Lectures 5-6: Taylor Series

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1 Math 1d Instructor: Padraic Bartlett Lectures 5-: Taylor Series Weeks 5- Caltech Taylor Polynomials and Series As we saw in week 4, power series are remarkably nice objects to work with. In particular, because power series uniformly converge within their radii of convergence R, we know that the equations ( a n x n = d n= ( a n x n = n= n= n= a n x n = C + d (a nx n = n= n a n x n 1 n=1 a n n + 1 xn+1, and hold and converge for any power series ( n= a nx n with radius of convergence R, and any x ( R, R. In other words, power series are easily differentiated and integrated as many times as we like! This is remarkably useful: as you probably remember from Math 1a, integration and differentiation are often really difficult tasks. A natural question is the following: how can we use this observation about power series to work with other kinds of functions? In other words: if I give you a function, is there any way to turn it into a power series, so that we can integrate and differentiate it easily? As it turns out, there is! As a motivating example, consider the function f(x = e x. How can we approximate this with a power series? Initially, we have no idea; so let s try something simpler! Specifically, let s try approximating it at exactly one point; say x =. Additionally, let s try not approximating it by a power series just yet, but instead try approximating it with just polynomials; maybe being finite will help illustrate what we re trying to do. Let s start with a polynomial of degree i.e. a constant. What constant best approximates e x at? Well: if we set our constant equal to e = 1, then the constant function will at least agree with e x at. That s about as good as we can expect to do: so let s call the th -order approximation to e x, T (e x, the constant function 1. Now, let s take a polynomial of degree 1: i.e. a line, of the form T 1 (e x = a + bx. What should our constants a and b be? Well: again, if we want this approximation to agree with e x at, we should set a = 1, so that e = 1 = 1+b. For b; visually, we d get an even better approximation of e x if we picked b such that the slope of a + bx and e x both agreed at! In other words, if we set b = 1, we would have d (ex = e x x= = 1 = d x= (1 + 1 x x=. So, let s define the 1 st -order approximation to e x, T 1 (e x, the line 1 + x. Similarly, to make the 2 nd -order approximation to e x, T 2 (e x, we want to take a degree-2 polynomial a + a 1 x + a 2 and find values of a, a 1, a 2 such that the th, 1 st, and 1

2 2 nd derivatives of e x agree with T 2 (e x at. Therefore, we have (e x = 1 = ( a + a 1 x + a 2 x= = a x= a = 1, d (ex = (e x = 1 = d ( a + a 1 x + a 2 x= x= x= = a 1 a 1 = 1, d 2 d (ex = (e x = 1 = d2 ( a x= x= d + a 1 x + a 2 x= = 2a 2 a 2 = 1 2, and therefore that the 2 nd -order approximation to e x, T 2 (e x, is 1 + x + x2 2. Using similar logic, we can see that because d n n (ex = (e x = 1, and x= x= d n n (a + a 1 x +... a n x n = dn 1 ( a1 x= n 1 + 2a 2 x +... na n x n 1 x= = dn 2 n 2 ( 2a2 x + 3 2a 3 x +... (n (n 1a n x n 2 x=... = a n, in general we have a n = 1, and thus that the nth -degree approximation to e x is n We graph some of these approximations here: n= xn. f(x = e x T 3 (e x = 1+x+ /2 + x 3 /3 T 2 (e x = 1+x+ /2 T 1 (e x = 1+x T (e x = 1 2

3 Surprisingly, these local approximations are getting close to e x not just at, but in fact in a decently-sized neighborhood of! In fact, if we take the limit of these approximations n= this power series turns out to be equal to e x everywhere! So: this idea of local approximations appears to be giving us a bunch of global data! This motivates the following definition of a Taylor polynomial and Taylor series, which we give here: Definition 1.1. Let f(x be a n-times differentiable function on some neighborhood (a δ, a + δ of some point a. We define the n th Taylor polynomial of f(x around a as the following degree-n polynomial: T n (f(x, a := n n= x n, f (n (a (x a n. Notice that this function s first n derivatives all agree with f(x s derivatives: i.e. for any k n, k x k (T n(f(x, a = f (k (a. a This motivates the idea of these Taylor polynomials as n th order approximations at a of the function f(x: if you only look at the first n derivatives of this function at a, they agree with this function completely. We define the n th order remainder function of f(x around a as the difference between f(x and its n th order approximation T n (f(x, a: R n (f(x, a = f(x T n (f(x, a. If f is an infinitely-differentiable function, and lim n R n (f(x, a = at some value of x, then we can say that these Taylor polynomials converge to f(x, and in fact write f(x as its Taylor series: T (f(x = n= f (n (a (x a n. Usually, we will assume that our Taylor series are being expanded around a =, and omit the a-part of the expressions above. If it is not specified, always assume that you re looking at a Taylor series expanded around. One of the largest questions, given a function f(x, is the following: at which values of x is f(x equal to its Taylor series? Equivalently, our question is the following: for what values of x is lim n R n (f(x =? Our only/most useful tool for answering this question is the following theorem of Taylor: Theorem 1.2. (Taylor s theorem: If f(x is infinitely differentiable, then d n+1 x (f(x R n (f(x, a = n+1 x=t (x t n dt. a 3

4 In other words, we can express the remainder (a quantity we will often not understand as an integral involving the derivatives of f divided by (which is often easily bounded and polynomials (which are also easy to deal with. The main use of Taylor series, as stated before, is our earlier observation about how easy it is to integrate and differentiate power series: Theorem 1.3. Suppose that f(x is a function with Taylor series T (f(x = n= f (n (a (x a n, and furthermore suppose that f(x = T (f(x on some interval ( a, a. Then we can integrate and differentiate f(x by just termwise integrating and differentiating T (f(x: i.e. d ( f(x = d f (n (a (x a n f (n (a = (n 1! (x an 1, and n= n= ( f (n (a f(x = (x a n f (n (a = (n + 1! (x an+1 + C. n= n= In the following section, we study several functions, find their Taylor series, and find out where these Taylor series converge to their original functions: 2 Taylor Series: Examples Proposition 2.1. The Taylor series for f(x = e x about is n= x n. Furthermore, this series converges and is equal to e x on all of R. Proof. From our discussion earlier, we know that T (e x = Furthermore, by using Taylor s theorem, we know that the remainder term R n (e x is just d n+1 x (e x R n (e x = n+1 x x=t (x t n e t dt = (x tn dt. Integrating this directly seems... painful. However, we don t need to know exactly what this integral is: we just need to know that it gets really small as n goes to infinity! So, instead of calculating this integral directly, we can just come up with some upper bounds on its magnitude. 4 n= x n.

5 Specifically: on the interval [, x], the function e t is bounded above by e x, and the function (x t n takes on its maximum at t =, where it s x n. Therefore, we have that x e t (x x tn dt e x x n dt. The function being integrated at the right is just a constant with respect to t: there aren t any t s in it! This makes integration a lot easier, as the integral of a constant is just that constant times the length of the interval: ( x e x x n e x dt = x n t x = e x x n+1. Again: to show that e x is equal to its Taylor series on all of R, we just need to show that the remainder terms R n (e x always go to as n goes to infinity. So, to finish our proof, it suffices to show that lim n e x x n+1 This is not hard to see: simply notice that because the ratio of successive terms is just e x x n+1 e x x n (n 1! =. = x n, as n goes to infinity the ratio between successive terms goes to. In specific, for n > 2 x, the ratio between any two consecutive terms is < 1 2 ; i.e. each consecutive term in our sequence is at most half as big as the one that preceeded it. Therefore, the sequence converges to, as any sequence consisting of numbers that you re repeatedly chopping in half must converge to. We work a second example here: Proposition 2.2. The Taylor series for f(x = sin(x about is ( 1 n x2n+1 (2n + 1! Furthermore, this series converges and is equal to sin(x on all of R. k= Proof. By induction, we know that for any n, 4n x 4n (sin(x = sin(x = 4n+1 x 4n+1 (sin(x = cos(x = 1 4n+2 x 4n+2 (sin(x = sin(x = 4n+3 x 4n+3 (sin(x = cos(x = 1. 5

6 Consequently, we have that the 2n + 1-st Taylor polynomial for sin(x around is just T 2n+1 (sin(x, = n ( 1 n x2n+1 (2n + 1!, k= because plugging in the derivatives of sin(x into the formula for Taylor polynomials just kills off every other term in the sum, and therefore that the Taylor series for sin(x is ( 1 n x2n+1 (2n + 1! k= Where does this series converge to sin(x? Well: if we look at the R n (sin(x terms, we have by Taylor s theorem that R n (sin(x = x d n+1 (sin(x n+1 x=t (x t n dt. Like before, we can simplify this expression by replacing several terms with upper bounds. For example, we can replace dn+1 (sin(x n+1 x=t with 1, because the derivatives of sin(x are all either ± sin(x or ± cos(x, and in either case are bounded above in magnitude by 1; as well, we can bound (x t n above by x n, for t [, x]. This gives us R n (sin(x x 1 x n dt = x n+1, which goes to as n goes to infinity, for any value of x. So sin(x is equal to its Taylor series on all of R, as claimed! Given our success rate so far with approximating functions by Taylor series, you might wonder why we bother with the R n (f(x-part of our proofs. The following example illustrates why it is that we *do* have to check that a Taylor series converges to its original function; as it turns out, it s possible for a function to have a Taylor series that is completely useless for approximating it away from! Example 2.3. Let f(x = { e 1/, x, x = Then T (f(x =, and T (f(x = f(x only at x =. Proof. So: derivatives of e 1/x2 are tricky to directly calculate. However, we don t have to! Simply notice that we have d ( e 1/x2 = 2 x 3 e 1/x2, and (( ( d polynomials polynomials ( polynomials 2 e 1/x2 = e 1/x2 + polynomials polynomials polynomials x 3 e 1/x2 ( polynomials = e 1/x2. polynomials

7 In other words, all of the derivatives of e 1/x2 look like some ratio of polynomial expressions, multiplied by e 1/x2. So, when we look at the limit as x goes to of derivatives of e 1/x2, we always have lim n ( polynomials e 1/x2 =, polynomials because the exponential term dominates (i.e. exponentials shrink faster than polynomials can grow. So, while we don t explicitly know the derivatives of f(x, we do know that they re all defined and equal to at x =! So, the Taylor series of f(x is just xn =, which is only equal to f(x at, because e 1/x2 for any x. k= Using similar techniques to the worked examples above, you can prove that the functions below have the following Taylor series, and furthermore that they converge to their Taylor series on the claimed sets: Proposition 2.4. T T (cos(x = ( 1 = 1 x ( 1 n x2n, and T (cos(x = cos(x whenever x R. (2n! ( 1 x n, and T = 1 whenever x ( 1, 1. 1 x 1 x k= k= In addition, by substituting terms like into the above Taylor series, we can derive Taylor series for other functions: Proposition 2.5. T (e x2 = T k= ( = ( 1 n x2n, and T (e x2 = e x2 whenever x R. ( 1 n n, and T k= ( = 1 whenever x ( 1, x2 2.1 Applications of Taylor Series Using Taylor series, we can approximate integrals that would otherwise be impossible to deal with. For example, consider the Gaussian integral: e x2. Somwhat frustratingly, there is no elementary antiderivative for e x2 : in other words, there is no finite combination of functions like sin(x, e x, x n that will give an antiderivative of e x2. Using Taylor series i.e. an infinite number of functions we can find this integral, to any level of precision we desire! We outline the method below: 7

8 Question 2.. Approximate e x2 to within ±.1 of its actual value. Proof. Above, we proved that T n (e x = n k= x k k!. Using this, we can write e x2 = T n (e x + R n (e x, and therefore write 2 e x2 = 2 T n (e x + 2 R n (e x Why is this nice? Well: the T n part is just a polynomial: specifically, we have T n (e x n ( 1 k k =, k! k= which is quite easy to integrate! As well, the R n -thing is something that should be rather small for large values of n: so in theory we should be able to make its integral small, as well! Specifically: using Taylor s theorem and the estimates we came up with earlier in our lectures, we have R n (e x x = e t (x tn dt R n(e x x x e x e x x t n dt x n dt = e x x n+1 ex2 x 2n+2.. 8

9 Using this, we can bound the integral of our remainder terms: R n (e x 2 e x2 n+2 2 e 1/4 x2n+2 ( = 2e 1/4 n+3 1/2 (2n + 3 = e 1/4 2 2n+2 (2n + 3 This quantity is.1 at n = 3. Therefore, we know that e x2 = T 3 (e x up to ±.1. So: to find this integral, it suffices to integrate T 3 (e x T 3 (e x = = 2, (1 + x4 2 x (1 + x4 2 x = 2 (x x3 3 + x5 1 x7 42 = = This is pretty easy: 1/2 So this is the integral of e x2, up to ±.1. Using Mathematica, we can see that the actual integral is =.92252, which verifies that our calculations worked! We close with one last application of Taylor series to the calculation of limits, as an alternate strategy to repeated applications of L Hôpital s rule: Example 2.7. Find the limit sin( + x lim n x 1 Proof. One way we could solve this problem would be to hit it with L Hôpital s rule ten times in a row, and get that we re looking at the limit as x goes to of the fraction 324 cos( 432x 4 cos( + 234x 8 cos( 324 sin( x sin( 124x 1 sin(, 1! 9

10 which is just 342 1! = And, you know, if you wanted to calculate ten derivatives in a row, you could do that. I... don t. So: what s a non-awful way to do this? Well: if we used sin(x s Taylor series, we can notice that sin( = ( 1 n x4n+2 (2n + 1! = x2 x + x1 5! n=..., and therefore that the fraction we re working with is just ( sin( + x x + x1 5!... x 1 = x 1 = 1 ( x 1 x 1 5! = 1 ( x 4 5! + x14 7! + x18 9! + x 7! x8 9! + x12 11! So: what does this do when x goes to? Well, the expression on the right is a power series that converges on all of R, if we use absolute convergence convergence along with the ratio test. Therefore, it s a continuous functio So, to find the limit, we just need to plug in : sin( ( ( + x 1 x 4 lim x x 1 = 5! + 7! x8 9! + x12 11!... x= = 1 5! = Which worked! And we didn t have to calculate ten derivatives. Nice. 1

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