Taylor Polynomials and Series

Slide 1: Sections 10.1–3 Taylor Polynomials and Series Math E-16 December 19, 2007 Announcements Matthew Leingang (leingang@math.harvard.edu) at your service Homework for next time: 10.1: #2,4,5,7,8,10-13,18,26-28,34 10.2: #1,6,16,17,21,22,28,32,40 10.3: #8-12,20,22,36

Slide 2: Taylor Polynomials Motivation Derivation Examples Taylor Series Definition Power Series and the Convergence Issue Famous Taylor Series New Taylor Series from Old

Slide 3: Motivation What is sin(91◦ )? √ What is 4.001?

Slide 4: How does your calculator work? Suppose you ask your calculator for sin(31◦ ). Does it construct a right triangle with one angle equal to 31◦ , then measure opposite-over-hypotenuse? Does it construct a unit circle, measure the arc length equal to 31◦ , then use the vertical coordinate? No, it uses a polynomial approximation. Deep down, calculators can only add and multiply anyway.

Slide 5: Constant Approximation If f is continuous at a, then f (x) ≈ f (a) when x is “close to” a. Example √ √ 4.001 ≈ 4=2 sin(91 ) ≈ sin(90◦ ) = 1 ◦ But we should be able to do better.

Slide 6: Linear Approximation How can we approximate a function with a line?

Slide 7: Linear Approximation How can we approximate a function with a line? Definition Let f be a function and a a • point at which f is differentiable. Then the linear approximation to f at a is L(x) = f (a) + f (a)(x − a)

Slide 8: Example Estimate these by linear approximation. √ (i) 4.001 (ii) sin(91◦ )

Slide 9: Example Estimate these by linear approximation. √ (i) 4.001 (ii) sin(91◦ ) Solution (i) √ 1 We use f (x) = x, a = 4. Then f (4) = 4 , so L(x) = 2 + 1 (x − 4) 4 This means √ 1 1 4.001 ≈ 2 + · = 2.00025 4 1000 Notice (2.00025)2 = 4.001000063

Slide 10: Solution (ii) We use f (x) = sin x, a = 90◦ = π/2. Then f (a) = 1, f (x) = cos x, so f (a) = 0. This means L(x) = 1 so the linear approximation is no better than the constant one.

Slide 11: Quadratic Approximation How can we approximate a function with a parabola?

Slide 12: Quadratic Approximation How can we approximate a function with a parabola? We are looking for a function Q(x) = c0 + c1 (x − a) + c2 (x − a)2 with Q(a) = f (a), Q (a) = f (a), Q (a) = f (a). What are c0 , c1 , c2 in terms of f ?

Slide 13: Quadratic Approximation How can we approximate a function with a parabola? We are looking for a function Q(x) = c0 + c1 (x − a) + c2 (x − a)2 with Q(a) = f (a), Q (a) = f (a), Q (a) = f (a). What are c0 , c1 , c2 in terms of f ? Since Q(a) = c0 , we need c0 = f (a). Since Q (x) = c1 + 2c2 (x − a) =⇒ Q(a) = c1 we need c1 = f (a). Since Q (x) = 2c2 = Q(a) we need c2 = 1 f (a). 2

Slide 14: Definition Let f be a function and a a • point at which f is twice differentiable. Then the quadratic approximation to f at a is Q(x) = f (a)+f (a)(x−a)+ 1 f (a)(x−a)2 2

Slide 15: Example Estimate these by quadratic approximation. √ 4.001 sin(91◦ )

Slide 16: Example Estimate these by quadratic approximation. √ 4.001 sin(91◦ ) Solution (i) √ 1 1 and f (4) = − 32 , so We use f (x) = x, a = 4. Then f (4) = 4 − 4)2 Q(x) = 2 + 1 (x − 4) − 1 64 (x 4 This means √ 1 1 11 4.001 ≈ 2 + · − = 2.000249984 103 64 106 4 √ This is the same answer my TI-83 gives for 4.001.

Slide 17: Solution (ii) We use f (x) = sin x, a = 90◦ = π/2. Then f (a) = 1, f (a) = 0, and f (x) = − sin x, so f (a) = −1. This means Q(x) = 1 − 2 (x − π)2 1 so π2 sin(91◦ ) ≈ 1 − 1 ≈ 0.9998476913 2 180 My TI-83 has sin(91◦ ) = 0.9998476952 which this agrees with up to the ninth place.

Slide 18: In General How can we approximate a function with a polynomial of degree n?

Slide 19: In General How can we approximate a function with a polynomial of degree n? Definition Let f be a function and a a point at which f is n times differentiable. The Taylor Polynomial of degree n for f centered at a is f (n) (a) f (a) (x − a)2 + · · · + (x − a)n Pn (x) = f (a) + f (a)(x − a) + 2! n! n f (k) (a) (x − a)k = k! k=0 (Convention: f (0) (x) = f (x))

Slide 20: Taylor Polynomials Motivation Derivation Examples Taylor Series Definition Power Series and the Convergence Issue Famous Taylor Series New Taylor Series from Old

Slide 21: Take it to the Limit Definition Let f be a function and a a point at which f is infinitely differentiable. The Taylor Series of for f centered at a is f (a) (x − a)2 + T (x) = f (a) + f (a)(x − a) + 2! f (n) (a) (x − a)n + . . . ··· + n! ∞ f (k) (a) (x − a)k = k! k=0

Slide 22: Example Compute the Taylor Series for f (x) = ln x centered at 1.

Slide 23: Example Compute the Taylor Series for f (x) = ln x centered at 1. Solution f (x) = ln x f (1) = 0 −1 f (x) = x f (1) = 1 −2 f (x) = −x f (1) = −1 −3 f (x) = 2x f (1) = 2 f (4) (x) = −6x −3 f (4) (1) = −6 ... ... ... ... f (k) (x) = (−1)k+1 (k − 1)!x −k f (k) (1) = (−1)k+1 (k − 1)! So ∞ ∞ (−1)k+1 (k − 1)! (−1)k+1 (x − 1)k = (x − 1)k T (x) = k! k k=1 k=1

Slide 24: Caution The infinite sum is dangerous! Sometimes it gives very good approximations quickly Sometimes it gives good approximations slowly Sometimes it doesn’t give anything. To see this, let Pn be the nth degree Taylor Polynomial of f (x) = ln x centered at 1. For which x is T (x) = lim Pn (x) n→∞ meaningful?

Slide 25: 1 n Pn ( 2 ) Pn (2) Pn (3) 1 -0.5 1. 2.0 2 -0.625 0.5 0.0 3 -0.666667 0.833333 2.66667 4 -0.682292 0.583333 -1.33333 5 -0.688542 0.783333 5.06667 6 -0.691146 0.616667 -5.6 7 -0.692262 0.759524 12.6857 8 -0.69275 0.634524 -19.3143 9 -0.692967 0.745635 37.5746 10 -0.693065 0.645635 -64.8254

Slide 26: Pn ( 1 ) n Pn (2) Pn (3) 2 10 -0.693065 0.645635 -64.8254 20 -0.693147 0.668771 -34359.7 -2.3593×107 30 -0.693147 0.676758 -1.81712×1010 40 -0.693147 0.680803 -1.49113×1013 50 -0.693147 0.683247 -1.27387×1016 60 -0.693147 0.684883 -1.11899×1019 70 -0.693147 0.686055 -1.00322×1022 80 -0.693147 0.686936 -9.13584×1024 90 -0.693147 0.687622 -8.42274×1027 100 -0.693147 0.688172

Slide 27: Pn ( 1 ) n Pn (2) Pn (3) 2 -8.42274×1027 100 -0.693147 0.688172 -5.34752×1057 200 -0.693147 0.690653 -4.52171×1087 300 -0.693147 0.691483 -4.30016×10117 400 -0.693147 0.691899 -4.36161×10147 500 -0.693147 0.692148 -4.60801×10177 600 -0.693147 0.692315 -5.00727×10207 700 -0.693147 0.692433 -5.55436×10237 800 -0.693147 0.692523 -6.25895×10267 900 -0.693147 0.692592 -7.14101×10297 1000 -0.693147 0.692647

Slide 28: Observations Pn ( 1 ) converges to f ( 1 ) = − ln 2 2 2 Pn (2) converges to f (2) = ln 2, but more slowly Pn (3) does not converge at all! So in examining this process of approximation by polynomials, we have to be a little bit careful about what numbers we plug in.

Slide 29: Definition (cf. §9.5) A power series centered at a is a sum of constants times powers of (x − a): c0 + c1 (xa ) + c2 (x − a)2 + · · · + cn (x − a)n + . . .

Slide 30: Theorem ∞ cn (x − a)n there are only three For a given power series n=1 possiblities: 1. There is a number R such that the series converges when |x − a| < R and diverges when |x − a| > R. 2. The series converges for all x (R = ∞) 3. The series converges only when x = a (R = 0). R is called the radius of convergence of the power series.

Slide 31: Why radius? An open interval is kind of like a one-dimensional circle: a−R a a+R

Slide 32: Why radius? An open interval is kind of like a one-dimensional circle: convergence on (a − R, a + R) a−R a a+R

Slide 33: Why radius? An open interval is kind of like a one-dimensional circle: divergence on (−∞, a − R) and (a + R, ∞) a−R a a+R

Slide 34: Why radius? An open interval is kind of like a one-dimensional circle: at the endpoints—??? a−R a a+R

Slide 35: Example Compute the radius of convergence of the power series ∞ xk f (x) = k=0

Slide 36: Example Compute the radius of convergence of the power series ∞ xk f (x) = k=0 Solution 1 This is a geometric series. We know it converges to when 1−x |x| < 1, and not when x = 1 or x = −1. So The radius of convergence is 1 The interval of convergence is the open interval (−1, 1)

Slide 37: Famous Taylor Series Example Compute Taylor series centered at zero for the following functions: ex sin x cos x (1 + x)p

Slide 38: Example Compute the Taylor series centered at zero for f (x) = e x

Slide 39: Example Compute the Taylor series centered at zero for f (x) = e x Solution f (x) = e x f (0) = 1 x f (x) = e f (0) = 1 x f (x) = e f (0) = 1 x f (x) = e f (0) = 1 ... ... ... ... (k) x (k) f (x) = e f (0) = 1 So ∞ xk T (x) = k! k=0

Slide 40: Fact The Taylor series for the function f (x) = e x converges for all x to ex .

Slide 41: Fact The Taylor series for the function f (x) = e x converges for all x to ex . The convergence is because the factorials k! grow much faster than the exponentials x k . It’s a little more work to say that it converges to e x .

Slide 42: Example Compute the Taylor series centered at zero for f (x) = sin x.

Slide 43: Example Compute the Taylor series centered at zero for f (x) = sin x. Solution f (x) = sin x f (0) = 0 f (x) = cos x f (0) = 1 f (x) = − sin x f (0) = 0 f (x) = − cos x f (0) = −1 (4) f (4) (0) = 1 f (x) = cos x And repeat! So ∞ ∞ ±x k (−1)m x 2m+1 x3 x5 =x− − ··· T (x) = = + k! (2m + 1)! 3! 5! m=0 k=0 k odd This turns out to converge for all x to sin x.

Slide 44: Example Compute the Taylor series centered at zero for f (x) = cos x.

Slide 45: Example Compute the Taylor series centered at zero for f (x) = cos x. Solution f (x) = cos x f (0) = 1 f (x) = − sin x f (0) = 0 f (x) = − cos x f (0) = −1 f (x) = sin x f (0) = 0 (4) f (4) (0) = 0 f (x) = sin x And repeat! So ∞ ∞ ±x k (−1)m x 2m x2 x4 =1− − ··· T (x) = = + k! (2m)! 2! 4! m=0 k=0 k even This turns out to converge for all x to cos x.

Slide 46: Example (The Binomial Series) Compute the Taylor series centered at zero for f (x) = (1 + x)p , where p is any number (not just a whole number).

Slide 47: Example (The Binomial Series) Compute the Taylor series centered at zero for f (x) = (1 + x)p , where p is any number (not just a whole number). Solution f (x) = (1 + x)p f (0) = 1 p−1 f (x) = p(1 + x) f (0) = p p−2 f (x) = p(p − 1)(1 + x) f (0) = p(p − 1) p−3 f (x) = p(p − 1)(p − 2)(1 + x) f (0) = p(p − 1)(p − 2) ... ... ... ... So p(p − 1) 2 T (x) = 1 + px + x + ... 2 ∞ p(p − 1)(p − 2) · · · (p − k + 1) k = x k!

Slide 48: New Taylor Series from Old Big Time Theorem We can integrate and differentiate power series, and the ROC stays ∞ ck (x − a)k has radius of convergence R, the same: If f (x) = k=0 that is, if ∞ ck (x − a)k when |x − a| < R, f (x) = k=0 then ∞ kck (x − a)k−1 when |x − a| < R, f (x) = k=1 and ∞ 1 ck (x − a)k+1 + C when |x − a| < R. f (x) dx = k +1

Slide 49: This is really saying two things: 1. The power series which is differentiated (or integrated) term-by-term has the same radius of convergence as the original power series. 2. It converges to the thing we want: the derivative or antiderivative of f

Slide 50: The other big theorem If f has any power series representation near a, then it is equal to the Taylor Series.

Slide 51: Example Compute the Taylor series centered at 0 for arctan x.

Slide 52: Example Compute the Taylor series centered at 0 for arctan x. Solution 1 First, we’ll find the Taylor Series for . It’s geometric: 1 + x2 ∞ ∞ 1 1 (−x 2 )k = (−1)k x 2k . = = 2 1 − (−x 2 ) 1+x k=0 k=0 This converges when x 2 < 1 ⇐⇒ |x| < 1, so the ROC is 1. Now arctan is the antiderivative: ∞ ∞ ∞ (−1)k x 2k+1 (−1)k x 2k dx = (−1)k x 2k dx = arctan x = (2k + 1) k=0 k=0 k=0 And the ROC of this 1, too.

Slide 53: Cool result This means if ∞ (−1)k 111 = 1 − + − + ··· (2k + 1) 357 k=0 converges (and it does), it converges to π arctan(1) = 4

Slide 54: Example Compute the Taylor series centered at 0 for f (x) = x 7 sin(x 3 ). Find f (2008) (0).

Slide 55: Example Compute the Taylor series centered at 0 for f (x) = x 7 sin(x 3 ). Find f (2008) (0). Solution ∞ ∞ (−1)m (x 3 )2m+1 (−1)m x 6m+3 7 3 7 = x7 x sin(x ) = x (2m + 1)! (2m + 1)! m=0 m=0 ∞ (−1)m x 6m+10 = (2m + 1)! m=0

Slide 56: To find f (2008) (0), note ∞ ∞ (−1)m x 6m+10 f (k) (0)x k = (2m + 1)! k! m=0 k=0 Equating the coefficients of x 2008 will get us what we want. If 6m + 10 = 2008, then m = 333. So f (2008) (0) (−1)333 = 2008! 667! and thus 2008! f (2008) (0) = − 667!