Lesson 1 Writing Equations Using Symbols: Fill & Download for Free

Because 1 + 1 = 2!TL;DRI’m not going to answer this short, however if you’re willing to follow, you’re going to watch probably the most beautiful and simplest idea of math. This gonna be a simple ride, but it would teach us the whole idea of math in just less than one hour. So, buckle up Dorothy!Counting stuffs, 1 + 1 = 2, yes, counting stuffs, by only and only this one simple rule, we built almost the entire foundation of the so called mathematics. Watch. We're not going to invent any new rule, NONE at all, to derive this result that - x - = + (minus times minus equals plus).Now, for the first step, let’s forget that you ever understood a number (1, 2, 3, etc), an operator (+, -, x, :, etc), or any thing else that you ever knew from mathematics. We’re going back to the time 5000 years ago when a number was barely invented (in Mesopotamia). Then, take a look at the stuffs at your desk. How are we going to count it? This is how we’re going to do it. For every one stuff at the desk, we put a mark (*) on a piece of wood, a clothes, a paper, or whatever, e.g.:* * * * *Those are 5 marks for probably: a laptop, a glass, a phone, a wallet, and a TV remote control.But soon, we're going into trouble. For example, how if we are going to also count all the stuffs in the room? Then we would end up with something like this:* * * * * * * * * * * * * * * * * * * * * * * * * * * * * *Oo sh#t, no way! So, we invent a symbol that we could agree together to write the result of that counting more practically. NOTE: there's no new rule here, no at all, we just agree on how to write the result of the counting. Here are the symbols that we invent:1 for *2 for * *3 for * * *4 for * * * *(and so on)Nice! We could stop at an arbitrary number of symbols, but we choose to stop at the symbol "9", and agree that for every * * * * * * * * * * (ten marks) we would write it to the left. Very nice! For example:* * * * * * * * * * * *Would be written as: 12. How if we have counted pass one hundred? We apply the same agreement, that is for every ten of ten marks, we write it to the left. E.g.: 112, and so on. With this simple agreement, we have had a powerful tool to write our counting.When we're in the middle of counting our stuffs, supposed there's a call from the nature, so we stop, and go to the bathroom. We're back and start to count again. So, on the paper that we write our marks, would be something like this:* * * * * * * * (nature calls, stop, and then continuing) * * * *But that is, of course, the same as writing it as:* * * * * * * * * * * *The fact that the nature calls, doesn't change the result of the counting. We just stop, and then continue to write our marks. However it does give us the idea of addition operation. It means to continue the counting, and that's just it! First, we invent a new symbol "=" to denote "the result is", and write the addition process as following:* * * * * * * * (nature calls, stop, and then continuing) * * * * = * * * * * * * * * * * *You would notice immediately that writing "nature calls, stop, and then continuing" is too cumbersome. So we invent again a new symbol "+". NOTE: again, there's no new rule, this is still just about counting. We just stop, then count again. It's still exactly the same process: counting stuffs! So we write it as following:* * * * * * * * + * * * * = * * * * * * * * * * * *As we have invented some symbols for those marks, we could choose to write it as following:8 + 4 = 12It's the same thing all over again.Are you still with me? Good. Then the scenario changes a little bit. Your Mom tells you something like this:"There must be 12 stuffs at your room, I've counted it up to 8, would you continue it for me?"Sure, no big deal. But then, we would think something like this. Well, if it's supposed to be 12 stuffs, then we must expect that we should count just another 4 stuffs. Note there's nothing new here, but this our thinking process is the same as writing it as following:(Mom has counted until) 8 + (the remaining stuffs to be counted) = 12Try to replace the "the remaining stuffs to be counted", following our counting rule, then it starts to look like a game. Indeed it's a game. Mathematics is no more than a game, and this game surprisingly has only one rule: COUNTING STUFFS. Of course, we could not pretend that we don't know the answer (it's 4), but pretend that we haven't used to the "new" symbols we just invented. So we do something like this:* * * * * * * * + (some marks) = * * * * * * * * * * * *Sure, we could count it one by one, but it would be easier and save time if we arrange it like this:* * * * * * * * * * * ** * * * * * * *We don't have to count it all! We just need to count the missing marks which is:* * * *And when we look at the table of our new symbols above, we would find that it could be written as number "4". Again, there's no new rule here. (Sorry to remind you again and again about this -- but this is the most important idea that I would like to convey.) It's just playing the same game: counting. But it gives us a new idea in the process. That's the so called subtraction operation. Imagine, our Mom mentioned 12 stuffs, then has counted it up to 8 stuffs, then our counting should give the result 4. If we write it, it would be something like this:12 (Mom have counted it until) 8 (so our counting should give the result) 4Remember that we have invented the symbol for "the result is" which is "=". But that "Mom have counted it to" is too cumbersome, because we haven't invented a symbol for it, so why not we invent another symbol for this process? Right, the symbol we choose is "-". So we write it again as:12 - 8 = 4It has the same meaning as we arrange our marks so 12 marks are put above and 8 marks are put below and then we count the missing marks which is 4. And it has the same meaning as:8 + 4 = 12Taking away? How if we take away 12 marks from 12 marks, how many marks left? Nothing. No new rule, because it's just the same game we play above:12 + (some count) = 12Or write it using the new symbol "-" as:12 - 12 = (nothing)But it's the same thing if we do it to all kind of numbers:5 - 5 = (nothing)16 - 16 = (nothing)31 - 31 = (nothing)(and so on)Well, it's too cumbersome. Of course, we invent another symbol: "0" (called zero). No new rule, it's just another agreement on a symbol. It's just about how to write it, instead to write 12 - 12, 5 - 5, 16 - 16, 31 - 31, or whatever, it's simpler to just write it with this new symbol "0":12 - 12 = 05 - 5 = 016 - 16 = 031 - 31 = 0(and so on)We have had some of our symbols, now we're going to do something ABSOLUTELY amusing, like creative human being does it all the time. How if we encounter something like this:8 - 12 = (what should we put here?)No! No, no, no: we could NOT just combine the symbol "-" and the symbol "4", and make "-4". There is a "hole" in this kind of thinking. You may not feel it because you use to the concept of negative number. But we haven't arrived at that concept yet! Be noted that we haven't had something like "-4" -- that is a bare minus sign without a number at its left.That's not how mathematics developed! This is a very important lesson in mathematics (and also in life!), nothing could be concluded from out of nowhere, that's called gambling, and we would never win! That's wrong, wrong, and wrong.( Edit : thanks to Gary Williams who pinpointed the previous logic I used is kinda of broken. )So, here is how we attack this problem properly. Notice that we could use the symbol “-” as following:12 - 3 = 99 - 5 = 4And shortcut the symbols by writing them as following:12 - 3 - 5 = 9 - 5 = 4No we are going to use this finding to solve the problem 8 - 12 = (what should we put here?). There's no jumping in logic because there's no new symbol (and no new rule). Alas, we have invented the symbol “0”, so let's also use it here:8 - 12 = 8 - 8 - 4 = 0 - 4It’s the same with what we are doing before: take 8, and take another 4, so it's the same as take 12 in one take.I know what you are thinking now, but stop it first!So, we have this: "0 - 4". What is it? (No, it's not a negative number -- we haven't invented it yet.) When we're facing this kind of question, and feel a little bit confused, always go back to what these symbols mean. It means the counting of our marks. So it's simply means that we have 8 marks, but we want to take away 12 marks, thus AFTER we take away 8 marks, we're short for 4 marks.* * * * * * * *(try to take away)* * * * * * * * * * * *(thus we're in short for) * * * *That should be easy to understand.From now on, I'm going to make an excuse from not always mentioning something antique, such our marks, but we are going to start to play this game differently. It's called the abstract thinking. It's how a mathematician runs his day. Don't worry, it's just a rather scary call, but what I mean, we would just be more sticking to the pattern of the game, WITHOUT looking back to where it comes from too often. You would find it very useful. Here it goes.Every time we subtract something bigger from something smaller, we could always write it... yes, just write it, like these:1 - 5 = 1 - 1 - 4 = 0 - 43 - 9 = 3 - 3 - 6 = 0 - 616 - 31 = 16 - 16 - 15 = 0 - 15Then, as you've guessed, we are going to invent another new symbol. Rather than writing "0 - ..." each time, we would just write "-...". We remove the symbol "0", because there's no use to write it every time. We still have to leave the "-" sign though, to differentiate it from the plain number. Then we call it as negative number. You could breath, now we have had the so called negative number! You may find it too artificial, but that's the correct way to do it; that's to ensure there's NO JUMPING in the process. Mathematicians call it rigorous.(Well… NOT really rigorous yet, but it’s always good to start somewhere with something more intuitive!)So now we could write it using the new symbol of negative number "-..." as:1 - 5 = 1 - 1 - 4 = -43 - 9 = 3 - 3 - 6 = -616 - 31 = 16 - 16 - 15 = -15(It means the same thing, the symbol "-" means "short for", so "-4" just means we're short for "4".)Then we notice some pattern! We could just run the subtraction backward and adding the symbol "-" to the result. That is if we get "1 - 5", we could just write it as "5 - 1" and then add the symbol "-" in front of it. No new rule, we just watch some pattern, and we SHORTCUT the thinking process to get the final process. That's just it. And that is my friend, what I mean by an abstract thinking in mathematics, and that's very useful (because it could shortcut our thinking process). So if we also cut out the writing, mimicking the way we think it now (in abstract), we would have something clearer:1 - 5 = -43 - 9 = -616 - 31 = -15(Subtract the smaller one from the bigger one, then add the symbol "-" to its left.)Wow, it has been quite long, I hope you got the idea already! Now I'm going to speed up this rather boring explanation, by thinking in more and more abstract. Yes, speed it up, Johnny!We're going to do something like this:2 + 2 + 2 = 6It's too cumbersome, so we invent a multiplication symbol "x". That's just counting the number of the same numbers then write it as following:3 x 2 = 6It means the same thing. Then we could have something like this:2 x 3 = 3 + 3 = 6Hmm... let's try with other numbers:3 x 4 = 4 x 3 = 125 x 6 = 6 x 5 = 30(and so on)It seems obvious that switching the numbers in a multiplication doesn't change the result. But we could not just go on and calculate it for all numbers! So we need another way to be really really sure that it is the way it is. Let's get back to what it means in the first place: counting number! For example if we have 3 x 2, it means:* * * * * *But we could write it as:* ** ** *If we turn our head to look at it from its left (or its right), or simply just to rotate it, of course the count of the marks won't change! But it would look like this:* * ** * *See, it's 2 x 3. Thus we could repeat the same procedure to any kind of number (draw the marks, then see it from its left / right), the result would be the same. Well, we have our first equation! So rather than writing it for all numbers, we invent a new kind of symbols. No new rule -- it's just about how we write it. This new kind of symbol works as following:For the same numbers, we denote it with the same letters.For different numbers, we denote it with different letters.This game is called algebra. With this we could write "2 x 3 = 3 x 2" as:a x b = b x aBut that symbol "x" looks confusing with the letter symbols, so we agree to throw it away, and write it more elegantly as following:ab = baNow how about "a x -b"? Because we may haven't used to our new symbols, let's put back our numbers, for example: "3 x -2"? What does it mean? According to our agreement, it's the same as:-2 + -2 + -2If you still feel a little bit confused, let's write it as:0 - 2 + 0 - 2 + 0 - 2Or in our old day marks notation, we're short for 2:* * ** * * * *Then short for another 2:* * ** * * * *Then, another 2:* * ** * * * *So if we combine them all:* * * * * * * * ** * * * * * * * * * * * * * *We're short for 6. Do the same procedure for all kind of number we would have result that "3 x -2" is the same as "3 x 2" with the symbol "-" at its left, or:3 x -2 = -3 x 2 = -6In our new algebra game, we could write it as:a x -b = - a x b = -abHow if we add "b" to "c" then multiply it by "a", that is:b + cThen multiply it by aTo avoid confusion, we invent two symbols "(" and ")" to mean that whatever inside those two symbols would be counted first. So for a such operation we could write it elegantly as:(b + c) a = (by switching the numbers, then we would get)a (b + c)Let's put some number back, e.g.:2 x (3 + 4)Which is of course: (no new rule -- still counting number)2 x (3 + 4) = 2 x 7 = 14But we notice something beautiful if we use back our marks notation, that is 2 x 7 is written as:* * * * * * ** * * * * * *If we arrange it a little bit differently, it would not change the result, but there's something interesting pattern there:* * * | * * * ** * * | * * * *Bingo! It's the same as:(2 x 3) + (2 x 4)So:2 x (3 + 4) = (2 x 3) + (2 x 4)We repeat the same procedure for all kind of numbers, surely we would get the same result, thus we write it as following:a (b + c) = ab + acThat, we got our second equation!This game is so fascinating that people play it for thousands of years, but we are going to stop it just here. We have had every thing we need to answer:WHY "-a x -b = ab"??(Or WHY "- x - = +"?)Here it goes... BECAUSE:-a -b =-a -b + 0 = (adding 0 would not change the counting)-a -b + ab + -ab = (subtracting something from the same thing equals 0)-a -b + -ab + ab = (shift the "-ab" term to the left)(Notice that (-a x -b) + (-a x b) has the same pattern with previous result (a x b) + (a x c) = a x (b + c), that we could write it as -a x (-b + b), so...)-a -b + -ab + ab =-a (-b + b) + ab =-a 0 + ab = (note that -a x 0 = -(a x 0) = -(0 + 0 + 0 + ...) = -0 = 0)abOr if you prefer to put our numbers back, e.g. the number "2", then we would have:-2 x -2 =-2 x -2 + 0 = (adding 0 would not change the counting)(-2 x -2) + (2 x 2) + (-2 x 2) = (subtracting something from the same thing equals 0)(-2 x -2) + (-2 x 2) + (2 x 2) = (shift the (-2 x 2) term to the left)-2 x (-2 + 2) + (2 x 2) =-2 x 0 + (2 x 2) = (note that -2 x 0 = -(2 x 0) = -(0 + 0) = -0 = 0)0 + 4 =4You may wonder why don't I just cut all the craps, and bring you directly to algebra! That's because algebra is not the most fundamental building block of math, and I think you would get much more of the fun by watching how, for 5000 years, we as human beings collectively build this wonderful tool called algebra, and thus math. (I hope!)Anyway, that's how mathematics developed, and would stay that way until unforeseeable future. So every time you see a mathematician, you could say to him / her confidently:"Whatever the result you get, I could tell you that's ONLY BECAUSE 1 + 1 = 2."Thank you.

Oh, I’d be happy to.Integration by partial fractions has a peculiar pedagogical pedigree. It is often presented in calculus courses as a way of integrating rational functions, but it is rarely accompanied by a complete theoretical description – it’s more like a messy recipe with various cases which suffice for the final exam. The presentation leaves many students unsure what exactly is it that they’re doing, what the general case is like, and why. This is usually left unresolved throughout undergrad and graduate studies.I can’t think of another instance where a basic result or technique is taught in such a half-assed way. Curriculum designers often choose to leave or include this or that in an undergrad program, but when something is included, it is usually included properly. The situation with partial fractions is akin to learning about the Jordan canonical form in the real case only, and merely through a few examples. It’s a travesty.In the paper by Bradley and Cook I cite below, they write:Analysts view this as pure algebra (which it is) so it should not be addressed in a course on analysis. Algebraists typically view partial fractions as a technique only good for integrating and thus a problem for analysts. So, outside of the realm of symbolic computation, the partial fraction decomposition tends to never be fully discussed.This doesn’t bother all students, perhaps not even most students, but it drives people like me insane. I can’t comprehend anything if I don’t see the general context and complete proofs. Scattered techniques like this make me feel like my knowledge has decreased, rather than increased.So, here’s my attempt at addressing this. I’ll try to explain the idea behind partial fraction decomposition in various contexts, as well as its (very legitimate) connection to integration, and perhaps a few other uses if we have time. I can’t do the topic full justice in an answer; I will merely be able to outline how I would teach this if I were to.Shall we?RepresentationsEveryone knows that natural numbers can be written down as usual, in decimal notation, like [math]196884[/math], or through their prime factorization, as in [math]2^2 \times 3^3 \times 1823[/math]. Most people (I think) also understand that various things are easier or harder to do in either form of notation.Many students of math also know that something very similar is true of polynomials. A (single variable) polynomial can be written down as [math]x^8 [/math][math][/math][math]+ 2 x^7 - 2 x^6 - 6 x^5 [/math][math][/math][math]+ 6 x^3 [/math][math][/math][math]+ 2 x^2 - 2 x - [/math][math]1[/math][math][/math], a sum of powers of [math]x[/math] with coefficients, or as a product of irreducible factors with possible multiplicity, as in [math](x^2+1)^2(x-1)^2[/math]. Once again, certain tasks become easier with either form of presentation.Now, what about rational numbers? And what about rational functions, which are the analogous thing for polynomials?We usually write rational numbers like [math]\frac{77}{12}[/math]: a numerator and a denominator expressed as decimals. Of course we can factor both numbers and write [math]\frac{7 \times 11}{2^2 \times 3}[/math]. Same thing.But there are other ways. One form is a continued fraction, as in[math]\displaystyle \frac{77}{12} = 6+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}[/math]This is also (extremely) useful in various ways, but we won’t go there now. Are there other ways?Well, yes there are.For reasons which, I think, aren’t fully understood, early texts from Egypt’s Middle Kingdom exhibit a strong preference to represent fractions as sums of distinct unit fractions, as in[math]\displaystyle \frac{29}{24} = \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{8}[/math].Such representations, dubbed “Egyptian Fractions”, do sometimes have a utility, but they aren’t important enough nowadays to be taught as a systematic way of representing rational numbers. We should note the idea, though: we are decomposing a rational number into a sum of fractions which have a certain simplicity. They all have a numerator of [math]1[/math][math][/math].Also, let’s note something about the denominators in that representations: [math]2, 3, 4[/math] and [math]8[/math] all divide [math]24[/math]. This isn’t surprising, when you think about what happens when you add up fractions. We’ll see how this plays out later.What would be the analogous thing for rational functions? And why would it be useful?Integrating Rational FunctionsFirst, a reminder: a rational function is a ratio of two polynomials, such as[math]\displaystyle \frac{x^5-2x^3+1}{x^2-3}[/math].In general, we write[math]\displaystyle R(x) = \frac{f(x)}{g(x)}[/math]where [math]f,g[/math] are polynomials.Rational functions come up in many different contexts, both in pure math and in applications. Just like rational numbers, they can be represented in different ways, so first let us motivate why we would go looking for a representation as a sum of simple fractions.When we learn about derivatives, we quickly see that differentiating any expression is a purely mechanical process. This is because, once we know to differentiate [math]f[/math] and [math]g[/math], we can also handle [math]f+g[/math], [math]fg[/math], and also [math]f \circ g[/math] which is the composition [math]f(g(x))[/math].Then we start going the other way, finding antiderivatives: given a function [math]f[/math], we seek a function [math]F[/math] with [math]F'=f[/math]. This is usually denoted as [math]F=\int f(x)\,\mathrm{d}x[/math], an indefinite integral. And here things get complicated.Indefinite integrals are linear, so they behave nicely with respect to addition and scalar multiplication. But unfortunately, that’s pretty much it. If you know the antiderivatives of [math]f[/math] and [math]g[/math], there’s generally no way to determine the antiderivatives of [math]fg[/math] or [math]f\circ g[/math] and so on.We do know how to integrate polynomials, and even some very simple rational functions, because we all know that [math]\frac{d}{dx}x^n = nx^{n-1}[/math], and so[math]\displaystyle \int x^k \,\mathrm{d}x = \frac{1}{k+1}x^{k+1}[/math]for any integer [math]k \ne -1[/math], while[math]\displaystyle \int x^{-1} \,\mathrm{d}x = \log(x)[/math]where [math]\log[/math] denotes the natural logarithm. As a result, we can also handle [math]\int (x-a)^k\,\mathrm{d}x[/math] for any number [math]a[/math] and integer [math]k[/math].Ok great. But what about more complicated rational functions, like [math]\frac{x^3+x+5}{x^2-1}[/math]?Well, obviously, all we need to do is manage to write every rational function as a sum of those simple ones, like [math]\frac{1}{(x-a)^k}[/math], or even [math]\frac{17}{(x-a)^k}[/math], since we aren’t worried about multiplicative constants. Integrating sums is easy, even with whatever multiplicative constants (aka linear combinations).Note well: the “simple” fractions we seek have very simple numerators, just numbers; and denominators which are just powers of a linear polynomial. As we shall see, this works perfect over the complex numbers, but another complication is introduced if we insist on working with real numbers only.Partial Fraction Decomposition: Fooling AroundBefore we do the general case, let’s look at some simple examples.[math]\displaystyle \frac{1}{x-1}[/math]This one is already fine. We know how to integrate [math](x-1)^{-1}[/math].[math]\displaystyle \frac{x}{x-1}[/math]Notice how merely changing the numerator from [math]1[/math][math][/math] to [math]x[/math] makes the integral non-obvious. However, there’s an easy fix:[math]\displaystyle \frac{x}{x-1} = \frac{x-1+1}{x-1}=1+\frac{1}{x-1}[/math]and similarly[math]\displaystyle \frac{x^2}{x-1} = \frac{(x+1)(x-1)+1}{x-1} =(x+1)+\frac{1}{x-1}[/math]what’s going on here? This should remind you of something: we’re simply replacing the numerator with a quotient and remainder upon division by the denominator.[math]\displaystyle \frac{19}{6} = \frac{3 \times 6 [/math][math][/math][math]+ 1}{6} = 3+\frac{1}{6}[/math]In a similar way, we re-wrote the numerator [math]x^2[/math] as [math]q(x)(x-1)+r(x)[/math], a quotient and a remainder. A basic skill you should have before tackling partial fractions is polynomial division with remainder, aka the Euclidean Algorithm for polynomials. It really works the exact same way as with numbers.The general conclusion here is that we can handle anything of the form[math]\displaystyle \frac{f(x)}{x-1}[/math]by simply dividing [math]f(x) = q(x)(x-1)+r(x)[/math], and so[math]\displaystyle \frac{f(x)}{x-1} = q(x)+\frac{r(x)}{x-1}[/math]In fact, since the denominator has degree [math]1[/math][math][/math], the remainder [math]r(x)[/math] is guaranteed to simply be a constant – just a number. We have no trouble integrating this.Ok, let’s move on.[math]\displaystyle \frac{1}{(x-1)^2}[/math]Still no problem. This is [math](x-1)^{-2}[/math]. Easy to integrate.[math]\displaystyle \frac{x}{(x-1)^2}[/math]Hmmm. Polynomial division won’t save us here, because the numerator already had a smaller degree than the denominator. What do we do?Here comes one of the basic ideas of Partial Fraction Decomposition. We can’t possibly hope to rewrite [math]\frac{x}{(x-1)^2}[/math] as simply [math]\frac{a}{(x-1)^2}[/math] for some number [math]a[/math], because [math]x[/math] isn’t a constant. But we might still hope to manage something like this:[math]\displaystyle \frac{x}{(x-1)^2} = \frac{a}{x-1}+\frac{b}{(x-1)^2}[/math]Why is that reasonable? Well, adding together the fractions on the right, we find[math]\displaystyle \frac{a}{x-1}+\frac{b}{(x-1)^2} = \frac{a(x-1)+b}{(x-1)^2}[/math]The denominator is simply [math](x-1)^2[/math], which is what we need, and the numerator looks darn close to a general polynomial of the first degree. Great: we can surely adjust [math]a[/math] and [math]b[/math] so they make [math]a(x-1)+b[/math] be whatever we want, such as our original numerator which happened to be [math]x[/math]. Clearly we want [math]a=1[/math], and then we get [math](x-1)+b[/math], so we want [math]b=1[/math] as well.[math]\displaystyle \frac{x}{(x-1)^2} = \frac{1}{x-1}+\frac{1}{(x-1)^2}[/math]The cool thing is, we could have done the exact same thing with [math]\frac{mx+n}{(x-1)^2}[/math] for any numbers [math]m,n[/math]. We just need to match [math]a(x-1)+b[/math] with [math]mx+n[/math], which is no trouble at all.And the coolest thing is, this lets us handle numerators up to degree [math]1[/math][math][/math], and numerators of degree [math]2[/math] and beyond can be handled as before, by first dividing by the denominator, being left with a remainder of lower degree – meaning, the first degree! We are able to exactly match the freedom we need.You may guess how this is going to go from here. For things like [math]\frac{f(x)}{(x-1)^3}[/math], we will first reduce [math]f(x)[/math] to a remainder of the second degree, and then handle it by finding appropriate constants [math]a,b,c[/math] to make [math]\frac{a}{x-1}+\frac{b}{(x-1)^2}+\frac{c}{(x-1)^3}[/math] do what we need it to do.Ok. So far the denominators were all [math](x-1)^k[/math] for some power [math]k[/math]. The same thing obviously works for [math](x-u)^k[/math] for any [math]u[/math], but what about more complicated denominators? For example, what are we supposed to do with[math]\displaystyle \frac{1}{x^2-1}[/math]?Well, [math]x^2-1[/math] factors as [math](x-1)(x+1)[/math]. Does this help? Can we hope to achieve[math]\displaystyle \frac{1}{x^2-1} = \frac{a}{x-1}+\frac{b}{x+1}[/math]?Why, sure. Again, let’s just add up those guys on the right, getting [math]\frac{a(x+1)+b(x-1)}{(x-1)(x+1)}[/math]. The numerator can be adjusted to be any linear polynomial, and that’s all we need since the denominator has degree [math]2[/math].What’s the general lesson? The general lesson is that the denominator [math]g(x)[/math] should be broken down into a product of [math](x-u)[/math]'s, and then we rewrite the fraction as a sum of simple things over these [math](x-u)[/math]'s individually. We’ve also learned that if any [math](x-u)[/math] appears with multiplicity, as in [math](x-u)^k[/math], we should allow for denominators of [math](x-u)^j[/math] for [math]j=1,2,\ldots,k[/math], to get enough freedom to match our numerators.Partial Fractions over the Complex NumbersBut… can we factor any polynomial [math]g(x)[/math] into linear factors, as in [math]g(x)=(x-u)^k(x-v)^l\ldots(x-w)^m[/math]? Can we?Well, sure we can. That’s just what complex numbers were made for. Over the complex numbers, any polynomial has just as many roots as its degree, or in other words, every polynomial factors completely into linear factors.Therefore, over the complex numbers, the random meanderings we just went through are completely sufficient to achieve a partial fraction decomposition of any rational function. It can have rational coefficients, it can have real coefficients, it can even have complex coefficients, doesn’t matter – it’s totally doable.Start with[math]\displaystyle R(x) = \frac{f(x)}{g(x)}[/math]If the degree of [math]f(x)[/math] is greater than that of [math]g(x)[/math], use long division to write [math]f(x) = q(x)g(x)+r(x)[/math] where [math]r[/math] has lower degree. Then[math]\displaystyle R(x) = q(x) [/math][math][/math][math]+ \frac{r(x)}{g(x)}[/math]and now the numerator [math]r(x)[/math] is smaller, in terms of degree, than the denominator.Next, factor [math]g(x) = (x-u_1)(x-u_2)\cdots(x-u_n)[/math]. The roots [math]u_1,u_2,\ldots,u_n[/math] may not be distinct, so a better way to write this is[math]\displaystyle g(x) = (x-u_1)^{m_1}\cdots(x-u_k)^{m_k}[/math]where [math]u_1,\ldots,u_k[/math] are the distinct roots and [math]m_1,\ldots,m_k[/math] are their multiplicities. We expect to need simple fractions that look like [math]\frac{a_{ij}}{(x-u_i)^j}[/math] where [math]a_{ij}[/math] are just some numbers, nothing fancy, and the denominators are all powers of [math](x-u_i)[/math], never exceeding the appropriate multiplicity, meaning [math]j =1,2,\ldots,m_i[/math].So the overall decomposition we expect looks like this:[math]\displaystyle \frac{r(x)}{g(x)} = \sum_{i=1}^k \sum_{j=1}^{m_i} \frac{a_{ij}}{(x-u_i)^j}[/math]This looks a bit scary, but seriously, it’s nothing. You merely write out the right hand side, add things up, and solve for the [math]a_{ij}[/math] to make them fit your actual numerator [math]r(x)[/math]. It’s nothing worse than solving linear equations, and you are guaranteed to succeed since there are just enough parameters [math]a_{ij}[/math] to fit the numerator [math]r(x)[/math].A formal proof of this can be written down in several ways. Since this answer is long enough as it is, let me point you to a wonderful paper by Bradley and Cook (from, incredibly, 2012!), which does everything with complete rigor.This Just Isn’t Hard Enough: Real NumbersThe way partial fractions are taught in most calculus or real analysis classes involves one more layer of complication, because teachers often insist on doing everything with real numbers. Honestly, in my mind, this adds unnecessary stress on top of something which is already notationally cumbersome, even if it’s not really complicated. Complex numbers are just awesome in how they let you factor any polynomial down to linear factors – why not use them?Here’s what actually happens in the classroom. Using the general method we just outlined, we find[math]\displaystyle \frac{1}{x^2-1} = \frac{1/2}{x-1}-\frac{1/2}{x+1}[/math]so we can integrate [math]\frac{1}{x^2-1}[/math] easily using logs.[math]\displaystyle \int \frac{1}{x^2-1}\,\mathrm{d}x = \frac{1}{2}\log(x-1)-\frac{1}{2}\log(x+1)[/math].Similarly,[math]\displaystyle \frac{1}{x^2+1} = \frac{-i/2}{x-i}+\frac{i/2}{x+i}[/math]so[math]\displaystyle \int \frac{1}{x^2+1}\,\mathrm{d}x = \frac{i}{2}\log(x+i)-\frac{i}{2}\log(x-i) [/math][math][/math][math][/math]but that’s not the answer most teachers want, unfortunately. They want [math]\text{arctan}(x)[/math] or [math]\text{atan}(x)[/math] or [math]\tan^{-1}(x)[/math] or whatever they call the inverse tangent function.Now of course, it’s the same thing. The expression we just wrote down with logs and [math]i[/math] is exactly the inverse tangent. An antiderivative is an antiderivative; there can’t be two of them, except for additive constants. But we’re supposed to fear the complex number, so instead of doing it the right and easy way, we complicate things more.Remember: partial fraction decomposition is really quite straightforward with complex numbers. You factor the denominator completely, you write down the required denominators to cover the multiplicities, done. Integrating is never more complicated then knowing how to integrate [math](x-u)^k[/math], which is really easy.But if you insist on doing things with real numbers only, you still can… at a cost. A polynomial with real coefficients cannot be factored into linear factors with real coefficients, but it can factored into linear and quadratic factors with real coefficients. (Why? Because the complex non-real roots of a real polynomial must come in conjugate pairs, and multiplying these pairs out gives you real quadratics).That’s not the end of the world. It just means that now, even in the absence of those annoying multiplicities, we will need to allow terms such as[math]\displaystyle \frac{ax+b}{x^2+ux+v}[/math]in our partial fraction decomposition. If we also have multiplicities, we will likewise need to allow[math]\displaystyle \frac{ax+b}{x^2+ux+v}+\frac{cx+d}{(x^2+ux+v)^2}[/math]or additional terms like this if the multiplicity is higher. The process of figuring out the coefficients now becomes correspondingly more cumbersome, but this is really just a nuisance.In terms of integration, it also means that we need to know how to integrate those simple fractions: linear numerators and quadratic denominators. This isn’t hard either, using a few standard trigonometric integrals.The general philosophy of partial fraction decomposition for rational functions over the real numbers is still the same: we write a general rational function as a sum of a polynomial and simple fractions, where “simple” now means having powers of linear of quadratic terms in the denominators, and nothing worse than linear polynomials in the numerators. Since we know how to integrate those, we can integrate any rational function.The General CaseIt is tempting to try and generalize everything we just did. We’ve handled rational functions over the real and complex numbers; what about other fields? And what if, instead of polynomials, we look at fractions made of elements of other rings?One of the simplest steps in our process was the reduction of the numerator into something of lower degree than the denominator. This concept of “degree” generalizes nicely into something called Euclidean domains, where you have a “degree” function and a guaranteed division with a remainder which has smaller degree than what you’re dividing by.In the general case, we cannot hope to control the degree of the irreducible factors of the denominators. Above, we made good use of the fact that complex numbers guarantee linear factors only, and real numbers guarantee linear or quadratic ones. In general, we don’t have a limit on the degrees of the irreducible factors, but that’s ok: we can still achieve a partial fraction decomposition where the denominators are powers of irreducible factors.This helps clarify things, I think. The denominators in the approach as it’s usually taught are a hodge-powers of linear and quadratic factors with various powers. The underlying mechanism is irreducibility: you factor the denominator as much as you can, and allow for power of these factors in the decomposition.One interesting quirk is that, in those more general cases, the decomposition ends up being non-unique. There doesn’t seem to be a simple recipe to define precisely the conditions on the terms that would guarantee uniqueness. The decomposition always exists, but it’s not necessarily unique in any obvious sense.For the purposes of finding antiderivatives, this really isn’t a concern at all. The decomposition we’ve described over [math]\C[/math] and [math]\R[/math] is, in fact, unique, but it doesn’t really need to be: we know from other considerations that an antiderivative must be unique up to an additive constant (since [math]f'=g'[/math] implies [math](f-g)'=0[/math], so [math]f-g[/math] is a constant). But it’s an interesting observation nonetheless. Uniqueness is often harder than mere existence, and it’s equally often very much necessary.To go back to the context we started with, way back at the top of this answer: what can we say about the partial fraction decomposition of ordinary rational numbers? Well, it does always exist, and it is actually unique if we insist on always taking positive remainders. We get things like[math]\frac{77}{12}=5+\frac{1}{2}+\frac{1}{2^2}+\frac{2}{3}[/math]note that this isn’t the “Egyptian Fraction” decomposition from ancient Egypt, but we do see how the numerators are smaller than the denominators. Also observe how we subtly needed to reduce the integer part from [math]6[/math] to [math]5[/math] in order for the decomposition to work. It’s a fascinating way of representing rational numbers, but unfortunately it’s not particularly useful (as far as we know), so we never teach it, which leaves the calculus technique sort of hanging there without context or precedent, which is really a shame. It’s a simple, and very useful, technique.

From the Common Core for Mathematics Grade 3 there are four focus areas for the year. The first is relevant to the test.(1) Students develop an understanding of the meanings of multiplication and division of whole numbers through activities and problems involving equal-sized groups, arrays, and area models; multiplication is finding an unknown product, and division is finding an unknown factor in these situations. For equal-sized group situations, division can require finding the unknown number of groups or the unknown group size. Students use properties of operations to calculate products of whole numbers, usingincreasingly sophisticated strategies based on these properties to solve multiplication and division problems involving single-digit factors. By comparing a variety of solution strategies, students learn the relationship between multiplication and division.The actual standards referenced at the top of the test sheet are:CCSS.Math.Content.3.OA.A.1Interpret products of whole numbers, e.g., interpret 5 × 7 as the total number of objects in 5 groups of 7 objects each. For example, describe a context in which a total number of objects can be expressed as 5 × 7.CCSS.Math.Content.3.OA.A.3.Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.The child’s version of the objectives are:“I can use multiplication strategies to help me multiply”And finally, we get to the discussion’s question:“Use the repeated addition strategy to solve : 5 x 3”What we notice is that the test question is an attempt to determine if the child retains the conceptual understanding of “5 x 3”, not just as symbols, but as a representation that could be of physical manipulatives, i.e., the objects arranged in groups in groups cited in the standard. This is important and should be validated.The problem begins with failure to cue the child to some physical objects. This is not a creative writing class. The question should suggest the manipulatives. Maybe: breadsticks and baskets.But, no. The question takes a shortcut to sneak in a little bit more testing, and immediately asks that a certain strategy be used – the repeated addition strategy. In the process, it fails an objective.A strategy is a template for action. The action follows a strategy in a context.The test has not provided the context, which would determine the action. It has only asked for an example of the strategy, and has skipped the anchoring to objects and groups for which the strategy can be applied in context.The author of the question took a shortcut and produced a defective question. The child produced a repeated addition expression, an otherwise arbitrary artifact with no requirement to relate to anything physical, at all.If the child had imagined 5 rows of three pennies, even though that not required, the child could simply rotate the mental image 90 degrees to get a simpler repeated sum, which was all teacher asked for.But the child was not asked to show the understanding that the symbols related to physical objects, even though that WAS part of the lesson objective.The answer was marked wrong because the teacher graded the test against expected answers to questions that were not actually asked.