Write Each Polynomial In Standard: Fill & Download for Free

If the polynomial is in a single variable, then standard form would require you to factor the variable out, leaving a parenthetical sum of constants as the single coefficient, e.g.:[math]3x^2-5x^2+\pi x^2-ex^2=(\pi-e-2)x^2[/math]If, on the other hand, the polynomial involves multiple variables but each term has equal degree (based on the sum of the exponents in each term), then you would arrange the terms first in alphabetical order by the variables, then in descending order of degree for the leading variable, e.g.:[math]4x^6–17x^4y^2+7x^3y^3–2xy^5+5y^6[/math]Note that each term is degree 6, but arranged in decreasing order of [math]x[/math] power.

The Laguerre polynomials are "orthogonal" with respect to the standard exponential distribution (or, measure). Fundamentally, the Laguerre polynomials are "good polynomials" in a few ways when we need to perform calculations which employ the exponential function. Two situations where this property is useful are in computing Gaussian quadratures, and in Polynomial Chaos expansions. Both examples involve an integral with respect to the standard exponential function.To elaborate on the first sentence, orthogonality with respect to the standard exponential distribution means the following; if we call the Laguerre polynomials [math] p_0,p_1,p_2,\dots [/math], then[math] \int_{\mathbb{R}_{\geq 0}} \, p_i(x) \, p_j(x) \, e^{-x} \, \mathrm d x = \delta_{i,j} \, c_j [/math]That is; if we integrate any of these two polynomials against each other, with respect to the exponential distribution, then the result is zero unless they are the same polynomial - in which case the result is some finite number (different for each polynomial).Gaussian quadrature is a well-known, specific method of computing integrals. In general, a quadrature rule is a formula for approximately integrating a function. This approximation is in the form of a sum - we can write as follows;[math] \int_a^b \, f(x) \, \mathrm d x \approx \sum_{i=1}^n \, w_i \, f(x_i) [/math]where we aim to choose some nodes [math]x_i[/math], and some weights [math]w_i[/math], which make the above approximation as accurate as possible. If our function [math]f[/math] can be written as the product of two functions [math]g(x) \, h(x)[/math] where [math]h(x) = e^{-x}[/math], then the Gaussian quadrature rule tells us to use the roots of the Laguerre polynomials as the nodes, and provides a suitable formula for the weights. Other polynomials are used for other choices of functions [math]h[/math].Polynomial Chaos expansions are similar to Fourier series, but (as the name suggests) use polynomials rather than trigonometric functions. For example, the following property of the sine function is hopefully familiar;[math] \int_0^{2 \pi} \, sin(m x) \, sin(n x) \, \mathrm d x = \delta_{m,n} [/math]See the similarity with the orthogonality relationship for Laguerre polynomials above? In fact, one way of expressing this property is to say that "[math]sin(n x)[/math] and [math]sin(m x)[/math] are orthogonal".The Fourier series for a (sufficiently nice and periodic) function [math]f[/math] is given by[math] f(x) = \sum_{k=1}^\infty \, \hat{f}(k) \, sin(kx) [/math]The "standard" method of computing the coefficients [math]\hat{f}(k)[/math], which is to compute the integral[math] \int_0^{2 \pi} \, f(x) \, sin(kx) \, \mathrm d x [/math]uses the orthogonality of the different sine functions, as above - if you integrate the right hand side similarly, then move the sum outside the integral, all but one of the integrals in this sum disappear by the orthogonality property.A Polynomial Chaos series for a (nice) function [math]f[/math] can be given by[math] f(x) = \sum_{k=1}^\infty \, \hat{f}(k) \, p_k (x) [/math]where the [math]p_k[/math] are the Laguerre polynomials. The formula for the coefficients can be found the same as above, by replacing [math]sin(kx)[/math] with the product [math]p_k(x) \, e^{x}[/math]. So this type of PC expansion is effectively just a Fourier series using the Laguerre polynomials.

Absolutely. In fact, the basic idea of modular arithmetic—that of quotienting—is completely general, and applies not just to things like the Gaussian or Eisenstein integers, but actually any ring whatsoever, including non-commutative examples, like the collection of [math]2 \times 2[/math] matrices with real coefficients.I thought for a while about whether to go into the general theory or not for this question. Ultimately, I decided that it was better to treat the general case with many examples, because I was going to need to talk about finite rings anyway, and if I was introducing rings, it made sense to introduce ideals while I was at it.What is a ring? A ring is any collection [math]R[/math] with two binary operations [math]+,\cdot[/math] (that is, these two operations take in any two members of [math]R[/math] as inputs, and they return a member of [math]R[/math] as an output) and two special elements called [math]0[/math] and [math]1[/math], satisfying the following properties:For all [math]a,b,c[/math] in [math]R[/math], [math]a + (b + c) = (a + b) + c[/math] and [math]a \cdot (b \cdot c) = (a \cdot b) \cdot c[/math].For all [math]a,b[/math] in [math]R[/math], [math]a + b = b + a[/math].For all [math]a[/math] in [math]R[/math], [math]a + 0 = 0 + a = a[/math], and [math]a \cdot 1 = 1 \cdot a = a[/math].For all [math]a[/math] in [math]R[/math], there exists a member which we call [math]-a[/math], defined by the property that [math]a + -a = -a + a = 0[/math].For all [math]a,b,c[/math] in [math]R[/math], [math]a \cdot (b + c) = a \cdot b + a \cdot c[/math].Common examples of rings include the integers, the integers modulo [math]N[/math] (where [math]N[/math] is some fixed positive integer), the Gaussian integers, the Eisenstein integers, the rational numbers, the real numbers, the complex numbers, polynomials with complex coefficients, [math]2 \times 2[/math] matrices with real coefficients, quaternions, etc., etc..Note that multiplication need not be commutative (e.g. matrix multiplication isn’t), nor does it need to be the case that a non-zero element has a multiplicative inverse (e.g. none of the integers other than [math]\pm 1[/math] have multiplicative inverses—or, if you prefer, their inverses are not integers). We will only consider commutative rings, which are precisely those rings for which multiplication is commutative. We don’t need to do this—you can define quotients even for general rings—but this restriction will make our life easier.Rings can be quite complicated, and so a common trick that one might want to employ is to figure out some way to simplify a given ring, so that it retains some information from the original ring, but other aspects of it are removed from play. This is precisely what modular arithmetic allows one to do with the integers—say that you want to determine whether a polynomial equation like [math]X^3 + Y^3 + Z^3 = 4[/math] has any integer solutions. This might be quite difficult to do with the integers themselves, so instead you try to figure out if this equation has solutions in the integers modulo [math]N[/math]—if you happen to find an [math]N[/math] for which there is no solution, then you know that there can be no integer solution. On the other hand, for a given [math]N[/math], determining whether a polynomial has solutions in the integers modulo [math]N[/math] is a finite computation, since there are only finitely many possibilities. And, indeed, in the given case, checking modulo [math]9[/math], we find that there are no solutions.The more general idea is precisely that of taking quotients, which will give us a means of taking a ring and producing a new ring that is simpler in some respects. Here is going to be the rough idea: given a ring [math]R[/math], we are going to construct a quotient [math]R/I[/math], where [math]I[/math] is some subset of [math]R[/math]. The intuitive meaning of this is that we going to start with elements of [math]R[/math], but if they differ by elements in [math]I[/math], we’ll consider them to be “the same.” This is precisely how the integer modulo [math]N[/math] work—we consider, say, [math]1[/math] and [math]9[/math] to be the same modulo [math]4[/math] because they differ by a multiple of [math]4[/math]. That is, for the integers modulo [math]N[/math], the set [math]I[/math] of interest is precisely the collection of all multiples of [math]N[/math].Another way to think about it is that everything in [math]I[/math] becomes [math]0[/math] inside of the quotient ring [math]R/I[/math]. So, [math]0,\pm 3, \pm 6, \pm 9, \ldots[/math] all get identified with [math]0[/math] in the integers modulo [math]3[/math]. However, if we have two things [math]x,y[/math] that will get identified with [math]0[/math], then [math]x + y[/math] should get identified with [math]0[/math] as well, as should [math]-x[/math]. Indeed, if [math]x[/math] will get identified with [math]0[/math], then [math]xy[/math] should get identified with [math]0[/math], regardless of what [math]y[/math] is.In short, not just any old subset [math]I[/math] will be suitable: what we need are ideals.What is an ideal? Given a ring [math]R[/math], a subset [math]I[/math] of it is an ideal if:For every [math]x,y[/math] in [math]I[/math], [math]x + y[/math] and [math]x - y[/math] is in [math]I[/math], andFor every [math]x[/math] in [math]I[/math], and for every [math]r[/math] in [math]R[/math], [math]rx[/math] is in [math]I[/math].In other words, an ideal is something such that we can freely take sums and differences of its members, and we can freely scale its members by elements of the ring. Note that these are exactly the properties that we requested in the previous section as what we would want from subsets that we could quotient by. Let’s give some examples of ideals.As noted previously, for any integer [math]N[/math], the collection of integer multiples of [math]N[/math] is an ideal of the integers.The collection of polynomials with no constant term is an ideal of polynomials with real coefficients and in one variable [math]X[/math]. (Another way to think about this is that this is the collection of all products of [math]X[/math] with another polynomial.)The collection of polynomials divisible by [math](X^2 + 1)[/math] is an ideal of polynomials with real coefficients and in one variable [math]X[/math].The collection of polynomials with no constant term is an ideal of polynomials with real coefficients and in two variables [math]X, Y[/math].More generally, given any ring [math]R[/math] and some elements [math]x_1, x_2, \ldots x_n[/math] of [math]R[/math], we can define an ideal [math](x_1, x_2, \ldots x_n)[/math], consisting of all sums and products of these elements with elements of [math]R[/math]. With this notation, note that we can concisely describe all of the preceding examples: they are [math](N)[/math] inside of the integers, [math](X)[/math] inside the polynomials with real coefficients and in one variable [math]X[/math], [math](X^2 + 1)[/math] inside the polynomials with real coefficients and in one variable [math]X[/math], and [math](X,Y)[/math] inside the polynomials with real coefficients and in one variable [math]X,Y[/math].For some rings, all ideals can be written down in the form [math](x)[/math] for some [math]x[/math] in the ring—the integers, Gaussian integers, Eisenstein integers, and real polynomials in one variable are actually all examples of this case, which is deeply connected to the fact that there is unique prime factorization in each of them. However, there are other rings for which some ideals require multiple, maybe even infinitely many, elements to fully describe—the collection of real polynomials in [math]X[/math] and [math]Y[/math] is an example, since the there is no way to express the ideal [math](X,Y)[/math] in terms of just one element. I leave a proof of this fact to the curious reader. Another example is the ring consisting of everything of the form [math]a + b\sqrt{-5}[/math], where [math]a,b[/math] are integers—it turns out that the ideal [math](2, 1 + \sqrt{-5})[/math] is necessarily generated by two elements.In any case, we can now define what we mean by a quotient ring. Given a ring [math]R[/math] and an ideal [math]I[/math], the quotient ring [math]R/I[/math] consists of all equivalence classes of [math]R[/math], where two elements [math]x,y[/math] are considered “equivalent” if and only if [math]x - y[/math] is in [math]I[/math]. Or, to put it in more layperson terms, we start with [math]R[/math], but then we’ll think of two such elements as being the same if they differ by an element of [math]I[/math], which is one of these elements that we crush to zero. One checks that addition and multiplication in [math]R/I[/math] is well-defined, making it into a proper ring. (The proof of this is exactly the same as the proof that adding or multiplying two integers modulo [math]N[/math] is well-defined.)The prototypical example of an ideal and a quotient ring is [math](N)[/math] and the integers modulo [math]N[/math]. Let’s see some less famous examples.Consider the real polynomials in one variable [math]X[/math]—these are typically denoted by [math]\mathbb{R}[X][/math] (since [math]\mathbb{R}[/math] denotes the reals, and the [math]X[/math] denotes that we have added on a variable). As we noted previously, [math](X)[/math] is an ideal of [math]\mathbb{R}[X][/math]. This begs the question: what is the quotient ring [math]\mathbb{R}[X]/(X)[/math]?Well, two polynomials [math]p(X), q(X)[/math] correspond to the same element of [math]\mathbb{R}[X]/(X)[/math] if and only if [math]p(X) - q(X)[/math] has no constant term. But that means that if [math]p(X) = a_0 + a_1 X + a_2 X^2 + \ldots + a_n X^n[/math], then [math]p(X)[/math] will be equivalent to just [math]a_0[/math] inside the quotient ring. In other words, quotienting by [math](X)[/math] makes it so that we “forget” everything about the polynomials other than their constant terms.This means that [math]\mathbb{R}[X]/(X)[/math] is nothing more than [math]\mathbb{R}[/math], and we can think of this quotient as being nothing more than what happens if we evaluate all of our polynomials at [math]X = 0[/math], or equivalently what happens if we just look at the constant terms.Consider again [math]\mathbb{R}[X][/math], but this time we’ll look at the ideal [math](X^2 + 1)[/math]. The quotient ring [math]\mathbb{R}[X]/(X^2 + 1)[/math] is going to be more complicated this time, but we can still get a handle on it. Note that I can write any real polynomial in the form [math]p(X) = a_0 + a_1 X + (X^2 + 1)q(X)[/math], for some other polynomial [math]q(X)[/math] (that this is true comes from polynomial long division).In the quotient ring, however, the [math](X^2 + 1)[/math] term is going to disappear, so we are just left with polynomials [math]a_0 + a_1 X[/math]. Addition of these polynomials behaves normally, but multiplication is a little weird.[math]\begin{align*} (a_0 + a_1 X)(b_0 + b_1 X) &= a_0 b_0 + (a_1 b_0 + a_0 b_1)X + a_1 b_1 X^2 \\ &= a_0 b_0 + (a_1 b_0 + a_0 b_1)X - a_1 b_1 \\ &= (a_0 b_0 - a_1 b_1) + (a_1 b_0 + a_0 b_1)X, \end{align*} \tag*{}[/math]where we have used the fact that since [math]X^2 + 1 \equiv 0[/math], it must be true that [math]X^2 \equiv - 1[/math]. In fact, if you stare at this, you’ll eventually see that these are exactly the rules for multiplying complex numbers together. In other words, what we have in effect done is set [math]X^2 = -1[/math], which leads it to behave like the complex numbers (in fact, this is one of the standard ways to define the complex numbers). In short, [math]\mathbb{R}[X]/(X^2 + 1)[/math] is nothing more than [math]\mathbb{C}[/math], the complex numbers.Consider [math]\mathbb{R}[X, Y][/math], the collection of real polynomials in two variables [math]X[/math] and [math]Y[/math]. The ideal [math](X,Y)[/math] will consist of all polynomials with no constant term—it’s comparatively easy to see that in fact [math]\mathbb{R}[X,Y]/(X,Y)[/math] will just give you [math]\mathbb{R}[/math] again, by an argument extremely similar to the one that I gave above.Okay. Let’s consider something a little more involved, shall we? Let’s consider the Gaussian integers [math]\mathbb{Z}[i][/math] (so written as they are what you get when you start with the integers [math]\mathbb{Z}[/math], and add to them the symbol [math]i[/math] defined by [math]i^2 = -1[/math]). Let’s start with a comparatively simple ideal: [math](3)[/math].What does [math](3)[/math] actually consist of? Well, the Gaussian integers consist of everything of the form [math]a + bi[/math], where [math]a,b[/math] are integers. The ideal [math](3)[/math] consists of all sums and products of [math]3[/math] with such elements—it isn’t too hard to see that this will precisely be the collection of everything of the form [math]3a + 3bi[/math]. That is, [math](3)[/math] is just a lattice of everything with both coordinates divisible by [math]3[/math], illustrated below.Now, what is [math]\mathbb{Z}[i]/(3)[/math]? Well, if you think about it, if I have a Gaussian integer [math]a + bi[/math], I can shift it over by [math]3[/math] and [math]3i[/math] however I wish, without actually changing what it will be inside of [math]\mathbb{Z}[i]/(3)[/math]. So, we can think of elements in [math]\mathbb{Z}[i]/(3)[/math] as being represented by Gaussian integers with both coordinates between [math]0[/math] and [math]2[/math]—these are precisely the elements in the following square.Explicitly, these are [math]0,1,2, i, 1 + i, 2 + i, 2i, 2i + 1, 2i + 2[/math]. All that remains is to work out how addition and multiplication of these things works. Addition is fairly straightforward: [math]a + bi + c + di = (a + c) + (b + d)i[/math], but with the stipulation that if either coordinate goes over [math]2[/math], you should “wrap around” in exactly the same way you do with the integers modulo [math]3[/math]. Multiplication is only slightly harder:[math]\displaystyle (a + bi)(c + di) = (ac - bd) + (ad + bc)i, \tag*{}[/math]where we understand that all coefficients should be reduced modulo [math]3[/math]. In short, what we have found is that [math]\mathbb{Z}[i]/(3) = \mathbb{Z}/(3) [i][/math], and that this is essentially just an extension of the usual modular arithmetic modulo [math]3[/math], except that we have an extra symbol [math]i[/math] with the property that [math]i^2 = -1 = 2[/math].Let’s keep thinking about [math]\mathbb{Z}[i][/math], but let’s change the ideal to [math](1 + 2i)[/math]. I claim that this ideal exactly consists of everything of the form [math]a(1 + 2i) + b(2 - i)[/math], where [math]a,b[/math] are integers. To see this, first note that certainly all expressions of this form are in [math](1 + 2i)[/math], since [math]-i(1 + 2i) = 2 - i[/math], and by definition ideals are closed under addition and scaling. To prove that that this encompasses everything in [math](1 + 2i)[/math], it suffices to prove that the given collection is closed under addition and scaling. Addition is easy to check. As for scaling:[math]\begin{align*} (c + di)\left(a(1 + 2i) + b(2 - i)\right) &= ac(1 + 2i) + bc(2 - i) - ad(2 - i) + bd (1 + 2i) \\ &= (ac + bd)(1 + 2i) + (bc - ad)(2 - i), \end{align*} \tag*{}[/math]which settles the matter. We conclude therefore that [math](1 + 2i)[/math] is a lattice inside of the Gaussian integers, which looks as follows.As before, quotienting out by this ideal allows us to move any and all points inside of some square of our choosing. For example, we might make the following choice.However, there is a more illuminating choice that can be made, because can actually translate all of our points onto the [math]x[/math]-axis.So, [math]\mathbb{Z}[i]/(1 + 2i)[/math] consists of the elements [math]0,1,2,3,4[/math]—in other words, it is just [math]\mathbb{Z}/(5)[/math], the integers modulo [math]5[/math].Here’s one final example with the Gaussian integers. Consider the ideal [math](1 + i)[/math]. I will leave it as an exercise for the reader to check that this consists of everything of the form [math]a(1 + i) + b(1 - i)[/math], where [math]a,b[/math] are integers (the proof is very similar to the preceding example), which is the following lattice.As before, we can translate Gaussian integers inside the minimal square—but that only leaves two such points, namely [math]0[/math] and [math]1[/math]!Ergo, [math]\mathbb{Z}[i]/(1 + i) = \mathbb{Z}/(2)[/math], the integers modulo [math]2[/math].In general, for every prime [math]p[/math], exactly one of the following happens:[math]p = 1 + 4k[/math] for some integer [math]k[/math], in which case there are integers [math]a,b[/math] such that [math]p = a^2 + b^2 = (a + bi)(a - bi)[/math], and [math]\mathbb{Z}[i]/(a \pm bi) = \mathbb{Z}/(p)[/math].[math]p = 3 + 4k[/math] for some integer [math]k[/math], in which case [math]p[/math] is not the sum of two squares, and [math]\mathbb{Z}[i]/(p) = \mathbb{Z}/(p)[i][/math], which will properly be an extension of the usual integers modulo [math]p[/math], with an extra symbol [math]i[/math] defined by [math]i^2 = -1 = p - 1[/math].[math]p = 2[/math], in which case [math]p = (1 + i)(1 - i)[/math], and [math]\mathbb{Z}[i]/(1 \pm i) = \mathbb{Z}/(2)[/math].Understanding why this is true requires more machinery than I am prepared to go through in this post, but if this is the sort of thing that you are interested in, consider taking a course in elementary number theory, or perhaps even a course in algebraic number theory.