# 1 Ordinary points and singular points

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1 Math 70 honors, Fall, 008 Notes 8 More on series solutions, and an introduction to \orthogonal polynomials" Homework at end Revised, /4. Some changes and additions starting on page 7. Ordinary points and singular points We are considering general second order linear homogeneous equations and looking for series solutions P (x) y 00 + Q (x) y 0 + R (x) y = 0; () y = a n (x x 0 ) n : We assume that P; Q and R all have power series expansions around x 0. (In most cases that we will consider, P; Q and R are polynomials. Any polynomial has a Taylor series around any point, and since eventually the derivatives of the function are zero, the Taylor series is a nite sum. This is the expansion we are referring to. Hence, Example: P (x) = x + x +. Find the Taylor series around x = : Method P 0 (x) = x + P 00 (x) = P (j) (x) = 0 for j > : P () = 7 P 0 () = 5 P 00 () = so P (x) = (x ) +! (x ) + 0 x 3 + ::

2 with all the remaining terms 0: Method x + x + = (x ) + 4x 4 + f(x ) + g + = (x ) + f4 (x ) + 8g 4 + (x ) + 3 = (x ) + 5 (x ) + 7 Many other ways of algebraic manipulation could be used. most straightforward, but not necessarily the quickest. The Taylor series is the Denition Assume that P; Q;and R have no common factor. Then a point x 0 is a \ordinary point" for () if P (x 0 ) 6= 0: Otherwise x 0 is a \singular" point for () : There are several examples given in the text. Be sure to reach Theorem It was illustrated at the end of the last section of notes. Hermite's equation Hermite's equation (problem, pg. 60 ) is y 00 xy 0 + y = 0; where y is a function of x. Notice that there are no singular points. We can nd a series solution by the usual way: y = a n x n n= y 0 = y 00 = na n x n n= n (n ) a n x n : Substituting this into Hermite's equation gives n (n ) a n x n na n x n + a n x n = 0 n= n=

3 Rearranging the indices, we let m = n to n; to get in the rst sum and then change m back (n + ) (n + ) a n+ x n na n x n + a n x n = 0 The coecient of x 0 then gives us n= a + a 0 = 0 and the coecient of x n for n > 0 gives Hence, the even terms are a n+ = a 4 = 4 (4) (3) a = a 6 = 8 (6) (5) a 4 = a = a 0 ( n) (n + ) (n + ) a n and so forth, while the odd numbered terms are ( 4) a 0 4! ( 4) ( 8) a 0 6! a 3 = a ( 6) ( ) a 5 = a 5! ( 0) ( 6) ( ) a 7 = a 7! and so forth. We get two linearly independent solutions by rst setting a 0 = ; a = 0; and then setting a 0 = 0; a =. Notice that y (0) = a 0 and y 0 (0) = a. The solutions are and y (x) = x y (x) = x 3 +! x + ( 4) 4! ( ) ( 6) 5! x 4 ( 4) ( 8) x 6 + 6! 3 x 5 ( ) ( 6) ( 0) x 7 + 7!

4 Since Hermite's equation has no singular points, Theorem 5.3. tells us that the series converges for all x: But the most interesting cases are for = ; 4; 6; 8; etc., that is, any even positive integer. In these cases we can see that one or the other of these functions is not an innite sum, but only a nite sum. For example, if = 0; then y (x) = x 8 x ! x5 and all the later terms are zero. The functions we get then are the so-called "Hermite polynomials", or multiples thereof. For a reason we will give below, the Hermite polynomial H n, where the highest order term is a multiple of x n, is chosen so the coecient of x n is n : For example, H 5 is supposed to start o with 3x n ; so we get H 5 (x) = 3x 5 8 5!x3 + 5!x = 3x 5 60x 3 + 0x: The rst four H n are (as given in the text), pg. 60 ) H 0 (x) = H (x) = x H (x) = 4x H 3 (x) = 8x 3 H 4 (x) = 6x 4 48x + I will now list some of the properties of the Hermite polynomials. We will not have time to prove these in general, but I will try to give some examples. You can nd these properties on Wikipedia, or on "MathWorld", another good math website.. They satisfy a "recurrence relation" This is a quick way of guring them out. H n+ (x) = xh n (x) nh n (x). They have a \generating function", obtained from the power series for e x : The relevant formula is e xt H t n (x) = t n : n! 4

5 This is seen from e xt t = +! xt t +! xt t + = + xt t + 4x t + t 4! = + x! t + 4x t +! 4xt 3 + :: Notice that the rst three terms agree with what we got earlier, and this formula gives a reason for choosing H n as we did, with n as the leading coecient. 3. They satisfy an \orthogonality relation": Z H m (x) H n (x) e x dx = 0: Recall from math 80 that polynomials can be orthogonal with respect to an inner product. Here the inner product is weighted by the factor e x ; which is necessary for the indenite integrals to converge. Many more properties are listed on the websites. The Hermite polynomials are examples of what are called \special functions", which are functions, usually dened by either polynomials or innite series, which are important in physics and other applications. Doing a Google search for "Hermite polynomials" yielded about 90,000 links. We shall run into several more special functions in this chapter. The next comes from: 3 Chebyshev's equation Chebyshev's equation (problem 0, pg. 65 ) is x y 00 xy 0 + y = 0; where y is a function of x. Now there are singular points at x =. However x 0 = 0 is an ordinary point. 5

6 We can nd these solutions in the usual way: y = a n x n ; etc. for the derivatives. Substituting this into Chebyshev's equation results eventually in a = a 0 The coecient of x gives a = a 0 6a 3 + a = 0 and the coecient of x n for n > 0 gives a 3 = a : a n+ = n (n ) + n a n = (n + ) (n + ) n (n + ) (n + ) a n: Hence, the even terms are a 4 = 4 (4) (3) a = (4 ) a 0 4! a 6 = 6 (6) (5) a 4 = (6 ) (4 ) 6! and so forth, while the odd numbered terms are a 0 and so forth. a 3 = a a 5 = (3 ) ( ) a 5! a 7 = (5 ) (3 ) ( ) a 7! We get two linearly independent solutions as usual: y (x) = (4 )! x x (4 ) (6 ) x 4! 4! 6

7 and y (x) = x + x 3 + (3 ) ( ) 5! x 5 + (5 ) (3 ) ( ) x 7 + 7! From Theorem 5.3. it follows that the radius of convergence in each case is at least ; because the series for has radius of convergence equal to. x But we can also see that if is an integer, then one of the series stops after a certain point, and we get a polynomial, as before, with Hermite's equation. In this case, there is no issue of convergence. We can say that r =. This emphasizes that Theorem 5.3. gives a minimum value for r: It could be larger, as it is in this example if is an integer. The Chebyshev polynomials are denoted by T n (x) : (Perhaps someone who has studied Russian can explain why.) They are dened to be the polynomial solution of Chebyshev's equation, with = n; normalized so that T n () = : They give another class of special functions, and have the same sorts of properties as the Hermite polynomials:. There is a \generating function", with t xt + t = T 0 (x) +. They have an orthogonality property Z Note the weight function. T n (x) T m (x) 3. There is a recurrence relation: T n (x) t n n= p dx = 0 if n 6= m: x T n+ (x) = xt n (x) T n (x) 4. Here is a direct formula for T n (x) : T n (x) = h x + p n x + x p x n i 7

8 This seems to involve complex numbers for jxj < ; but the complex parts will cancel. 5. Here is another neat formula: If x = cos, then In other words, T n (x) = cos (n) : T n (x) = cos (n arccos (x)) Notice that this formula cannot be used if jxj > ; since then we can't have x = cos. (Or can we? It turns out that we can if we allow to be complex, but we are not considering that here.) The Chebyshev polynomials turn out to be important in numerical analysis, because when they are used to approximate non-polynomial functions, they are ecient in giving a good approximation..we can see that there is something unusual about them from the two formulas involving trig functions. Their values lie in [ ; ] if jxj. (\Obviously", this cannot be true for all x: Why not?) In this sense, they resemble trigonometric functions, over this interval, and yet they are polynomials, not innite sums. Here is the plot of T 0 (x):; 8

9 Another quite surprising property is that T n has all of its maxima and minima in [ ; ], and the maximum and minimum values are. It is perhaps surprising that any polynomial could have such a property. Plus, as property () above states, they are orthogonal with respect to a particular weight function. 4 Homework pg. 59, # pg. 66, # 3. (You can use the result of # ) pg. 67, # 6 (See # 4 for the denition of Legendre polynomial. Problem is of no help here. Don't attempt to use the solution. Work with the equation.) 9

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