N. This problem can therefore be broken down into three possibilities:(Z-Y) is a factor of (Z2+Y2). If this is true then there is a simple way to show that Z, Y and Z must have a factor in common. Therefore you end up with a situation of Reductio Ad Absurdem because once you divide out the common factor, you are still left with a FLT equation.(Z-Y) or (Z2-Y2) is a number (N1) raised to the power of N and (Z2+Y2) is another number (N2) raised to the power of N, and they have no factors in common. Therefore X^N = (Z-Y) x (Z2+Y2) = N1^N * N2^N where (Z-Y) = (Z2-Y2) = N1^N. The difference between (Z2-Y2) and (Z2+Y2) is 2Y^2, and therefore N1^N + 2Y^2 = N2^N(Z-Y) or (Z2-Y2) shares a factor in common with (Z2+Y2). This is most probably a situation leading to Reductio Ad Absurdem also, as it shows that (Z-Y) and the product of division, once X^N is divided by (Z-Y), still contains a factor of (Z-Y).Consider the following observation: The number that immediately precedes a perfect square is never prime with the single exception of 4 - 1 = 3.The huge surprise is that the rule holds true for every perfect square. In every case - with the important and irregular exception of the first subtraction the numbers produced by subtraction of perfect squares are composite numbers.It turns out that 9 - 4 = 5 is a prime number of course. In other words, this is a single exception (like 4 - 1 = 3) to the otherwise compositeness of all perfect squares minus 4. Similarly, 36 - 25 = 11 is also an exception to the otherwise uniform compositeness.We can now state a "fundamental theorem of perfect square subtraction" such that every odd natural number is the subtractive product of two, and only two, adjacent perfect squares.Fermat's factorization method says that by adding N to perfect squares (PSs) one will find eventually a sum N + PS that is itself a perfect square. When that happens one can know the factors of N by adding and subtracting the square roots of PS and N + PS.First, one must ask "why use perfect squares at all?" if the factoring is ultimately done with the square roots of the perfect squares.So let's consider for a moment how we apply this method to a composite with two prime factors (a semiprime):Say N = 32951If we begin with 1 + 32951 and continue through all the perfect squares (2*2 , 3*3, 4*4 and so on), we will find that 24649 (157*157) + 32951 finally equals another perfect square: 57600 (240*240).So 24649 + 32951 = 57600at which point we know that240 + 157 = 397240 - 157 = 83and that83 * 397 = 32951All well and good, but factoring 32951 this way requires evaluating 157 perfect squares.Using the same principle as for perfect square subtraction, the addition of perfect squares to any odd number provides a simple deterministic test for compositeness.I am calling this a test for compositeness, rather than primality, for a couple of reasons:The test produces a positive proof of compositeness but a negative proof of primality (by elimination).(... and from this it follows that...)It will always prove compositeness faster than primality.The basis for a deterministic test predicated solely on the addition of perfect squares can be stated as follows:N is composite if there exists a perfect square that when added to N creates another perfect square and providing that the two perfect squares are not adjacent.By adjacent, I am referring to two perfect squares whose roots are in counting order (so that Ra = Rb + 1). So, for example, 25 and 36 are adjacent because their respective roots - 5 (5*5=25) and 6 (6*6=36) - are separated by 1.In number theory, the fundamental theorem of arithmetic (or unique-prime-factorization theorem) states that every natural number greater than 1 can be written as a unique product of prime numbers.So, for example, the unique prime factorization of 999 is: 3 * 3 * 3 * 37.We can now state a "fundamental theorem of perfect square subtraction" such that every odd natural number is the subtractive product of two, and only two, adjacent perfect squares.You will see that to obtain through subtraction, say, 999, you will need to subtract some relatively large perfect squares. The unique perfect square subtraction that produces 999 is: 250000 (500*500) - 249001 (499*499).There is a further statement we can make about composite numbers that is of great significance because we cannot make it about prime numbers:*For every composite number that is not itself a perfect square there exists a pair of nonadjacent perfect squares whose subtractive product is equal to the composite.**If only it were easier to find the nonadjacent squares than it is to find their factors! If such a method exists, it has never been disclosed - and would, in fact, have epic consequences for the theory of complexity classes and the secure application of information technology.**If you've been paying attention, you may notice a very beautiful fact - namely that for any N the square roots of the nonadjacent squares are the progenitors of its factors. If the number is a semiprime, they are the progenitors of its prime factors.What do I mean by progenitor?The difference between these two square roots gives one factor; the sum of these two square roots gives the other factor.It's that simple - and it's true for every composite that's not a perfect square.Thus, for 21, the square roots of the nonadjacent squares are;sqr(4) = 2 and sqr(25) = 5.5 - 2 = 35 + 2 = 73 * 7 = 21How simple and satisfying is that!There are a few statements that can be made about the nonadjacent perfect squares:We can say with certainty that the nonadjacent perfect squares cannot be greater than the adjacent perfect squares because that would mean the subtractive product would be greater than the composite being considered.It would appear that, if one of the adjacent squares is close to the upper bound, the other should be close to the lower bound (that is, 1).There is the very important caveat that the rule of nonadjacent perfect squares does not apply to odd perfect squares. For example, the odd perfect square 9 (3*3) cannot be created by nonadjacent squares less than its adjacent squares (16, 25). This is a property that the odd squares share with all primes. It needs to be noted and understood otherwise, any algorithm we attempt to construct from our observations so far will incorrectly find these odd perfect squares to be primes.Semiprimes in PThirdly and finally, aside from the single-digit prime numbers 2 and 5, all other prime numbers can only end in one of four digits: 1, 3, 7, or 9. (If a number ends in 2, 4, 6, 8 or 0, it will be divisible by 2. If it ends in 5, it will be divisible by 5.) Thus, if they were truly random, a prime number that ends in 1 should be followed by another prime number ending in 1 about 25% of the time. That is, this kind of pairing should occur at least one in four times.In 2016 Kannan Soundararajan and Robert Lemke Oliver checked the first billion prime numbers, they found that a prime number ending in 1 is followed by another also ending in 1 about 18% of the time. That is, this kind of pairing occurred only one in five times. Instead, that prime number was followed by a prime number ending in 3 or 7 about 30% of the time and by 9 about 22% of the time. This result holds true for prime numbers ending in 3, 7 or 9 too, but with slightly less bias.Numbers ending in a 1, 3, 7, or 9 can also be formed by multiplying two numbers with different combinations of the ending digit.The inability to separate out some of these combinations from others has meant that no simplification of multiplication has been achieved since prime numbers were first observed. This demonstrates that a number ending in 1, 3, 7, or 9 can be formed in 4 different ways, that is for 1→ (1, 1) (3, 7) (7, 3) (9, 9), 3→ (1, 3) (3, 1) (7, 9) and (9, 7), 7→ (1, 7) (7, 1) (3, 9) and (9, 3), 9→ (1, 9) (9, 1) (3, 3) (7, 7). Using the Prime number channels, some of these multiplication products are located in some channels, while others resided in different channels. For example: channel 7 could be formed by a multiple of 30 + 1 (or known as 30Mod1) multiplied by a number that is a multiple of 30 + 7 (or 30Mod7). The other possibilities were 30Mod11 x 30Mod17, 30Mod13 x 30Mod19 and 30Mod23 x 30Mod29. If you wanted to test a channel 7 number, you now only needed to test 30Mod7, 30Mod13, 30Mod17 and 30Mod23 numbers as possible factors. No longer is it necessary to test multiples of 3, 5, or all of the channel numbers.A Deeper Analysis of Goldbach's Strong ConjectureThe Prime Number Channels give one way of gaining a deep understanding of this conjecture. There is a deeper level that operates on a more independent basis and this relies on the attribute of Prime Numbers presented above which shows how both prime numbers and the squares of prime numbers do not exist as the difference between non-consecutive squares.To prove this conjecture one must first show that there is a reason why prime numbers align at equal distances in great numbers either side of the mid-point of any even number.The historical prime number definition is of little value because it doesn’t suggest anything remotely useful. But the difference between squares gives rise to the powerful notion that prime numbers naturally occur between consecutive squares. It also leads to a sense of why there are prime numbers when one attempts to put pressure on the equation for the differences between non-consecutive squares, that is, by subtracting a prime number or the square of a prime number from a square creates Virgin Prime Number positions because it forces the squares to be consecutive. For example, 3^2 is the difference between 5^2 and 4^2. The pair of squares are of course the two consecutive numbers that sum together to form 3^2. Because they are consecutive numbers, they share no factors in common and therefore they cannot be recombined into any other square. Also because a prime or prime number squared cannot be recombined into another combination, they are only the difference between consecutive squares.Consider 10^2 as an example.Subtracting 3^2 or 7^2 leaves 91 and 51 remainder respectively. These numbers have only two factors each; 7 x 13 and 3 x 17.Both 7+13 and 3+17 produce the same sum of 20, and the member of each pair is the same distance from 10 as the other member of each pair.In both examples, a prime subtracted from 10 produces a prime number, and a prime number added to 10 produces a prime number.The relevance of 10^2 is that the sum of prime number factors equidistant by another prime number from 10 is 2x10 or 20 i.e. 10-P + 10+P = 2 x 10Instead of considering any even number N, this problem now shifts to considering both N^2 and 2N.If instead of an even number, an odd number squared is used, something similar happens except in this case the prime number subtracted is multiplied by a multiple of 2.This is clearly not a proof by any means but shows that when an additional behaviour of prime numbers is applied to this problem, a prime number building mechanism is revealed that depends on lower prime numbers. It also shows that this conjecture is not purely finding enough aligned prime numbers, but about what happens at the level 2N that leads to the generation of the second pair of prime numbers which is used for N^2.The following table shows 2N, N and N^2 for the even integers between 4 and 30. One is generally not regarded by modern convention as a prime number but is never-the-less the product of a square (1) and can therefore be used in the following table. The second reason I have used it, is that it falls in Prime Number Channel 1.I have excluded pairs which are a straight doubling of a prime number.Deducting the square of a prime number from the square of an even number, particularly if the prime number is large compared to the even number, reduces the probability that the remainder will have multiple factors. Because the difference cannot be a prime number, it therefore has to be a number that has fewer factors. Many of these will be just two factors which is why prime number pairs do tend to form even integers. This is by no means any claim to a proof, but gives a simple description of why this trend occurs. The other factors that combine to strengthen this trend are that the prime number being deducted as a square and the even square cannot share common factors with difference between their squares.The difference of squares shows that there is a hidden symmetry to the placement or distribution of primes. It also lends great weight to the concept that prime numbers are naturally created between adjacent or consecutive squares.The symmetry breaking mechanism shows that prime numbers and the squares of primes can never be the difference between nonadjacent or non-consecutive squares.NP vs PAll problems are hard unless proven otherwise, not vice versa. To clarify, a combinatorial task is called "hard" if it can be shown that no computer algorithm can solve it in "polynomial time." This results from the fact that as a certain parameter of the problem increases, the time required to solve the problem grows at an exponential, or "nonpolynomial," rate. Studies of such problems belong to a branch of mathematics and computer science called "complexity theory." A great deal of work has been done and is continuing on a special class of problems called NP-complete (NP for nondeterministic-polynomial). There are now hundreds of such problems, all believed to be hard (though no proof has yet been found), and all related so that if an algorithm is found for solving one of them in polynomial time, it will at once solve all of them.While factoring large numbers is not provably difficult, it is a well-known problem that has been worked on for the last three hundred years by many famous mathematicians. Fermat (1601?-1665) and Legendre (1752-1833) developed factoring algorithms; some of today’s more efficient algorithms are based on the work of Legendre, however, no one has yet found an algorithm which can factor a 200-digit number in a reasonable amount of time. The factoring of large numbers is not in the NP-complete family, but it is thought to be hard, (By “hard” I mean that the number does not have a small prime factor that could be discovered by trial division or the elliptic curve method, nor does the number succumb easily to other special methods such as the p - 1 factoring method, quadratic sieve or the special number field sieve.) and so far no one has found a way to factor large numbers in polynomial time. However, such techniques have been steadily improving along with methods of testing the primality of big numbers.Fermat’s ‘little theorem’ and Carmichael NumbersOn October 18th 1640, Fermat wrote in a letter to Fenicle, that whenever p is prime, p divides X^P-1 - 1 for all integers p divisible by p, a result now known as Fermat’s ‘little Theorem’. An equivalent formulation is the assertion that p divides X^P - x for all integers x, whenever p is prime. The question naturally arose whether the primes are the only integers exceeding 1 that satisfy this criterion, but Carmichael pointed out in 1910 that 561 (=3 x 11 x 17) divides X^561 - x for all integers x.Noting this phenomenon above, if there are an infinite number of primes p, q and r, of the form p = 6N + 1, q = 12N + 1, r = 18N + 1 are all prime then, pqr is a Carmichael number. According to Hardy and Littlewoods widely believed prime k-tuplets conjecture, these should simultaneously be prime infinitely, which would tell us that they are infinitely many Carmichael numbers.Primes in PThat’s where things stood until 2002. Then “AKS”—Manindra Agrawal, Neeraj Kayal, and Nitin Saxena, from the Indian Institute of Technology at Kanpur—announced a fast primality test that always works under no assumptions and is not randomized. Not only does the test never fail, but it’s considerably simpler than older tests (particularly Adleman- Huang). On the other hand, the AKS test is also slower than older tests. Originally the number of steps in AKS grew like the number of digits of N raised to the 12th power. Lenstra improved the results to the number of digits raised to roughly the 6th power. For older tests, by comparison, the growth rate is like the number of digits cubed.So how does the AKS test work? The key is the Pascal’s Triangle;If N is prime, then the Nth row contains only multiples of N, ignoring the 1’s on the left and right. But if N is composite, then the Nth row has numbers other than 1 that are not multiples of N—for instance, 15 and 20 in the 6th row.Let’s try to see why that is. In the Nth row of Pascal’s Triangle, the Kth number from the left (starting at 0) isSuppose we pick two huge prime numbers, say P and Q. (A technical requirement is that neither P −1 nor Q−1 is a multiple of 3.) We know the product of P and Q, call it N.To clarify let us define Y =X^3modNwhere “X^3 modN” just means the remainder when N goes into X^3modNGiven Y , we need to find X.We not only know N, but the prime factors P and Q of N. Remember Fermat’s Little Theorem says if P is prime and X is not a multiple of P , then P divides X^P −1 − 1. It turns out that this can be generalized to composite numbers as follows:If N = PQ where P and Q are primes, and X is relatively prime to N, then N divides X^(P−1)(Q−1) − 1. It follows that N divides X^1+(P−1)(Q−1) − X, or in other words, X^1+(P−1)(Q−1) modN equals X. Now think about this sequence:For a fixed N, each value in the sequence depends only on the last one. Therefore, since there are N possible values, (0 to N − 1), the sequence has to repeat eventually. What’s more, by the above, the period of the sequence (the number of values until it starts repeating itself) has to divide (P − 1)(Q − 1). For observe that it makes no difference whether we take the remainder mod N at every step, or whether we wait until the end to do so. In fact, we could imagine an ‘alternate universe’ of arithmetic, where everything is a remainder mod N! Let’s do an example. Take N = 15 and any X relatively prime to N, say X = 2. The prime factors of 15 are P=5 and Q=3. So (P−1)(Q−1)=8, and we predict that the period of the sequence should divide 8. Does it?Indeed! The period is 2 which divides 8. So the general law is:Our goal is to find X, given Y = X^3 modN. Recall that (P −1)(Q−1) is not divisible by 3, since neither P −1 nor Q−1 is. It’s easy to check, then, that either 1+(P −1)(Q−1) or 2+(P −1)(Q−1) must be divisible by 3. Suppose 1 + (P − 1)(Q − 1) is, and letThenAll we need to do is raise Y to the C power and take the remainder mod N, and we get X. Only one question remains: how do we raise Y to the C power? One way would be just to multiply Y by itself C times. But if C is large, then this method takes way too long. A better method uses what’s called repeated squaring. Starting from Y , it’s easy to produce Y raised to any power of 2:Since we take the remainder mod N after each squaring step, the numbers never get too big to handle. Now, we can write any whole number C as a sum of powers of 2—for example, if C = 25, then C = 16+8+1. Then (in this example) Y^C = Y^16+8+1 = Y^16 ×Y^8 ×Y^1. So we can write Y^C as a product of terms, each of which we can compute using repeated squaring.The security of this system rests on the assumption anyone who knows Y and N but not the prime factors P and Q of N, would have an extremely hard time figuring out X. So it goes without saying that, if there’s a fast method for factoring N into P × Q, then the whole scheme is broken.If N is prime, then every number in the Nth row of Pascal’s Triangle (besides 1) is a multiple of N. If N is composite, this is no longer true, there are numbers besides 1 in the Nth row that are not multiples of N. A corollary of this fact (and of Fermat’s Little Theorem) is that, for any A that’s relatively prime to N,for all values of X, if and only if N is prime.Let’s do two examples. Take N = 3 and A = 2. Thenwhich is (X^N + A). So the identity holds, and we conclude that 3 is prime. Now take N = 4 and A = 3. ThenBecause of the extra 2X^2 term, this is not equal to X^N + A = X^4 + 1 for all values of X. So we conclude that 4 is composite, even though we haven’t learned what its prime factors are.This means that, if only we could examine the whole binomial expansion of (X + A)^N , we’d have an infallible test for whether N is prime! If all the “middle” terms have coefficients divisible by N, then N is prime; otherwise N is composite. What’s the trouble with this binomial test? Same as with the original Pascal’s Triangle test: for huge N, there are too many terms in the binomial expansion to check them all! The idea of AKS was to speed up the binomial test, while still being able to prove that it gives the right answer whenever N is composite. The way they speed things up is to work, not only mod N, but also mod a polynomial X^R − 1 (where R is a reasonably small prime). That is, instead of computing (X + A)^N , they compute the remainder when (X + A)^N is divided by X^R − 1. That’s done using the same method you learn in algebra class for dividing one polynomial by another. What’s nice about the remainder is that it has only R + 1 terms, as opposed to N + 1 for the expansion of (X + A)^N . So if R is small enough, you can actually compute the remainder—using the squaring trick just shown above, except with polynomials instead of whole numbers.Now, since the original identity was always true when N is prime, it’s obvious that the new identityis also always true when N is prime. If two things are equal, then the remainder when you divide X^R − 1 into the first thing equals the remainder when you divide X^R − 1 into the second! What isn’t obvious is that the new identity is always false when N is composite. For even though,it’s possible that two different polynomials could leave the same remainder when we divide X^R −1 into both of them. What AKS showed was this: if N is composite, and if you choose the “right” value of R, then you need to try only a small number of A’s until you find one such thatOnce you find such an A, you’ve proved that N is composite. What’s more, you don’t have to pick the A’s at random; there’s a deterministic way to do it.Throughout their proof of this claim, AKS use “heavy-duty” math in only one place: they appeal to a theorem of Fouvry to show that the “right” value of R can be found efficiently. Aside from that, the proof rests entirely on facts that any undergrad math major would know. I’ll end on that note.My thumbnail thesis provides evidence that integrating the AKS algorithm into the prime number channels equation for very large numbers with more than 200 digits. Factorization can be done efficiently, and "efficiently" means "one that requires shockingly less time than you would expect for a problem of this size" because no matter how many channels we apply the AKS algorithm to, we are simultaneously finding the factors with no additional time penalties. It’s like parallel processing because the complete predictability of the product location account for the polynomial speed of this algorithm.A further investigation into semiprimes could yield very interesting results within the framework of perfect square boundaries. For example:Let's consider all integers less than 50000 and identify how many are semiprimes. First we will eliminate all evens and all primes. That leaves 24330 composite candidates. Next we must eliminate composites with more than two factors. This leaves 11740 candidate semiprimes. Finally, we will eliminate the perfect squares (though technically semiprimes). This leaves a total of 11576. Of these semiprimes, how many conform to the type that the next perfect square minus N equals any perfect square (x^2)? It doesn't matter how large N is in this special case - it could be 200 or 2000 digits - if it conforms to this simple formula, it can be factored instantly. There are 228 such semiprimes under 50,000.…Not suprisingly, these semiprimes are the symmetrical kind - in fact, they would appear to be the 'hardest' kind for oversized Ns - where the factors are fairly close to the square root.The table above also shows that the two factors F1 and F2 combine differently in each boundary forming the composite number in the 1st column.1*1 F1*F2 form the composite that can only end in 3 or 92*2 (1 or 7) 47*43 =2021 or 113*109=123173*3 (1, 7)4*4 (3, 9)5*5 (1, 9)6*6 (3, 9)7*7 (1, 7)8*8 (3, 9)9*9 (3, 9)10*10 (1, 9)11*11 (3, 9)12*12 (1, 7)13*13 (1, 7)14*14 (3, 9)15*15 (1, 9)16*16 (3, 9)17*17 (1, 7)18*18 (1, 7)19*19 (3, 9)20*20 (1, 9)25*25 (1, 9)30*30 (1, 9)It would be interesting to note the only exceptions are boundaries of 5*5, 10*10, 15*15 and 20*20 etc.. whose factors F1*F2 are the same, that is either both factors end in (1, 1), (3, 3), (7, 7) or (9, 9) to form composites ending in (1, 9). Also composites ending in (1 or 3), (3, 7) or (7, 9) never pair together within square boundaries to form these kind of semiprimes. Instead of testing all composites, we only need to test two types of composites in each square boundary for semiprime factors. Prime number channels 1, 11, 19 and 29 form all composites ending in (1 or 9).We can now state the twin-prime composite conjecture, that is:Of the list of composite numbers between any square interval there is at least one semiprime 2y and one semiprime 3z.Researching prime numbers is a trying and frustrating occupation, so much work for so little return. You have to be very persistent in something you believe in, that no one else does. This is precisely the basis on which I proceeded and the following observations were a direct result of this approach;Prime Number ChannelsEquations governing non-prime numbers in these channelsThe repeating patterns of multiplication using consecutive blocks of 30 Phases of multiplicationA structural basis for understanding Goldbach's ConjectureAlternative Prime Number definitionsDiscovery of natural boundaries for groups of Prime NumbersDiscovery of the relationship between boundaries and Prime Number densityThere are many such avenues of research I have briefly highlighted, which may suggest some more discoveries that may have been left out, however, the symmetry of the Prime Number Channels suggests very strongly to me that some further understanding lies in these particular channels and the equations that populate them.Dirichlet’s PrincipleSimply, Dirichlet observed that linear polynomials of the type ax + b may provide infinite prime numbers when x runs through all the positive integers and a, b were positive integers with no prime factor in common (like 2x + 1, 4x + 1, 4x + 3 and so on); it also follows that there exist infinite primes in the arithmetical progression k, k + l, k + 2l, k + 3l, .... Dirichlet, in proving this his outcome, went outside the realm of integers, introducing tools of analysis to pursue this. Furthermore, Dirichlet's work contained two radically new ideas: to be precise, Dirichlet's ideas gave birth to the modern theory of duality on locally compact Abelian groups, while Dirichiet's second great idea opened the way to the well-known Dirichlet L-function theory, as it nowadays is called, thanks to which he proved a notable theorem on primes in arithmetic progressions, which states that for any two positive coprime integers n and m ( if the only positive integer that equally divides both of them, is 1), there are infinitely many primes of the form n + lm where l is a non-negative integer, that analytic methods (in this case the study of the Dirichlet L-series) could be fruitfully applied to arithmetic problems (in this case the problem of primes in arithmetic progressions).Dirichlet's idea in proving the infinitude of primes in the arithmetic progression a, a + n, a + 2n, ..., (a, n) = 1, was to imitate, somehow, Euler's proof of the infinitude of primes, by studying the function L(s, χ) for s near 1. The first giant strides toward a proof of the prime number theory were taken by B. Riemann in a memoir written in 1859. Riemann followed Dirichlet in connecting problems of an arithmetic nature with the properties of a function of a continuous variable. However, where Dirichlet considered L-functions L(s, χ) as functions of a real variable s, Riemann took the decisive step in connecting the arithmetic with theory of functions of a complex variable introducing the function which has come to be known as the celebrated Riemann zeta function in relation to the conjecture about the location of its non-trivial zeros.Very roughly speaking, such a conjecture, the Riemann Hypothesis (RH), is an outgrowth of the Pythagorean tradition in number theory. It determines how the prime numbers are distributed among all the integers, raising the possibility that there is a hidden regularity amid the apparent randomness. The key question turns out to be the location of the zeros of a certain function, that is to say, the Riemann zeta function.Riemann’s HypothesisBernhard Riemann lived in seventeenth century Germany and developed a complex tool to create an artificial landscape in which prime numbers seem to exist at specific positions. He postulated that this is the case for all prime numbers, and this became known as the Riemann’s Hypothesis.Riemann’s zeroes are connected with prime numbers through a complicated formula called “the explicit formula”. It is an improvement on the prime number theorem, a more accurate guess at how many primes occur up to and equal to a number x.1. The first term, “the principle term”, is the logarithmic integral Li(x), which is the better estimation on the prime counting function π(x) in the prime number theorem. It is by far the largest term, an overestimate on how many primes there are up to a given value x.2. The second term, “the periodic term”, is the sum of the logarithmic integral of x to the power p, summed over p, which are the non-trivial zeros of the Riemann Zeta function:It is the term that adjusts overestimate of the principle term.3. The third term is the constant log 2 = 0.693147...The fourth term is an integral. It is zero for x < 2 because there are no primes smaller than two. It has its maximum value at x = 2 , when the integral isLet’s take the imaginary parts of the Riemann zeta function’s zeroes:Zn = 14.134725142, 21.022039639, 25.010857580, 30.424876126..all the way upto the 100th value of 236.524229665. Let Zn be the nth entry in this list, and now define the function F(x) according to the following rule:Note that the vertical axis is at x=5, not x=0.Graph of F(x) for N=100Notice where those major dips occur: at x = 2, 3, 5, 7, etc. We’ve found some prime numbers!Notice that while the primes cause the most pronounced dips, other numbers produce more moderate dips—look at 4, 9, 8, and 16 for the best four examples of this. And some numbers don’t correspond to anything significant in the graph—see 6, 10, 12, 14, 15, and 18.Take a second to reflect on this: we’ve used the zeroes of the Riemann zeta function to find the sequence of prime numbers. And along the way, we’ve found a correspondence between the analytic properties of F(x) and some non-prime numbers! And, as you might imagine, the graph of F(x) will produce more accurate values as N increases.The non-prime dips appear to happen at powers of primes!Riemann noted that the increases and decreases of density in the prime numbers due to the periodic terms had already been observed in previous counts, but it had escaped notice that it is regulated by a certain law.The exact formula for π(x) obtained by Riemann involves the function J(x) defined below, which is a step function:The first thing to notice about this sum is that it is not infinite. At some term, the prime counting function will be zero because there are no primes for x < 2 . So, taking J (100) for example, will contain seven terms, because the eight term will include an eighth root of 100, which is approximately equal to 1.778279..., so this prime counting term becomes 0, and the sum J(100) = 28.5333..Like the prime counting function π(x), Riemann’s prime counting function J(x) is a step function, which jumps by:We can recover the prime counting function π(x) from J(x) by doing Möbius inversion on it. This means that we can now write the prime counting function as:Our new expression for π(x) is still a finite sum because J(x) is zero when x < 2, because there are no primes less than 2 .If we now look at π(100), we get the sumWhich we know is the number of primes less than 100.Using Riemann’s explicit function, one can approximate the number of primes up to and including a given number x to a very good accuracy. Using the zeros of the Riemann hypothesis to error correct the logarithmic integral function is equivalent to the “best possible” bound for the error term in the prime number theorem.Riemann’s zeta function gives an approximating expression for the density of the prime number + 1/2 the density of the squares of the prime numbers + 1/3 of the density of the cubes of the prime numbers etc. at any given magnitude of xThe power series of 2 forms an arithmetic-geometric mean bridge between all prime numbers. This series can be further composed only of numbers that could be divided into the difference of two squares.The Prime number channel sieve is the source of two efficient prime number factorization algorithms and is easily the fastest algorithm that completely deconstructs the twin prime sequence.To bridge the axiomatic gap between the quadratic growth of square numbers with the asymptotic growth of prime numbers, we need to understand it is not primes, but pairs of primes, that are the building blocks of all numbers.Consider the following way the twin primes are distributed:3(5+7)=36 or (3^2 x 2^2)6(11+13)=144 or (3^2 x 4^2)9(17+19)=324 or (3^2 x 6^2)Multiples of 3 multiples itself through the system as a perfect square multiplying all even squares.There are numerous ways to extend the above observation, take 3 and 5 consecutive numbers in the following set:5+6+7=18 and 4+5+6+7+8=3011+12+13=36 and 3+4+5+6+7+8+9=6017+18+19=54 and 16+17+18+19+20=90(18 and 30)(36 and 60)(54 and 90)Extending the 3, 4, 5 rule to (4N)^2 = (5N)^2 - (3N)^2We can find any N^2 given we already know (3N and 5N)This was a problem (Diophantine Book 2 problem 8) Fermat was working on where he wrote down his idea:“It is impossible for a cube to be the sum of two cubes, or for a fourth power to be the sum of two other fourth powers, or, in general, for any power other than two to be the sum of two similar powers.”“I have discovered a truly marvelous demonstration of this proposition, but this margin is too narrow to contain it.”

Introduction A Bode plot is a graph of the transfer function of a linear, time-invariant system versus frequency, plotted with a log-frequency axis, to show a system's frequency response. It is usually a combination of a Bode magnitude plot, expressing the magnitude of the frequency response gain, and a Bode phase plot, expressing the frequency response phase shift. Among his several important contributions to circuit theory and control theory, engineer Hendrik Wade Bode (1905–1982), while working at Bell Labs in the United States in the 1930s, devised a simple but accurate method for graphing gain and phase-shift plots. These bear his name, Bode gain plot and Bode phase plot. The magnitude axis of the Bode plot is usually expressed as decibels of power, that is by the 20 log rule: 20 times the common (base 10) logarithm of the amplitude gain. With the magnitude gain being logarithmic, Bode plots make multiplication of magnitudes a simple matter of adding distances on the graph (in decibels), since Most Motion Control Manufacturer’s provide a Bode Tool with their Programming and/or Tuning Software. A Bode Tool's main function is to make it easy to get frequency response data for your system. For frequency response data to be useful, you need to know what it means. This White Paper is an introduction to understanding frequency response plots. We will try to keep math to a minimum while still demonstrating the concepts involved. Keep in mind, this page is intended as a guide to understanding how to read frequency response plots of your system, not a definitive dissertation on frequency analysis or a control loop tuning guide. Background This section will lay some of the foundation for how non - sine wave based signals can be read in terms of their sine components. If you are comfortable with this concept, or want to skip it, go straight to the section Frequency Response. Making a Square Wave From Sine Waves A linear function (i.e. a square or triangle wave, or the movement of your servo system) can be represented by a series of sine waves. This is a critical point, so let's take a look at a couple of examples. The first example is as easy as it gets. A single sine wave can be represented by y = sin(x). This is more a definition than a discoveryOK, that was easy. Let's see what we can do with a square wave. A square wave that goes from 1 to -1 at the same frequency as the sine wave above can be represented by the series of sine waves represented byNow don't worry about being able to come up with this equation. Let's just focus on the idea that you can make a square wave with just a bunch of sine waves. Another important point is that we need to add an infinite number (all the odd values of n) to make a perfect square wave out of a bunch of sine waves. We'll get to the ramification of this later. Right now, let's take a look at a square wave made using the above equation using n = 1, n = 1 to 11, and all the odd n's.A couple of important notes here:With an infinite number of n's, the square wave made of sine waves is perfect.With only one sine wave (n = 1), the square wave turns into a sine wave.With n = 1 to 11, the square wave is close to, but not exactly a square wave.The more n's you use, the better your square wave. This means that you get ripples in your square wave if you can't reproduce high frequencies (think about how high a frequency your machine can reproduce).

NDA examination is conducted twice a year by UPSC. Basic requirements as obtained from UPSC website:CONDITIONS OF ELIGIBILITY :(a) Nationality : A candidate must either be :(i) a citizen of India, or(ii) a subject of Bhutan, or(iii) a subject of Nepal, or(iv) a Tibetan refugee who came over to India before the 1st January, 1962 with the intention of permanently settling in India, or(v) a person of Indian origin who has migrated from Pakistan, Burma, Sri Lanka and East African Countries of Kenya, Uganda, the United Republic of Tanzania, Zambia, Malawi, Zaire and Ethiopia or Vietnam with the intention of permanently settling in India.Provided that a candidate belonging to categories (ii), (iii), (iv) and (v) above shall be a person in whose favour a certificate of eligibility has been issued by the Government of India.Certificate of eligibility will not, however, be necessary in the case of candidates who are Gorkha subjects of Nepal.(b) Age Limits, Sex and Marital Status :Only unmarried male candidates of age between 16-1/2 - 19–1/2 are eligible. (Confirm the exact dates on UPSC website: UPSC)The date of birth accepted by the Commission is that entered in the Matriculation or Secondary School Leaving Certificate or in a certificate recognised by an Indian University as equivalent to Matriculation or in an extract from a Register of Matriculates maintained by a University which must be certified by the proper authority of the University or in the Higher Secondary or an equivalent examination certificates. These certificates are required to be submitted only after the declaration of the result of the written part of the examination. No other document relating to age like horoscopes, affidavits, birth extracts from Municipal Corporation, service records and the like will be accepted. The expression Matriculation/Higher Secondary Examination Certificate in this part of the instruction includes the alternative certificates mentioned above.NOTE 1 : Candidates should note that only the date of birth as recorded in the Matriculation/Higher Secondary Examination Certificate available or an equivalent certificate on the date of submission of applications will be accepted by the Commission and no subsequent request for its change will be considered or granted.NOTE 2 : Candidates should also note that once a date of birth has been claimed by them and entered in the records of the Commission for the purpose of admission to an Examination, no change will be allowed subsequently or at any subsequent examination on any ground whatsoever.NOTE 3 : The candidates should exercise due care while entering their date of birth in the respective column of the Online Application Form for the Examination. If on verification at any subsequent stage any variation is found in their date of birth from the one entered in their Matriculation or equivalent Examination certificate, disciplinary action will be taken against them by the Commission under the Rules.NOTE 4 : Candidates must undertake not to marry until they complete their full training. A candidate who marries subsequent to the date of his application though successful at this or any subsequent Examination will not be selected for training. A candidate who marries during training shall be discharged and will be liable to refund all expenditure incurred on him by the Government.(c) Educational Qualifications:(i) For Army Wing of National Defence Academy :—12th Class pass of the 10+2 pattern of School Education or equivalent examination conducted by a State Education Board or a University.(ii) For Air Force and Naval Wings of National Defence Academy and for the 10+2 Cadet Entry Scheme at the Indian Naval Academy :—12th Class pass of the 10+2 pattern of School Education or equivalent with Physics and Mathematics conducted by a State Education Board or a University.Candidates who are appearing in the 12th Class under the 10+2 pattern of School Education or equivalent examination can also apply for this examination.Such candidates who qualify the SSB interview but could not produce Matriculation/10+2 or equivalent certificate in original at the time of SSB interview should forward duly self-attested Photocopies to ‘Directorate General of Recruiting, Army HQ, West Block.III, R.K. Puram, New Delhi-110066’ and for Naval Academy candidates to ‘Naval Headquarters, DMPR, OI&R Section, Room No. 204, ‘C’ Wing, Sena Bhavan, New Delhi-110011’ by date mentioned in the respective advertisement failing which their candidature will be cancelled. All other candidates who have produced their Matriculation and 10+2 pass or equivalent certificates in original at the time of attending the SSB interview and have got the same verified by the SSB authorities are not required to submit the same to Army HQ or Naval HQ as the case may be. Certificates in original issued by the Principals of the Institutions are also acceptable in cases where Boards/Universities have not yet issued certificates. Certified true copies/photostate copies of such certificates will not be accepted. In exceptional cases the Commission may treat a candidate, who does not possess any of the qualifications prescribed in this rule as educationally qualified provided that he possesses qualifications, the standard of which in the opinion of the Commission, justifies his admission to the examination.NOTE 1 : Candidates appearing in the 11th class exam are not eligible for this examination.NOTE 2 : Those candidates who have yet to qualify in the 12th class or equivalent examination and are allowed to appear in the UPSC Examination should note that this is only a special concession given to them. They are required to submit proof of passing the 12th class or equivalent examination by the prescribed date (i.e. 24th June, 2017) and no request for extending this date will be entertained on the grounds of late conduct of Board/University Examination, delay in declaration of results or any other ground whatsoever.NOTE 3 : Candidates who are debarred by the Ministry of Defence from holding any type of Commission in the Defence Services shall not be eligible for admission to the examination and if admitted, their candidature will be cancelled.NOTE 4 : Those candidates who have failed INSB/PABT earlier are not eligible for Air Force.(d) A candidate who has resigned or withdrawn on disciplinary grounds from any of the training academies of Armed Forces is not eligible to apply.(e) Physical Standards:Candidates must be physically fit according to physical standards for admission to National Defence Academy and Naval Academy Examination as per the following guidelines:GUIDELINES FOR PHYSICAL STANDARDS FOR ADMISSION TO THE NATIONAL DEFENCE ACADEMY.NOTE : CANDIDATES MUST BE PHYSICALLY AND MENTALLY FIT ACCORDING TO THE PRESCRIBED PHYSICAL STANDARDS. THE GUIDELINES FOR THE SAME ARE GIVEN BELOW.A NUMBER OF QUALIFIED CANDIDATES ARE REJECTED SUBSEQUENTLY ON MEDICAL GROUNDS. CANDIDATES ARE THEREFORE ADVISED IN THEIR OWN INTEREST TO GET THEMSELVES MEDICALLY EXAMINED BEFORE SUBMITTING THEIR APPLICATIONS TO AVOID DISAPPOINTMENT AT THE FINAL STAGE.Candidates are also advised to rectify minor defects/ailments in order to speed up finalisation of medical examination conducted at the Military Hospital after being recommended at the SSB.Few of such commonly found defects/ailments are listed below :(a) Wax (Ears)(b) Deviated Nasal Septum(c) Hydrocele/Phimosis(d) Overweight/Underweight(e) Under Sized Chest(f) Piles(g) Gynaecomastia(h) Tonsillitis(i) VaricoceleNOTE : Permanent body tattoos are only permitted on inner face of forearm i.e. from inside of elbow to the wrist and on the reverse side of palm/back (dorsal) side of hand/Permanent body tattoos on any other part of the body are not acceptable and candidates will be barred from further selection. Tribes with tattoo marks on the face or body as per their existing custom and traditions will be permitted on a case to case basis. Comdt Selection Centre will be competent auth for clearing such cases.Civilian candidates appearing for all types of commission in the Armed Forces will be entitled to out-patients treatment from service sources at public expense for injuries sustained or diseases contracted during the course of their examination by the Selection Board. They will also be entitled to in-patient treatment at public expense in the Officer’s ward of a hospital provided—(a) the injury is sustained during the tests or,(b) the disease is contracted during the course of the examination by selection board and there is no suitable accommodation in local civil hospital or it is impracticable to remove the patient to the civil hospital; or,(c) the medical board requires the candidate’s admission for observation.NOTE : They are not entitled to special nursing.A candidate recommended by the Services Selection Board will undergo a medical examination by a Board of Service Medical Officers. Only those candidates will be admitted to the academy who are declared fit by the Medical Board. The proceedings of the Medical Board are confidential and will not be divulged to anyone. However, the candidates declared unfit will be intimated by the President of the Medical Board and the procedure for request for an Appeal Medical Board will also be intimated to the candidate. Candidates declared unfit during Appeal Medical Board will be intimated about the provision of Review Medical Board.(a) The candidate must be in good physical and mental health and free from any disease/disability which is likely to interfere with the efficient performance of Military duties.(b) There should be no evidence of weak constitution, bodily defects or underweight. The Candidate should not be overweight or obese.(c) The minimum acceptable height is 157 cms(162.5 cms. for Air Force). For Gorkhas and individuals belonging to hills of North-Eastern regions of India, Garhwal and Kumaon, the minimum acceptable heights will be 5 cms. less. In case of candidates from Lakshadweep the minimum acceptable height can be reduced by 2 cms. Height and weight standards are given below :HEIGHT/WEIGHT STANDARDS FOR ARMY/AIR FORCETABLE-IHEIGHT/WEIGHT STANDARDS FOR NAVYTABLE-II“Individual’s weight is considered normal if it is within ±10% departure from average weight given in the table I and II”. However, in individuals with heavy bones and broad build as well as individuals with thin build but otherwise healthy this may be relaxed to some extent on merit.NOTE 1 : Height relaxable upto 2.5 cm. (5 cm. for Navy) may be allowed where the Medical Board certifies that the candidate is likely to grow and come up to the required standard on completion of his training.NOTE 2 : To meet special requirement as a pilot in the Air Force the acceptable measurements of leg length, thigh length and sitting height will be as under :(d) Chest should be well developed. Fully expanded chest should not be less than 81 cms. The minimum range of expansion after full inspiration should be 5 cms. The measurement will be taken with a tape so adjusted that its lower edge should touch the nipple in front and the upper part of the tape should touch the lower angle of the shoulder blades behind. X-Ray of the chest is compulsory and will be taken to rule out any disease of the chest.(e) There should be no maldevelopment or impairment of function of the bones or joint.Spinal Conditions-(f) Past medical history of diseases or injury of the spine or sacro iliac joints, either with or without objective signs which have prevented the candidate from successfully following a physically active life, is a cause for rejection for commissioning in IAF. History of spinal fracture/prolapsed intervertebral disc and surgical treatment for these conditions will entail rejection. The following conditions detected radiologically during medical exam will disqualify a candidate for Air Force service:(i) Granulomatous disease of spine(ii) Arthritidies/spondylosis— Rheumatoid arthritis and allied disorders— Ankylosing spondylitis— Osteoarthrosis, spondylosis and degenerative joint disease— Non articular rheumatism (e.g. lesions of the rotator cuff, tennis elbow, recurrent lumbago etc.)— Miscellaneous disorders including SLE, , polymyositis, vasculitis.(iii) Spondylolisthesis/spondylolysis.(iv) Compression fracture of vertebrae.(v) Scheuerman's disease (Adolescent kyphosis)(vi) Loss of cervical lordosis when associated with clinically restricted movements of cervical spine.(vii) Unilateral/Bilateral cervical ribs with demonstrable neurological or circulatory deficit.(viii) Scoliosis more than 15 degree as measured by Cobb's method.(ix) Degenerative Disc. Disease.(x) Presence of schmorl's nodes at more than one level.(xi) Atlanto-occipital and atlantoaxial anomalies.(xii) Hemi vertebrae and/or incomplete block (fused) vertebrae at any level in cervical, dorsal or lumbar spine and complete block (fused) vertebrae at more than one level in cervical or dorsal spine.(xiii) Unilateral Sacralisation or lumbarisation (Complete or incomplete) at all levels and bilateral incomplete sacralisation or lumbarisation.(xiv) Any other abnormality if so considered by the specialist.(g) Mild Kyphosis or Lordosis where deformity is barely noticeable and there is no pain or restriction of movement will not preclude acceptance.(h) In case of noticeable Scoliosis or suspicion of any other abnormality or spinal deformity, more than mild, appropriate X-rays of the spine are to be taken and the Examinee referred for specialist’s advice.(i) The following conditions detected on X-ray examination will be disqualifying for entry to Armed Forces :(i) Granulomatius disease of spine.(ii) Arthritidies/spondylosis(iii) Scoliosis more than 15 degree as measured by Cobb’s Method (10 degree for Army and Navy).(iv) More than mild Kyphosis/Lordosis(v) Spondylolisthesis/Spondylosis/Spondylolysis(vi) Herniated nucleus pulposes.(vii) Compression fracture of Vertebra.(viii) Sacaralisation Disease(ix) Cervical ribs with demonstrable neurological or Circulatory deficit.(x) Presence of Schmorl’s node at more than one level.(xi) Atlanto-occipital and atlanto-axial anomalies.(xii) Incomplete Sacaralisation Unilateral or Bilateral(xiii) Spina Bifida other than SV 1 and LV 5 if completely Sacralised(xiv) Any other abnormality, if so considered by specialist.(j) A candidate should have no past history of mental breakdown or fits.(k) The hearing should be normal. A candidate should be able to hear a forced whisper with each ear at a distance of 610 cms. in a quiet room. There should be no evidence of present or past disease of the ear, nose and throat. Audiometric test will be done for AF. Audiometric hearing loss should not be greater than 20 db in frequencies between 250 and 8000 Hz. There is no impediment of speech.(l) There should be no signs of functional or organic disease of the heart and blood vessels. Blood pressure should be normal.(m) There should be no enlargement of liver or spleen. Any evidence of disease of internal organs of the abdomen will be a cause for rejection.(n) Un-operated hernias will make a candidate unfit. In case of Hernia which has been operated, a minimum of 6 months must have passed prior to final medical examination before commencement of the course.(o) There should be no hydrocele, varicocele or piles.(p) Urine examination will be done and any abnormality if detected will be a cause for rejection.(q) Any disease of skin which is likely to cause disability or disfigurement will also be a cause for rejection.(r) Visual standards:- The distance vision (corrected) should be 6/6 in better eye and 6/9 in worse eye. Myopia should not be more than 2.5 D and hypermetropia not more than 3.5 D including Astigmatism. Internal examination of the eye will be done by means of opthalmoscope to rule out any disease of the eye. A candidate must have good binocular vision. The colour vision standard will be (CP-III) for Army. A candidate should be able to recognise red and green colours. Candidates will be required to give certificates that neither he nor any member of his family has suffered from congenital night blindness. Candidates who have undergone or have the evidence of having undergone Radial Keratotomy, to improve the visual acuity will be permanently rejected for all the Services. Candidates who have undergone Laser Surgery for correction of refractive error are also not acceptable to defence services.Vision standard for Naval candidatesVisual Standards for Air ForceCandidates who habitually wear spectacles are not eligible for Air Force, Minimum distant vision 6/6 in one eye and 6/9 in other, correctable to 6/6 only for Hypermetropia. Colour vision CP-I Hypermetropia : +2.0 D Sph Manifest Myopia : Nil Retinoscopic Myopia : 0.5 in any Meridian permitted Astigmatism : + 0.75 D Cyl (within + 2.0 D.Max) Maddox Rod TestHand held Stereoscope–All of BSV grades Convergence–Up to 10 cm Cover test for distant and near–Lateral divergence/convergence recovery rapid and complete Radial Keratotomy, Photo Refractive Keratotomy/laser in Situ, Keratomileusis (PRK/LASIK) surgeries for correction of refractive errors are not permitted for any Air Force duties. Candidates having undergone cataract surgery with or without IOL implants will also be declared unfit.Binocular vision must possess good binocular vision (fusion and stereopsis with good amplitude and depth).Candidates who have undergone LASIK surgery are not considered fit for permanent commission in flying branch in IAF.(s) USG abdomen examination will be carried out and any congential structural anomaly or disease of the abdominal organs will be a cause for rejection in Armed Forces.(t) The candidates should have sufficient number of natural and sound teeth. A minimum of 14 dental points will be acceptable. When 32 teeth are present, the total dental points are 22. A candidate should not be suffering from severe pyorrhoea.(u) Routine ECG for Air Force candidates must be within normal limits.(v) Physical conditioning : Prospective candidates are advised to keep themselves in good physical condition, by following the undermentioned routine :—(a) Running 2.4 km in 15 minutes(b) Skipping(c) Pushups and sit ups (minimum 20 each)(d) Chin ups (minimum 08)(e) Rope climbing 3-4 metres.ANNOUNCEMENT OF THE RESULTS OF THE WRITTEN EXAMINATION, INTERVIEW OF QUALIFIED CANDIDATES, ANNOUNCEMENT OF FINAL RESULTS AND ADMISSION TO THE TRAINING COURSES OF THE FINALLY QUALIFIED CANDIDATES :The Union Public Service Commission shall prepare a list of candidates who obtain the minimum qualifying marks in the written examination as fixed by the Commission at their discretion. Such candidates shall appear before a Services Selection Board for Intelligence and Personality Test where candidates for the Army/Navy wings of the NDA and 10+2 Cadet Entry Scheme of Indian Naval Academy will be assessed on Officers Potentiality and those for the Air Force in Computerised Pilot Selection System (CPSS) and Pilot Aptitude Test (which forms the whole PAB Test) and for Officers Potentiality. PABT applicable to candidates with Air Force as First choice would also be conducted for all SSB qualified candidates with one of the choice as Air Force subject to their eligibility and if they are so willing.TWO-STAGE SELECTION PROCEDURETwo-stage selection procedure based on Psychological Aptitude Test and Intelligence Test has been introduced at Selection Centres/Air Force Selection Boards/Naval Selection Boards. All the candidates will be put to stage-one test on first day of reporting at Selection Centres/Air Force Selection Boards/Naval Selection Boards. Only those candidates who qualify at stage one will be admitted to the second stage/remaining tests. Those candidates who qualify stage II will be required to submit the Original Certificates along with one photocopy each of : (i) Original Matriculation pass certificate or equivalent in support of date of birth, (ii) Original 10+2 pass certificate or equivalent in support of educational qualification.Candidates who appear before the Services Selection Board and undergo the test there, will do so at their own risk and will not be entitled to claim any compensation or other relief from Government in respect of any injury which they may sustain in the course of or as a result of any of the tests given to them at the Services Selection Board whether due to the negligence of any person or otherwise. Parents or guardians of the candidates will be required to sign a certificate to this effect.To be acceptable, candidates for the Army/Navy/Naval Academy and Air Force should secure the minimum qualifying marks separately in (i) Written examination as fixed by the Commission at their discretion and (ii) Officer Potentiality Test as fixed by the Services Selection Board at their discretion.Over and above candidates for the Air Force, and all the SSB qualified candidates as per their willingness, eligibility and preference for Air Force, should separately qualify the PABT.Subject to these conditions the qualified candidates will then be placed in a single combined list on the basis of total marks secured by them in the Written Examination and the Services Selection Board Tests. The final allocation/selection for admission to the Army, Navy, Air Force of the National Defence Academy and 10+2 Cadet Entry Scheme of Indian Naval Academy will be made upto the number of vacancies available subject to eligibility, medical fitness and merit-cum-preference of the candidates. The candidates who are eligible to be admitted to multiple Services/Courses will be considered for allocation/selection with reference to their order or preferences and in the event of their final allocation/ selection to one Service/Course, they will not be considered for admission to other remaining Services/Courses.N.B.: EVERY CANDIDATE FOR THE AIR FORCE IS GIVEN COMPUTERISED PILOT SELECTION SYSTEM (CPSS) AND PILOT APTITUDE TEST (WHICH FORMS THE WHOLE PAB TEST) ONLY ONCE. THE GRADE SECURED BY HIM AT THE FIRST TEST WILL THEREFORE HOLD GOOD FOR EVERY SUBSEQUENT INTERVIEW HE HAS WITH THE AIR FORCE SELECTION BOARD. A CANDIDATE WHO FAILS IN THE FIRST PILOT APTITUDE TEST CANNOT APPLY FOR ADMISSION TO THE NATIONAL DEFENCE ACADEMY EXAMINATION FOR THE AIR FORCE WING OR GENERAL DUTIES (PILOT) BRANCH OR NAVAL AIR ARM.Candidates who have been given the Computerised Pilot Selection System (CPSS) and Pilot Aptitude Test (which forms the whole PAB Test) for any previous N.D.A. course should submit their application for this examination for the Air Force Wing only if they have been notified as having qualified in the Pilot Aptitude Test.The form and manner of communication of the result of the examination to individual candidates shall be decided by the Commission at their discretion and the Commission will not enter into correspondence with them regarding the result.Success in the examination confers no right of admission to the Academy. A candidate must satisfy the appointing authority that he is suitable in all respects for admission to the Academy.THOROUGH MEDICAL EXAMINATION WILL BE CONDUCTED FOR THE SUCCESSFUL CANDIDATES AFTER SSB INTERVIEW AND BEFORE DECLARING THE FINAL RESULT.SCHEME OF EXAMINATION1. The subjects of the written examination, the time allowed and the maximum marks allotted to each subject will be as follows :—2. THE PAPERS IN ALL THE SUBJECTS WILL CONSIST OF OBJECTIVE TYPE QUESTIONS ONLY. THE QUESTION PAPERS (TEST BOOKLETS) OF MATHEMATICS AND PART “B” OF GENERAL ABILITY TEST WILL BE SET BILINGUALLY IN HINDI AS WELL AS ENGLISH.3. In the question papers, wherever necessary, questions involving the metric system of Weights and Measures only will be set.4. Candidates must write the papers in their own hand. In no circumstances will they be allowed the help of a scribe to write answers for them.5. The Commission have discretion to fix qualifying marks in any or all the subjects at the examination.6. The candidates are not permitted to use calculator or Mathematical or logarithmic table for answering objective type papers (Test Booklets). They should not therefore, bring the same inside the Examination Hall.B. SYLLABUS OF THE EXAMINATIONPAPER-IMATHEMATICS(Code No. 01)(Maximum Marks-300)1. ALGEBRAConcept of set, operations on sets, Venn diagrams. De Morgan laws, Cartesian product, relation, equivalence relation. Representation of real numbers on a line. Complex numbers—basic properties, modulus, argument, cube roots of unity. Binary system of numbers. Conversion of a number in decimal system to binary system and vice-versa. Arithmetic, Geometric and Harmonic progressions. Quadratic equations with real coefficients. Solution of linear inequations of two variables by graphs. Permutation and Combination. Binomial theorem and its applications. Logarithms and their applications.2. MATRICES AND DETERMINANTS :Types of matrices, operations on matrices. Determinant of a matrix, basic properties of determinants. Adjoint and inverse of a square matrix, Applications-Solution of a system of linear equations in two or three unknowns by Cramer’s rule and by Matrix Method.3. TRIGONOMETRY :Angles and their measures in degrees and in radians. Trigonometrical ratios. Trigonometric identities Sum and difference formulae. Multiple and Sub-multiple angles. Inverse trigonometric functions. Applications-Height and distance, properties of triangles.4. ANALYTICAL GEOMETRY OF TWO AND THREE DIMENSIONS:Rectangular Cartesian Coordinate system. Distance formula. Equation of a line in various forms. Angle between two lines. Distance of a point from a line. Equation of a circle in standard and in general form. Standard forms of parabola, ellipse and hyperbola. Eccentricity and axis of a conic. Point in a three dimensional space, distance between two points. Direction Cosines and direction ratios. Equation two points. Direction Cosines and direction ratios. Equation of a plane and a line in various forms. Angle between two lines and angle between two planes. Equation of a sphere.5. DIFFERENTIAL CALCULUS :Concept of a real valued function–domain, range and graph of a function. Composite functions, one to one, onto and inverse functions. Notion of limit, Standard limits—examples. Continuity of functions—examples, algebraic operations on continuous functions. Derivative of function at a point, geometrical and physical interpretation of a derivative—applications. Derivatives of sum, product and quotient of functions, derivative of a function with respect to another function, derivative of a composite function. Second order derivatives. Increasing and decreasing functions. Application of derivatives in problems of maxima and minima.6. INTEGRAL CALCULUS AND DIFFERENTIAL EQUATIONS :Integration as inverse of differentiation, integration by substitution and by parts, standard integrals involving algebraic expressions, trigonometric, exponential and hyperbolic functions. Evaluation of definite integrals—determination of areas of plane regions bounded by curves— applications. Definition of order and degree of a differential equation, formation of a differential equation by examples. General and particular solution of a differential equations, solution of first order and first degree differential equations of various types—examples. Application in problems of growth and decay.7. VECTOR ALGEBRA :Vectors in two and three dimensions, magnitude and direction of a vector. Unit and null vectors, addition of vectors, scalar multiplication of a vector, scalar product or dot product of two vectors. Vector product or cross product of two vectors. Applications—work done by a force and moment of a force and in geometrical problems.8. STATISTICS AND PROBABILITY :Statistics : Classification of data, Frequency distribution, cumulative frequency distribution—examples. Graphical representation—Histogram, Pie Chart, frequency polygon—examples. Measures of Central tendency—Mean, median and mode. Variance and standard deviation—determination and comparison. Correlation and regression.Probability : Random experiment, outcomes and associated sample space, events, mutually exclusive and exhaustive events, impossible and certain events. Union and Intersection of events. Complementary, elementary and composite events. Definition of probability—classical and statistical— examples. Elementary theorems on probability—simple problems. Conditional probability, Bayes’ theorem—simple problems. Random variable as function on a sample space. Binomial distribution, examples of random experiments giving rise to Binominal distribution.PAPER-IIGENERAL ABILITY TEST(Code No. 02)(Maximum Marks—600)Part ‘A’—ENGLISH (Maximum Marks—200)The question paper in English will be designed to test the candidate’s understanding of English and workman like use of words. The syllabus covers various aspects like : Grammar and usage, vocabulary, comprehension and cohesion in extended text to test the candidate’s proficiency in English.Part ‘B’—GENERAL KNOWLEDGE (Maximum Marks—400)The question paper on General Knowledge will broadly cover the subjects : Physics, Chemistry, General Science, Social Studies, Geography and Current Events.- The syllabus given below is designed to indicate the scope of these subjects included in this paper. The topics mentioned are not to be regarded as exhaustive and questions on topics of similar nature not specifically mentioned in the syllabus may also be asked. Candidate’s answers are expected to show their knowledge and intelligent understanding of the subject.Section ‘A’ (Physics)Physical Properties and States of Matter, Mass, Weight, Volume, Density and Specific Gravity, Principle of Archimedes, Pressure Barometer. Motion of objects, Velocity and Acceleration, Newton’s Laws of Motion, Force and Momentum, Parallelogram of Forces, Stability and Equilibrium of bodies, Gravitation, elementary ideas of work, Power and Energy.Effects of Heat, Measurement of Temperature and Heat, change of State and Latent Heat, Modes of transference of Heat. Sound waves and their properties, Simple musical instruments. Rectilinear propagation of Light, Reflection and refraction. Spherical mirrors and Lenses, Human Eye. Natural and Artificial Magnets, Properties of a Magnet, Earth as a Magnet. Static and Current Electricity, conductors and Non-conductors, Ohm’s Law, Simple Electrical Circuits, Heating, Lighting and Magnetic effects of Current, Measurement of Electrical Power, Primary and Secondary Cells, Use of X-Rays.General Principles in the working of the following:Simple Pendulum, Simple Pulleys, Siphon, Levers, Balloon, Pumps, Hydrometer, Pressure Cooker, Thermos Flask, Gramophone, Telegraphs, Telephone, Periscope, Telescope, Microscope, Mariner’s Compass; Lightening Conductors, Safety Fuses.Section ‘B’ (Chemistry)Physical and Chemical changes. Elements, Mixtures and Compounds, Symbols, Formulae and simple Chemical Equations, Law of Chemical Combination (excluding problems). Properties of Air and Water. Preparation and Properties of Hydrogen, Oxygen, Nitrogen and Carbondioxide, Oxidation and Reduction. Acids, bases and salts. Carbon—different forms. Fertilizers—Natural and Artificial.Material used in the preparation of substances like Soap, Glass, Ink, Paper, Cement, Paints, Safety Matches and Gun-Powder. Elementary ideas about the structure of Atom, Atomic Equivalent and Molecular Weights, Valency.Section ‘C’ (General Science)Difference between the living and non-living. Basis of Life—Cells, Protoplasms and Tissues. Growth and Reproduction in Plants and Animals. Elementary knowledge of Human Body and its important organs. Common Epidemics, their causes and prevention. Food—Source of Energy for man. Constituents of food, Balanced Diet. The Solar System—Meteors and Comets, Eclipses. Achievements of Eminent Scientists.Section ‘D’ (History, Freedom Movement etc.)A broad survey of Indian History, with emphasis on Culture and Civilisation. Freedom Movement in India. Elementary study of Indian Constitution and Administration. Elementary knowledge of Five Year Plans of India. Panchayati Raj, Co-operatives and Community Development. Bhoodan, Sarvodaya, National Integration and Welfare State, Basic Teachings of Mahatma Gandhi. Forces shaping the modern world; Renaissance, Exploration and Discovery; War of American Independence. French Revolution, Industrial Revolution and Russian Revolution. Impact of Science and Technology on Society. Concept of one World, United Nations, Panchsheel, Democracy, Socialism and Communism. Role of India in the present world.Section ‘E’ (Geography)The Earth, its shape and size. Lattitudes and Longitudes, Concept of time. International Date Line. Movements of Earth and their effects. Origin of Earth. Rocks and their classification; Weathering— Mechanical and Chemical, Earthquakes and Volcanoes. Ocean Currents and Tides Atmosphere and its composition; Temperature and Atmospheric Pressure, Planetary Winds, Cyclones and Anti-cyclones; Humidity; Condensation and Precipitation; Types of Climate, Major Natural regions of the World. Regional Geography of India—Climate, Natural vegetation. Mineral and Power resources; location and distribution of agricultural and Industrial activities. Important Sea ports and main sea, land and air routes of India. Main items of Imports and Exports of India.Section ‘F’ (Current Events)Knowledge of Important events that have happened in India in the recent years. Current important world events. Prominent personalities—both Indian and International including those connected with cultural activities and sports.NOTE :Out of maximum marks assigned to part ‘B’ of this paper, questions on Sections ‘A’, ‘B’, ‘C’, ‘D’, ‘E’ and ‘F’ will carry approximately 25%, 15%, 10%, 20%, 20% and 10% weightages respectively.Intelligence and Personality TestThe SSB procedure consists of two stage Selection process - stage I and stage II. Only those candidates who clear the stage I are permitted to appear for stage II. The details are :(a) Stage I comprises of Officer Intelligence Rating (OIR) tests are Picture Perception * Description Test (PP&DT). The candidates will be shortlisted based on combination of performance in OIR Test and PP&DT.(b) Stage II Comprises of Interview, Group Testing Officer Tasks, Psychology Tests and the Conference. These tests are conducted over 4 days. The details of these tests are given on the website Join Indian Army | Government of India.The personality of a candidate is assessed by three different assessors viz. The Interviewing Officer (IO), Group Testing Officer (GTO) and the Psychologist. There are no separate weightage for each test. The mks are allotted by assessors only after taking into consideration the performance of the candidate holistically in all the test. In addition, marks for Conference are also allotted based on the initial performance of the Candidate in the three techniques and decision of the Board. All these have equal weightage.The various tests of IO, GTO and Psych are designed to bring out the presence/absence of Officer Like Qualities and their trainability in a candidate. Accordingly candidates are Recommended or Not Recommended at the SSB.(For more details, check UPSC website: UPSC)SSB interview is the world's toughest army test and typically these TOP 15 OFFICER LIKE QUALITIES (OLQs) are tested during the entire interview process:Effective Intelligence.Reasoning Ability.Power of Expression.Self Confidence.Determination.Organizing Ability.Initiative.Courage.Cooperation.Sense of Responsibility.Stamina.Group Influencing Ability.Liveliness.Social Adaptability.Speed of Decision“Origin of this type of testing belongs to first world war era. In the First World War, large numbers of casualties on battlefields necessitated recruitment of best fighting talent in armed forces. For this need, psychologists designed some scientific tests which assess candidate's Officer Like Qualities (OLQs). Over the years SSB testing has been improved based on feedback and has been proved to be an effective and accurate way of selecting candidates who are capable of being at least an average officer after training.”