CAN WE INTEGRATE x 2 e x2 /2? 1. THE PROBLEM. xe x2 /2 dx


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1 CAN WE INTEGRATE x 2 e x2 /2? MICHAEL ANSHELEVICH ABSTRACT. Not every nice function has an integral given by a formula. The standard example is e x 2 /2 dx which is not an elementary function. On the other hand, xe x2 /2 dx = e x2 /2. What about x 2 e x2 /2 dx? Is this a calculus integral? What about x 3 e x2 /2 dx? In this talk, I will give a complete answer to this question. The answer involves Hermite polynomials. The arguments do not use anything beyond calculus, but connect to a number of more advanced topics.. THE PROBLEM Every calculus textbook (for example Stewart, page 493) points out that not every integral can be done, that is, can be expressed in terms of the usual ( elementary ) functions. A standard example is e x2 /2 dx, which cannot be computed by calculus methods, no matter how clever you are, despite its importance in probability and statistics (this is the famous Bell Curve). On the other hand, the integral xe x2 /2 dx is easy: substitution shows that it equals What about u = x 2 /2, du = x dx e u du = e u + C = e x2 /2 + C. x 2 e x2 /2 dx? I claim that it is not elementary. But note that d ( ) xe x2 /2 = (x 2 )e x2 /2, dx so (x 2 )e x2 /2 dx = xe x2 /2 + C. Also x 3 e x2 /2 dx = (x 2 + 2)e x2 /2 + C. Date: March 5, 200.
2 2 How did I guess that? 2. THE SOLUTION USING HERMITE POLYNOMIALS Definition. For each n, define the Hermite polynomial H n (x) by Example. For example, so so so etc. d n dx n e x2 /2 = ( ) n H n (x)e x2 /2. H 0 (x) =, d /2 dx e x2 = xe x2 /2 H (x) = x, d 2 dx 2 e x2 /2 = e x2 /2 + x 2 e x2 /2 H 2 (x) = x 2, d 3 /2 dx 3 e x2 = (x + 2x x 3 )e x2 /2 H 3 (x) = x 3 3x Clearly H n (x) is a polynomial of degree n, with the highest term x n. Hermite polynomials appear in many contexts. If you know Linear Algebra: Hermite polynomials are orthogonal polynomials. If we define the inner product between two functions then f, g = H n, H k = f(x)g(x) e x2 /2 dx, H n (x)h k (x) e x2 /2 dx = 0 for n k, so that H n and H k are orthogonal to each other. In quantum mechanics, Hermite polynomials are closely connected to the eigenfunctions for the harmonic oscillator. How do they help us? Let us compute the derivative of a Hermite polynomial times the exponential function. d ( ) H n (x) e x2 /2 = d ) (( ) n dn /2 = ( ) n dn+ /2 = H dx dx dx n e x2 dx n+ e x2 n+ (x) e x2 /2! In other words, we proved
3 3 Lemma. H n+ (x) e x2 /2 dx = H n (x) e x2 /2 + C. Example 2. For example, etc. xe x2 /2 dx = e x2 /2 + C, (x 2 )e x2 /2 dx = xe x2 /2 + C, (x 3 3x)e x2 /2 dx = (x 2 )e x2 /2 + C, Exercise. What about the derivative of H n itself? One can check that in fact, d dx H n(x) = nh n (x) Note that this is the same rule as d dx xn = nx n! Are there any other polynomials with this property? What about our original question, for x n e x2 /2 dx? Well, e x2 /2 dx cannot be done (so says Stewart, so it must be true). We know how to do xe x2 /2 dx. Also, by our calculation x 2 e x2 /2 dx = e x2 /2 dx xe x2 /2 + C. Since the righthandside cannot be done, neither can the lefthandside. For general n? We can write = = H 0 (x), x = x = H (x), x 2 = (x 2 ) + = H 2 (x) + H 0 (x), x 3 = (x 3 3x) + 3x = H 3 (x) + 3H (x),
4 4 and in general x n = H n (x) + a n H n (x) + a n 2 H n 2 (x) + + a H (x) + a 0 H 0 (x). And we know H k (x) e x2 /2 dx = H k (x)e x2 /2 + C. Not quite: only for k. So: can integrate x n e x2 /2 if and only if x n contains no H 0, that is, if a 0 = 0. How to find a 0? In Linear Algebra: {H 0, H, H 2, H 3, } form an orthogonal basis. It is not orthonormal: So H 0, H 0 = a 0 = xn, H 0 H 0, H 0 = 2π e x2 /2 dx = 2π. x n e x2 /2 dx. Note this is a number, not a function. We do not know this number. But: If n is odd, x n e x2 /2 is an odd function, so that so can integrate. a 0 = 2π If n is even, x n e x2 /2 > 0 is a positive function, so that and we cannot integrate. a 0 = 2π x n e x2 /2 dx = 0 x n e x2 /2 dx > 0 Theorem 2. We can integrate x n e x2 /2 if n is odd, and cannot integrate it if n is even. More generally, we can integrate P (x)e x2 /2 if and only if P n (x) is a linear combination of {H, H 2, } (excluding H 0 ). We also discussed that the moments of the Gaussian distribution are zero for odd n and REFERENCE: 2π x 2n e x2 dx = (2n )!! = (2n ) (2n 3) 5 3. Persi Diaconis and Sandy Zabell, CLOSED FORM SUMMATION FOR CLASSICAL DISTRIBUTIONS: VARIATIONS ON A THEME OF DE MOIVRE, Statistical Science 6 n. 3 (Aug. 99)
5 5 3. MORE ON HERMITE POLYNOMIALS 3.. Solution of Exercise. For the first part: look at the generating function F (x, z) = n! H n(x)z n. Using Taylor series f (n) (a)b n = f(a + b), n! F (x, z) = n! H n(x)z n = We showed = e x2 /2 Lemma 3 (Generating function). d n /2 n! dx n e x2 ( z) n = e x2 /2 e (x z)2 /2 = e xz z2 /2. F (x, z) = n! ( )n e x2 /2 dn dx n e x2 /2 z n n! H n(x)z n = e xz z2 /2. Differentiating with respect to x, we get Thus x exz z2 /2 = ze xz z2 /2. n! H n(x)z n = F (x, z) = zf (x, z) x = n! H n(x)z n+ = (n )! H n (x)z n = Comparing coefficients of z n, we get Lemma 4 (Differential recursion). n= H n(x) = nh n (x). n= n n! H n (x)z n. Exercise 2. Use the lemma just above to show that the Hermite polynomials are orthogonal. In fact, use induction to compute H n, H k = H n (x)h k (x) e x2 /2 dx.
6 6 For the second part of Exercise (other polynomials with P n(x) = np n (x), look up terms such as Bernoulli polynomials, Euler polynomials, and Appell polynomials. Try to write out polynomials {P n (x)} such that n! P n(x)z n = e xz e f(z) for some simple functions f. Dave asked: what about polynomials with some other relation P n(x) = b n P n (x), for example P n = n 2 P n? Note that the Hermite polynomials are monic, that is, their leading coefficient is. For monic polynomials, we can only hope to have P n = np n (why?) But there is a class of polynomials satisfying a similar more general property, called the BoasBuck polynomials (Boas is Ralph Boas, father of our Harold Boas). This is class is far from completely understood, and in fact I am interested in it for my research Discrete Math. How to compute H n quickly? Recursively! Differentiating F (x, z) with respect to z, we get Thus or So z exz z2 /2 = xe xz z2 /2 ze xz z2 /2. n! (xh n(x))z n = Lemma 5 (Threeterm recursion). Exercise 3 (Discrete math). Show that xf (x, z) = F (x, z) + zf (x, z), z = n! nh n(x)z n + n! H n+(x)z n + xh n (x) = H n+ (x) + nh n (x). n! H n(x)z n+ n n! H n (x)z n. H n (x) = c n,n x n c n,n 2 x n 2 + c n,n 4 x n 4 c n,n 6 x n 6 +, where c n,k is the total number of partitions of n elements into pairs and singletons, with k singletons. In fact, show that n! c n,n 2k = (n 2k)!2 k k!.
7 Differential Equations. We compute H n xh n = n(n )H n 2 nxh n = n[(n )H n 2 xh n ] = nh n. So Lemma 6 (Differential equation). H n is a solution of the secondorder linear differential equation H n xh n + nh n = 0, or: H n is an eigenfunction of the differential operator y xy with eigenvalue n Linear Algebra. Look at the matrix M n = n n
8 8 What are the eigenvalues of M n? Expand with respect to the last row:. λ λ det(m n+ λi) = 0 λ n λ n λ. λ 0 0 λ λ = λ 0 λ. λ λ n λ n. λ λ λ = λ 0 λ. λ n 0 λ.... n.. n λ λ Recall So So Lemma 7. Exercise 4. The Hermite polynomials = λ det(m n λi) n det(m n λi). H n+ (x) = xh n (x) nh n (x). det(m n λi) = H n ( x). eigenvalues of M n = ( ) zeros of H n. {H 0 (x), H (x), H 2 (x), } form a basis for the vector space of polynomials P. In this basis, the matrix M is a matrix of some linear operator. What is that operator on P? Lemma 8. Each H n has n real, simple zeroes. Moreover, these zeros are interlacing: if H n has zeros x < x 2 < < x n and H n has zeroes y < y 2 < < y n,
9 9 then x < y < x 2 < y 2 < < x n < y n < x n. Proof. We can factor H n (x) = (x a ) k (x a 2 ) k 2 (x a m ) km Q(x) where a i are real roots, and Q does not have any real roots. Say the product is written in such an order that k, k 2,, k j are odd and k j+,, k m are even. Denote P (x) = (x a )(x a 2 ) (x a j ). Then the polynomial P (x)h n (x) never changes sign (since all its real roots have even multiplicity), so it either always positive or always negative. So in particular, P (x)h n (x)e x2 /2 dx 0. But since the Hermite polynomials are orthogonal, such a product is zero for any polynomial of degree less than n. So deg P = n, which means that all the roots of P are real and different. Why interlacing: H n(x) = nh n (x). So roots of H n are maxima and minima of H n. Zero of the Hermite polynomials also appear in numerical integration in Numerical Analysis (the term is Gaussian quadrature). For other orthogonal polynomials: 4. GENERAL ORTHOGONAL POLYNOMIALS {P 0 (x), P (x), P 2 (x), }, P n (x) = x n +, Orthogonal with respect to w(x), meaning that if n k. P n, P k = P n (x)p k (x)w(x) dx = 0 Examples: w(x) = e x on [0, ), w(x) = on [, ]. There are also discrete weights, such as the Poisson distribution. Which of the properties of Hermite polynomials hold for more general orthogonal polynomials? Recursion: always xp n (x) = P n+ (x) + β n P n (x) + γ n P n (x), for some β i and some nonnegative γ i. Derivative P n = np n : only Hermite. Secondorder differential equation: Hermite, Laguerre, Jacobi. The eigenvalue problem is (p(x)y (x)) + q(x)y (x) = λy(x). The operator on the lefthandside is called a SturmLiouville operator; such operators are studied in many differential equations books.
10 0 Exercise 5. Check that for this differential operator to have polynomial eigenfunctions, we for sure need that p is a polynomial of degree 2, q is a polynomial of degree. One can show that if the polynomial eigenfunctions are orthogonal, then w w = q(x) p(x) (One checks that this is the condition for the operator to be symmetric with respect to the inner product given by w(x).) Use partial fractions to find w(x) for different types of q(x) and p(x). Characteristic polynomial of a matrix: always true, use β i, γ i as above. Zeros: always true, even though cannot use the derivative.
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