MATH 118, LECTURES 27 & 28: TAYLOR SERIES

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1 MATH 8, LECTURES 7 & 8: TAYLOR SERIES Taylor Series Suppose we know that the power series a n (x c) n converges on some interval c R < x < c + R to the function f(x). That is to say, we have f(x) = a 0 + a (x c) + a (x c) + a 3 (x c) 3 + () on c R < x < c + R. Thus far, we have been interested in this problem: given a series of the form a n(x c) n, determine the interval and limit of convergence of the series (i.e. find f(x)). Suppose now that we ask the question in the other direction. Given a function f(x), can we find a power series which converges to this function? That is to say, given a function f(x), can we determine the values of a n so that the power series with these coefficients converges to f(x)? Let s revisit equation () and see if we can solve for the first value, a 0. This seems at first glance like an ominous task there are, after all, an infinite number of terms on the right-hand side, so how can we possibly eliminate all of these terms except for a 0? The answer is surprisingly simple, since we see all but the first term vanish when we substitute x = c. This gives f(c) = a 0. So the first term in the power series is a 0 = f(c) but how do we go about solving for the other terms? We need to isolate a in the same way that a 0 was isolated in the previous derivation. We can do this by differentiating the series to get f (x) = a + a (x c) + 3a 3 (x c) + 4a 4 (x c) 3 +.

2 Again, substituting x = c we get f (c) = a. A pattern is clearly emerging to get each successive term a n we differentiate and then substitute x = c. We carry this out one more time for clarity, to get which gives f (x) = a + 3a 3 (x c) + 3 4(x c) + 4 5(x c) 3 + f (c) = a = a = f (c). We can see that the derivation process leads to a factorial term when solving for a n at each step. It should not be difficult to convince yourself that this process gives rise to the general form a n = f(n) (c). n! This is exactly what we set to out find it is exactly the answer to the question we original posed. Given a function f(x), the power series which converges (on some interval) to f(x) is given by f(x) = f (n) (c) (x c) n. () n! A series of the form () is called a Taylor series. While it may not look like much, it is in fact one of the most versatile and powerful tools in applied mathematical analysis. As we will see in the following weeks, Taylor series can be differentiated, integrated, added, multiplied, and have a variety of other operations performed to it even when these operations cannot be performed explicitly, or performed easily, to the original function f(x)! Since a Taylor series consists only of polynomial terms, these operations can be carried out trivially. Many functions of applied interest can, in fact, only be represented by their Taylor series expansions (such as Bessel s functions). A Taylor series centred at c = 0 is called a Maclaurin series. We notice that the formula () gives an explicit method for determining the desired series. So long as the function f(x) is infinitely differentiable, we can determine the coefficients a n. We might wonder if this is a unique representation that is to say, could there be another series b n(x c) n such that f(x) is the limit of this series as well. This possibility is eliminated by the following result:

3 Proposition.. If two power series a n(x c) n and b n(x c) n with positive radii of convergence have identical sums then a n = b n for all n. a n (x c) n = b n (x c) n, So the Taylor series representation of f(x) is unique up to the choice of c. That is to say, if we choose a particular centre c, there is only one possible set of {a 0,a,a,...} so that a n(x c) n = f(x) on any non-trivial interval. Let s use this fact to find the Taylor series expansion of the well-known functions e x, ln(x), sin(x), and a/( rx). Taylor series expansion of e x We will centre the series about c = 0 for reasons which will be obvious in a moment. In order to determine the Taylor series expansion of the function, we need to find f (n) (c) for all values of n. We evaluate the first few terms to get f(x) = e x = a 0 = f(0) = e 0 = f (x) = e x = a = f (0) = f (x) = e x = a = f (0) = f (n) (x) = e x = f (n) (0) =. We substitute into the formula () to get e x = n n! xn. We have not yet determined the radius of convergence for this series in fact, we have not discussed convergence for Taylor series yet at all. In order to determine the radius of convergence, we need to rely on our power series results. For this series, we have R = lim n a n a n+ = lim n (n + )! n! = lim n n + =. 3

4 So the series converges on < x <. Taylor series expansion of ln(x) We will centre the series about c = since we cannot evaluate f(0). The first few derivatives at are f(x) = ln(x) = a 0 = f() = ln() = 0 f (x) = x = a = f () = f (x) = x = a = f () = f (n) (x) = ( )n (n )! x n = f (n) () = ( ) n (n )!, n. We have to specify n because for n = 0 we have f() = 0 which does not fit this form. We substitute into the formula () to get ln(x) = n= ( ) n (n )! n! (x ) n = ( ) n (x ) n. n n= We still have to determine where this series converges. We have R = lim c n n = lim ( ) n n n + n ( ) n =. c n+ The series converges on the open interval 0 < x <. At x = 0 we have n= /n which diverges to (negative harmonic series). At x = we have n= ( )n /n which converges (alternating harmonic series). We therefore have the interval of convergence 0 < x. Taylor series expansion of sin(x) We will centre the series about c = 0 for simplicity. We have f(x) = sin(x) = a 0 = f(0) = 0 f (x) = cos(x) = a = f (0) = f (x) = sin(x) = a = f (0) = 0 f (x) = cos(x) = a 3 = f (0) = 4

5 for n =,5,9,... a n = f (n) (0) = for n = 3,7,,... 0 otherwise. We can reindex and substitute into the formula () to get sin(x) = ( ) n (n + )! xn+ = x x3 3! + x5 5! x7 7! +. We will consider the interval of convergences in a later section; however, we can take it for granted now that this converges on < x <. It can be derived similarly that cos(x) = on < x <. ( ) n (n)! x n = x! + x4 4! x6 6! + Taylor series expansion of a/( rx) We will centre at c = 0. We have f(x) = a = a 0 = f(0) = a rx f ar (x) = ( rx) = a = f (0) = ar f (x) = f (n) = ar ( rx) 3 = a = f (0) = ar n! arn ( rx) n = a n = f (n) (0) = n! ar n. We substitute into the formula () to get a rx = a(rx) (= n ) a(rx) n. We have already seen this series before: it is the geometric series with initial term a and common ratio rx. It converges for xr < which implies /r < x < /r. n= 5

6 . Finding Taylor Series The procedure outlined so far is sufficient for determining the Taylor series expansion for a wide variety of functions however, given the very common functions we have determined the expansion for already, it is often easier just relate functions to these known functions. We can then immediately use our previous results to obtain the desired series. Example : Determine the Taylor series expansion of e x about the point c = 0. Since the Taylor series expansion of e x is e x = n! xn = + x + x! + x3 3! + we have that e x = n! ( x ) n = ( ) n n! x n = x + x x6 +. The series converges on < x <. Example : Determine the Taylor series expansion of about the point c = x We need to change this function into the form of a/( rx). We can accomplish this by dividing top and bottom by 5 to get 5 + 3x = x = ( 3 ) n 5 5 x. This converges on the interval (3/5)x < which is 5/3 < x < 5/3. 6

7 Example 3: Determine the Taylor series expansion of sin(x) about the point c = π/. We could do this directly by finding a formula for f (n) (π/); however, it is easier in this case to notice that sin(x) = sin(x π/ + π/) = sin(x π/)cos(π/) + cos(x π/)sin(π/) = cos(x π/). This implies immediately that sin(x) = cos(x π/) = which converges for < x <. Example 4: Determing the Taylor series expansion of 3x ( ) n (x π/)n (n)! about the point c =. We need to change this function so that it has a factor of x + in it. This can be accomplished by noticing This implies that 3x = 4 3(x + ) = (x + ). 3x = ( ) 3 n (x + ) 4 4 which converges for (3/4)(x + ) < which is 7/3 < x < /3. 7

8 . Operations on Taylor Series So far we have considered how to rearrange functions into the form of series for which we already know the Taylor series expansion. This can be extended to include such operations as addition, subtraction, differential, integration, multiplication, and division. Consider two convergent series, f(x) = a n(x c) n and g(x) = b n(x c) n. The following results guarantee the convergence of the Taylor series expansions in a natural way. Proposition.. If f(x) = a n(x c) n and g(x) = b n(x c) n have positive radii of convergence R and R, respectively, then f(x) ± g(x) = (a n ± b n )(x c) n, and where f(x)g(x) = d n (x c) n, d n = a i b n i = a 0 b n + a b n + + a n b + a n b 0, and the radius of convergence is the smaller of R and R. Furthermore f(x)±g(x) converges at the endpoints of the common interval of convergence if both f(x) and g(x) do. Proposition.3. If f(x) = a n(x c) n, and the radius of convergence R is greater than zero, then each of the following series has radius of convergence R: f (x) = na n (x c) n, Example : f(x) dx = a n n + (x c)n+ + C. Determine the Taylor series expansion of ln(x) + /x about c =. 8

9 We have that ln(x) = on 0 < x. We also have ( ) n (x ) n n n= x = + (x ) = ( ) n (x ) n = + ( ) n (x ) n which converges on x < which is 0 < x <. We have taken out the first term n = 0 so that we can combine the two series, since they both now start at n =. We have ln(x) + x = + [ ] ( ) n + ( ) n (x ) n = + n= n= The series converges on 0 < x <. Example : n ( n Determine the Taylor series expansion of ( )( ). x 3x We have n= ) ( ) n (x ) n. x = x n = + x + x + x 3 + converging on < x < and 3x = (3x) n = ( + 3x + 3 x x 3 + ) converging on /3 < x < /3. Therefore, we have ( )( ) = ( + x + x + x 3 + )( + 3x + 3 x x 3 + ) x 3x = + ( + 3)x + ( )x + ( )x

10 We can see that the coefficients satisfy a n = n (3) k. k=0 This is a partial sum of a geometric series. Even though the sum does not converge, we can determine the partial sum by our partial sum formula to get n ( 3 (3) k n+) =. 3 This implies that k=0 ( )( ) = x 3x ( ) 3 n+ x n 3 which converges on the smaller of the two intervals, which is /3 < x < /3. Example 3: Determine the first three non-zero terms in the Maclaurin series sec(x). We have sec(x) = cos(x) = a n x n. Expanding cos(x) now would not benefit us, so we multiply across to get ) = cos(x) a n x n = ( x! + x4 (a0 3! + a x + a x + ) ( = a 0 + a x + a a ) ( 0 x + a 3 x 3 + a 4 a!! + a ) 4 x ! This equality can be satisfied if and only if all coefficient on the left- and right-hand sides match up. For the left-hand side, this is extremely easy since = +(0)x+(0)x + so that all coefficients except the first are zero. We can solve for the coefficients of our series by solve systematically for the a n. We notice first of all that for the odd powers we have a = a 3 = a 5 = = 0 0

11 so that the first three non-zero terms are a 0, a, and a 4. We have 0 = a a 0! 0 = a 4 a! + a 0 4! a 0 =, = a =!, = a 4 = (!) 4!. Therefore, to the first three non-zero terms, we have sec(x) = + (! x + (!) ) x 4 + 4! Example 4: = + x x4 +. Determine the Taylor series expansion of x ( + x). We cannot directly use any of the formulas for known Taylor series which we have at our disposal; however, we have a result relating the derivatives and integrals of functions of known functions. This opens things up, since we immediately recognize /( + x) as the derivative of /( + x). We start with f(x) = + x = ( ) n x n. This converges on < x <. By the derivative result for power series, we have f (x) = ( + x) = n( ) n x n which converges also on < x <. Lastly, this implies that x f (x) = x ( + x) = n( ) n+ x n+ = n= (n )( ) n x n. Example 5:

12 Determine the sum of the power series (n + ) x n. Now that we are familiar with the form of several Taylor series expansions, we can use the various properties of Taylor series to manipulate general series into these forms, and thus allow ourselves to evaluate the sum. We have f(x) = (n + ) x n. This would be a geometric series if there were no (n + ) term. We notice, however, that we can eliminate this term by integrating. This gives f(x) dx = x n+ + C = x x n + C = x x + C. We now differentiate to obtain d f(x) dx = f(x) = dx ( x)() x( ) ( x) = ( x). We still need to determine the radius of convergence, however. We have R = lim n + n n + =. We therefore have that (n + ) x n = Example 6: Determine the sum of the power series ( x), < x <. ( ) n n xn. n= We notice that if the n in the denominator was not present we would have a geometric series in x. We differentiate to obtain f (x) = ( ) n x n = x n= ( x ) n. n=

13 We know that ( x ) n = /( + x ). In order to compensate for the missing n = 0 term, we need to subtract it from the right-hand side, so that f (x) = [ ] x + x = [ ] x x + x We integrate to get f (x) dx = f(x) = = x + x. x + x dx = ln( + x ) + C. We have that f(0) = 0 so that C = 0. The radius of convergence is R = lim ( ) n n + n n ( ) n+ =. Notice that the series also converges at the endpoints x = and x =. This implies ( ) n n= n xn = ln( + x ), x. 3