Solutions to Problem Set 12

MAT 127: Calculus C, Spring 2017 Solutions to Problem Set 12

Section 8.7, Problem 55 (webassign 8pts)

Use multiplication and division to ﬁnd the ﬁrst three nonzero terms in the Taylor series

− 2 y =e x cos x at x=0

Since ∞ ∞ ∞ ∞ n 2 n n 2n n 2n x − 2 ( x ) ( 1) x ( 1) x ex = , e x = − = − , cos x = − , n! n! n! (2n)! n=0 n=0 n=0 n=0 X X X X we obtain ∞ ∞ n 2n n 2n − 2 ( 1) x ( 1) x 1 1 1 1 e x cos x = − − = 1 x2 + x4 ... 1 x2 + x4 ... n! (2n)! − 1! 2! − − 2! 4! − n=0 n=0 X X 1 1 1 1 1 1 =1+ 1 x2 + x2 1 + 1 x4 + x2 x2 + x4 1 + ... · − 2! − 1! · · 4! − 1! − 2! 2! · ! ! 1 1 1 1 3 25 = 1 + 1 x2 + + + x4 + ... = 1 x2 + x4 + ... − 2 24 2 2 − 2 24 Note: if one of the above coeﬃcients turned out to be 0, we would have to expand each factor another step to get three nonzero terms of the product.

Section 8.7, Problem 22 (5pts)

1 Use the binomial series to expand the function f(x) = as a power series centered at x = 0 (1+x)4 and state the radius of convergence.

By the binomial theorem:

∞ 1 − 4 = (1+x) 4 = − xn , where (1+x)4 n n=0 X 4 ( 4)( 4 1) ... ( 4 n + 1) ( 1)n4 5 ... (n+3) (n+1)(n+2)(n+3) − = − − − − − = − · · =( 1)n . n n! n! − 6 ∞ ∞ 1 (n+1)(n+2)(n+3) n3 + 6n2 + 11n + 6 So = ( 1)n xn = ( 1)n xn (1+x)4 − 6 − 6 n=0 n=0 X X The radius of convergence is 1 by the binomial series theorem. Section 8.7, Problem 42 (webassign 18pts)

(a) Expand 1/√4 1+ x as a power series. (b) Use it to estimate 1/√4 1.1 correct to 3 decimal place.

(a) By the binomial theorem:

∞ 4 − 1/4 1/√1+ x = (1+x) 1/4 = − xn , where n n=0 X − 1/4 ( 1/4)( 1/4 1) ... ( 1/4 n + 1) ( 1)n 1 5 ... 4n 3 1 5 ... (4n 3) − = − − − · − − = − 4 · 4 · · 4 =( 1)n · · · − . n n! n! − 4nn! ∞ 4 1 5 ... (4n 3) So 1/√1+ x = ( 1)n · · · − xn By the binomial series theorem, this holds if x <1. − 4nn! | | n=0 X (b) Since .1 < 1, by part (a) | | ∞ ∞ 4 4 1 5 ... (4n 3) 1 5 ... (4n 3) 1/√1.1 = 1/√1+ .1= ( 1)n · · · − .1n = ( 1)n · · · − . − 4nn! − 4nn! 10n n=0 n=0 X X · This inﬁnite sum is to be estimated by the sum of the ﬁrst m terms with m chosen so that

∞ n=m 1 5 ... (4n 3) 1 5 ... (4n 3) 1 − ( 1)n · · · − ( 1)n · · · − < 10 3 . − 4nn! 10n − − 4nn! 10n 2 · n=0 n=0 X · X ·

This series is alternating, with

1 5 ... (4n 3) (4(n+1) 3) 4n + 1 b b = · · · − · − = b < n . n+1 4n+1(n+1)! 10n+1 4(n+1)10 n 10 · Thus, b

So it is enough to choose m so that

1 5 ... (4m 3) (4m+1) 1 · · · − · < 4m+1(m+1)! 10m+1 2 103 · · Considering the exponents of 10 above, the first m to try is m=2: 1 5 9 45 1 · · = < 42+1(2+1)! 102+1 64 6 103 2 103 · · · · So m=2 works. We could also try m=1, but 1 5 5 1 · = > . 41+1(1+1)! 101+1 16 2 102 2 103 · · · · 2 So, we must use the finite sum estimate with m=2: n=2 4 1 5 ... (4n 3) 1 1 5 1 5 1/√1.1 ( 1)n · · · − = 1 + · = 1 + ≈ − 4nn! 10n − 411! 101 422! 102 − 4 10 32 102 n=0 X · · · · · 1 1 640 16 + 1 625 125 = 1 + = − = = − 4 10 32 2 10 640 640 128 · · · This is an over-estimate because the last term used is positive (this reasoning is only for alternating series; for sums with only positive terms, a finite-sum estimate is always an under-estimate).

Remark: 1/√4 1.1 .9765, while 125/128 .9766, so our estimate is indeed accurate to 3 decimal ≈ ≈ places and is an over-estimate. Section 8.7, Problem 69 (10pts) ∞ k (a) Show that the function g(x)= xn satisﬁes n n=0 X ′ kg(x) g (x)= 1

(a; 6pts) Since the g series converges on ( 1, 1), by the theorem on p600 on ( 1, 1) ∞ ∞ − − ′ k − k (k 1) ... (k n+1) − g (x)= nxn 1 = · − · · − nxn 1 n n! n=1 n=1 X X∞ ∞ (k 1) ... (k n+1) − (k 1) ... (k n) = k − · · − xn 1 = k − · · − xn (n 1)! n! n=1 n=0 X − X This gives ∞ ∞ (1+x) ′ (k 1) ... (k n) (k 1) ... (k n+1) − g (x)= − · · − xn + x − · · − xn 1 k n! (n 1)! n=0 n=1 − X ∞ X (k 1) ... (k n+1) k n =1+ − · · − − + 1 xn (n 1)! n n=1 − X∞ ∞ (k 1) ... (k n+1)k k =1+ − · · − = xn = g(x). (n 1)!n n n=1 n=0 X − X This proves the required identity.

(b; 3pts) By the product rule and part (a), on ( 1, 1) − ′ − ′ − ′ − − − kg(x) h (x)= (1 + x) k g(x)+(1+ x) kg (x)= k(1 + x) k 1g(x)+(1+ x) k = 0. − 1+ x (c; 1pt) By part (b), h(x) = C for some constant C and so g(x)= C(1 + x)k on ( 1, 1). Since k − g(0) = =1 and (1+0)k = 1, C = 0. 0

3 Section 8.8, Problem 24 (webassign 9pts) Use the Alternating Series Test to estimate the range of values of x for which the approximation x2 x4 cos x 1 + ≈ − 2 24 is accurate to within .005.

∞ x2n For each ﬁxed x, the inﬁnite series cos x = ( 1)n is alternating, − (2n)! n=0 X x2(n+1) x2n x2 x2n = < if 2n + 1 x , (2(n+1))! (2n)! · (2n+1)(2n+2) (2n)! ≥| | and x2n/(2n)! 0 because the ratio of two consecutive terms approaches 0. Thus, by the Alter- −→ nating Series Estimation Theorem on p587, n=2 x2 x4 x2n x2·3 x6 cos x 1 + = cos x ( 1)n < b = = if x 7. − − 2 24 − − 2n! 3 (2 3)! 720 | |≤ n=0 X · 6 So, the estimate is accurate to within .005 if x6/720 1 /200 or x √3.6 ≤ | |≤

Problem K (35pts) Power series are typically used to “break” a function into a sequence of numbers (the Taylor coeﬃ- cients of the function). However, sometimes it is useful to go in the opposite direction, assembling a sequence of numbers into a function.

Let fn be the n-th Fibonacci number of Example 3c in 8.1, k=n k=n 2 2 2 2 An = k =1+2+ ... + n, Bn = k = 1 + 2 + ... + n ; Xk=1 Xk=1 by deﬁnition f0 = A0 = B0 = 0.

(a; 3pts) Give a recursive deﬁnition of the numbers f , A , B with n 0. n n n ≥ f = 0, f = 1, f = f + f n 0, 0 1 n+2 n+1 n ≥ A = 0, A = A + n+1 n 0, B = 0, B = B +(n+1)2 n 0. 0 n+1 n ≥ 0 n+1 n ≥ (b; 9pts) Use mathematical induction and only part (a) to show that f , A ,B 5n for all n 0. n n n ≤ ≥ This is true for n = 0, 1: (f = A = B = 0, f = A = B =1). Suppose this is true for some n 1. 0 0 0 1 1 1 ≥ Then, n n−1 n n n n+1 f = f + f − 5 + 5 5 + 5 = 5 2 5 n+1 n n 1 ≤ ≤ · ≤ A = A + n + 1 A + A + A 3 5n 5n+1 n+1 n ≤ n n n ≤ · ≤ B = B +(n+1)2 B + n2 + 2n + 1 B + B + 2B + B 5 5n = 5n+1 . n+1 n ≤ n ≤ n n n n ≤ · So, if f , A ,B 5n, then f , A ,B 5n+1. Thus, this is the case for all n. n n n ≤ n+1 n+1 n+1 ≤

4 (c; 3pts) Use the Absolute Convergence and Comparison Tests and only part (b) to show that the power series ∞ ∞ ∞ n n n f(x)= fnx , A(x)= Anx , B(x)= Bnx , n=0 n=0 n=0 X X X converge for if x < 1/6 (and thus determine smooth functions near x=0). | | Since f , A ,B 0, by the Absolute Convergence Test it is enough to show that the series n n n ≥ ∞ ∞ ∞ ∞ ∞ ∞ f xn = f x n , A xn = A x n , B xn = B x n n n| | n n| | n n| | n=0 n=0 n=0 n=0 n=0 n=0 X X X X X X converge if x <1/6. Since | | 5 n 5n 1 0 f x n, A x n,B x n = if x ≤ n| | n| | n| | ≤ 6 6n | |≤ 6 ∞ 5 n and the geometric series converges, the claim follows from the Comparison Test. 6 n=0 X (d; 10pts) Using only part (a), show that

x x 2x2 f(x)= x + xf(x)+ x2f(x), A(x)= xA(x)+ , B(x)= xB(x)+ + . (1 x)2 (1 x)2 (1 x)3 − − − Since f0 =A0 =B0 =0, ∞ ∞ ∞ ∞ n n n−1 2 n−2 f(x)= f1x + fnx = f1x + fn−1 +fn−2 x = x + x fn−1x + x fn−2x n=2 n=2 n=2 n=2 X X∞ ∞ X X n 2 n 2 = x + x fnx + x fnx = x + xf(x)+ x f(x) n=1 n=0 ∞ ∞ X ∞ X ∞ n n n−1 n−1 A(x)= Anx = (An−1 + n)x = x An−1x + x nx n=1 n=1 n=1 n=1 X X ∞ X∞ ′ X ′ 1 1 = x A xn + x xn = xA(x)+ x = xA(x)+ x n 1 x (1 x)2 n=0 n=0 − − ∞ ∞ X X∞ ∞ ∞ n 2 n n−1 n−1 2 n−2 B(x)= B x = (B − + n )x = x B − x + x nx + x n(n 1)x n n 1 n 1 − n=1 n=1 n=1 n=1 n=1 X X ∞ ∞ X ′ ∞ X′′ X ′ ′′ 1 1 = x B xn + x xn + x2 xn = xB(x)+ x + x2 n 1 x 1 x n=0 n=0 n=0 X X X − − 1 2 = xB(x)+ x + x2 (1 x)2 (1 x)3 − −

5 (e; 10pts) Using part (d), express fn, An, and Bn explicitly in terms of n.

By part (b),

∞ ′′ ∞ ′′ ∞ x x 1 x x − A(x)= A xn = = = xn = n(n 1)xn 2 n (1 x)3 2 · 1 x 2 2 − n=0 − − n=0 n=2 X∞ ∞ ∞ X X n(n 1) − (n+1)n (n+1)n = − xn 1 = xn = xn 2 2 2 n=2 n=1 n=0 X∞ X X ′′ ′′′ x 2x2 x 1 2x2 1 B(x)= B xn = + = + n (1 x)3 (1 x)4 2 · 1 x 6 · 1 x n=0 − − − − X ∞ ∞ ∞ ∞ ′′ 2 ′′′ 2 x 2x x − 2x − = xn + xn = n(n 1)xn 2 + n(n 1)(n 2)xn 3 2 6 2 − 6 − − n=0 n=0 n=2 n=3 ∞ X ∞X X ∞ X ∞ n(n 1) − 2n(n 1)(n 2) − (n+1)n 2(n+1)n(n 1) = − xn 1 + − − xn 1 = xn + − xn 2 6 2 6 n=2 n=3 n=1 n=2 X∞ ∞X ∞ X X (n+1)n 2(n+1)n(n 1) (2n+1)(n+1)n = xn + − xn = xn 2 6 6 n=0 n=0 n=0 X X X Comparing the left and right-most power series, we obtain

(n+1)n (2n+1)(n+1)n A =1+2+ ... + n = , B = 12 + 22 + ... + n2 = n 2 n 6

Also, by part (b), x 1 f(x)= = x , 1 x x2 − (x r−)(x + r ) − − − + 1 √5 2 where r± = − ± are the two roots of x + x 1 = 0; thus, r−r+ = 1 and r+ r− =√5. Since ∞ 2 − − − 1 = xn if x <1, 1 x | | n=0 − X ∞ 1 1 r+ n = = = r+ ( r+x) ; x r− −r−(1 x/r−) 1 ( r x) − − − − − + n=0 X∞ 1 1 r− n = = = r− ( r−x) . x r −r (1 x/r ) 1 ( r−x) − + + + n=0 − − − − X We use these two Taylor series to get a Taylor series for f(x)=x/(1 x x2). This can be done using − −

6 multiplication of power series:

∞ ∞ ∞ n 1 1 n n f(x)= fnx = x = x r+ ( r+x) r− ( r−x) − · x r− · x r − − − n=0 − − + n=0 n=0 X ∞ X X n n−1 n−1 n = xr−r 1 ( r−x) +( r x)( r−x) + ... +( r x) ( r−x)+( r x) 1 − + · − − + − − + − − + · n=0 ∞ X 0 n n−1 n 0 n = x r r− + r r− + ... + r r− + r r− ( x) + · + + + · − n=0 X ∞ n+1 n+1 ∞ n+1 n+1 ∞ n n r r− r r− r r− = x + − ( x)n = + − ( x)n+1 = + − ( x)n r+ r− − − √5 − − √5 − n=0 − n=0 n=1 ∞X X X ( r−)n ( r )n = − − − + xn . √ n=0 5 X An easier alternative is to use partial fractions and addition of power series:

∞ n 1 1 1 1 1 f(x)= fnx = x = x − · x r− · x r − · ( r ) ( r−) x r− − x r n=0 + + + X − − − − − − − ∞ ∞ ∞ n+1 n+1 ∞ n n x n n r+ r− n+1 r+ r− n = r+ ( r+x) r− ( r−x) = − ( x) = − ( x) √5 − − − − √5 − − √5 − n=0 n=0 n=0 n=1 ∞ X X X X ( r−)n ( r )n = − − − + xn . √ n=0 5 X In either case, we ﬁnd that

n n n n ( r−) (r+) 1 1+ √5 1 √5 fn = − − = − = fn √5 √5 2 − 2 !

This was the formula in Problem F on PS5.

Remark: In principle, there is a simpler way of obtaining the above formula for the sum of squares: see Solution 1 on page A39. The book’s approach can be used to compute sums of higher powers as well, but recursively (to get a formula for the sum of 5th powers, you’ll ﬁrst need to get formulas for the sums of smaller power). The above approach via power series can be used to compute sums of higher powers also, but more directly.