Math 112 Rahman. Week Taylor Series Suppose the function f has the following power series:

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1 Math Rahman Week Taylor Series Suppose the function f has the following power series: fx) c 0 + c x a) + c x a) + c 3 x a) 3 + c n x a) n. ) Can we figure out what the coefficients are? Yes, yes we can. Notice that fa) c 0, so that gives us the first coefficient. For the second one lets differentiate to get f x) c + c x a) + 3c 3 x a) +. Now, if we plug in a we get f a) c. How about the third? Well, f x) c + c 3 x a) +, so f a) c. Can we figure out what c n should be? Well we see that if we keep taking derivatives and evaluating them at the center, we get f n) x) c n +, so c n f n) x)/. We have just derived a general formula for finding the coefficients of our series. Theorem. Suppose fx) c nx a) n converges for x a < R. Then, c n f n)a) and fx) f n) a) x a) n. Definition. The series representation f n) a) fx) x a) n fa) + f a)x a) + f a) x a) + + f n) a) x a) n + ) is called a Taylor series of f at x a. If a 0 we simply call this the Taylor series of f at x 0 or the McLaurin series of f - both are used interchangeably. Ex: Find the Taylor series of fx) e x and its radius of convergence. Solution: This is easy because we can find the n th derivative of e x straightaway, i.e. f n) x) e x, hence f n) 0). So e x xn /. Now, this is still a power series so like any other power series we can find the radius of convergence by using either root or ratio test. Lets apply ratio test, a n+ a n x n+ n + )! x n x n +. Taking the limit gives us lim n a n+ /a n 0, so R. Therefore, the Taylor series converges everywhere and it is an exact representation of e x. Definition. Let ξ a, b), and f have k derivatives on a, b), then for any n < k positive, P n x) fξ) + f ξ)x ξ) + f ξ) is the n th order Taylor polynomial of f at x ξ. x ξ) + + f n) ξ) x ξ) n Notice that the Taylor polynomial is just the truncated Taylor series. Lets do some problems that we did in class 0.8.7) We first evaluate the derivatives up to order n 3: fπ/4) /, f π/4) /, f π/4) /, and f π/4) /. Then the Taylor polynomial is fx) P 3 x) [ ) We just convert this into a power series, x π 4 ) x π 4 ) fx) + x x) x) n. 0.8.) Here we just use the common Taylor series, x) n e x. x π ) ] 3. 4

2 0.8.9) For this problem we converted cosh x e x + e x )/, then we can use the Taylor series for e x that we derived in the first example, [ cosh x ex + e x ) ] x n xn x n + )n n)! ) Here we have a case where the Taylor series is about a point other than zero. f n) ) e x x e e x e x )n. Ex: The derivatives of sine at x 0 are as follows, f0) 0, f 0), f 0) 0, f 0) If we continue this we see the pattern gives us sin x ) n xn+ n + )! x x3 3! + x5 5! + 3) Ex: Similarly cosine has the following derivatives at x 0, f0), f 0) 0, f 0), f 0) 0 If we continue this we see the pattern gives us cos x ) n xn n)! x + x4 4! + 4) 0.8.7) For this we just plug in x into the cosine equation above ) n 7 cos x) 7 n)! xn ) We want up to order so lets find the respective polynomials up to order and then just subtract them, fx) cos x ) x x + + x x + ) x 5x ) If we differentiate we notice d ln + x) dx + x x) x) n ) n x n ; x < To get back to ln + x) all we do is integrate the RHS, ln + x) ) n xn+ ; x <. n + This means sin x)ln+x)) x x3 3! + ) ) ) x x + x3 3! + x x3 + x4 3 x4 + x x3 Since Taylor series are just power series, we will have the usual convergence results and remainders. Theorem Taylor). If f has derivatives up to order n at a neighborhood of x a, then where for some ξ between x and a. fx) fa) + f a)x a) + f a) +x4 ; x <. x a) + + f n) a) x a) n + R n x) 5) R n x) f n+) ξ) x a)n+ n + )!

3 The fact that we have a remainder is cool and all, but we really need a way to bound it. Theorem 3 Remainder). If f n+) x) M for x a d, then Lets do one remainder example before going back to it at the end, Ex: M n + )! x a n+ ) a) Find the series representation of e x dx. Solution: We know what the Taylor series of e x is about x 0, so e x ) n x ) n ) n xn We can integrate this term-by-term, e x dx ) n x n+ n + ) b) Find the value of n that approximates 0 e x to Solution: Here we use the alternating series remainder, x n+3 n + )!n + 3) max R nx) ) n + )!n + 3) < 000 and we ll see that n 4 does the trick, 0 e x dx Notice that there are some Taylor series that are extremely common and useful, Common Taylor Series x x n + x + x + 7) + x + x + x3 3! + 8) sin x ) n xn+ n + )! x x3 3! + x5 5! + 9) cos x ) n xn n)! x + x4 4! + 0) Now lets do a bunch of assorted problems that we did in class, 0.9.) If we have a single term multiplying a series, we can simply multiply through, x x x n+. Ex: Similarly, x cos x x ) n xn n)! ) n xn+ n)! 0.9.0) Here we use the geometric series technique to derive a power series, x / x/ x ) n

4 0.9.9) Note that this technique tends to get convoluted, so if you don t feel comfortable using it then you should just take the derivatives and derive the Taylor series as you usually would. But lets try getting the first few terms of a Taylor series via multiplication ) ) e x sin x + x + x x + x3 + x3 + x + x + 3 x3 + Ex: This next example will show how to use Taylor series to simplify limits with indeterminate form, e x x lim x 0 x lim x 0 + x + x / + x 3 / + ) x x lim x 0 + x +. Ex: We could also do long division to find the first few terms of series, but that s not recommended at all, tan x sin x cos x x x3 / + x 5 /5! + x / + x 4 /4! + x + 3 x3 + 5 x5 + Ex: Find the first two terms of fx) x sin x. Solution: Lets take derivatives and plug in the center, f0) 0, f 0) sin x + x sin x cos x x0 0, f 0) 4 sin x cos x + x cos x x sin x 0, x0 f 0) cos x sin x 8x cos x sin x, x0 f 4) 3 cos x sin x + 8x sin x 8x cos x 0, x0 f 5) 40 sin x 40 cos x + 3x sin x cos x 40 x0 Then we plug into the formula for Taylor series to get x sin x 3! x3 40 5! x5 x 3 3 x ) a) This is a bit tricky for an exam, but there are a lot of pieces that will be useful for exam type problems. The original problem is difficult, but we notice that if we take two derivatives we get a geometric series like sum. Then we can work backwards to get our series. x + x ) n x n tan x ) n n + xn+ Now we look at the remainder and bound it, 0 tan xdx x n+4 n + 3)n + 4) max R nx) ) n + 3)n + 4) < 000 We did this on the calculator and got n 5 that satisfies the above condition. b) For this the maximum changes since the domain changes. 0.5) n+4 n + 3)n + 4) < 000 n ) Here we find the sum of the series using geometric series, x 3 + x 4 + x 5 + x + x n+3 x ) As we did above, x 3 x 5 + x 7 x 9 + x + ) n x n+3 x 3 x n x3 x. x ) n ) n x n+ n + )n + ). x3 + x.

5 0.9.35) We first write down the Taylor series of sine and then us the remainder formula to find the maximum error for our point. sin x ) n xn+ n + )! R nx) x 5 R 0.) 0.)5 5! 5! 0.0.4) Here instead of a point we have a domain to find our maximum error in, R x) f 3) x) x 3 ex x 3 R x) e0. 0.) 3 3! on the domain x < 0.