Kenneth Kuttler

5.16. EXERCISES 145

This is pretty easy if |x| ≤ 1. This is because∣∣∣ex −∑

nk=0

∣∣∣ ≤ eξ

(n+1)! ≤e

(n+1)! . Clearlythe term at the end converges to 0 as n → ∞. Now suppose you have arbitrary x. This works

out exactly the same for arbitrary x if it can be shown that limn→∞e|x||x|n+1

(n+1)! = 0. This is

because x → ex is an increasing function. It suffices to show that for r > 0, limn→∞rn

n! = 0.

Why would this be so? It is because (rn+1/(n+1)!)rn/n! = r

n+1 < 12 for all n large enough. Say

this happens for all n ≥ N. Then, letting an =rn

n! to save space, this has shown that there issome N such that ak+1/ak ≤ 1

2 for all k ≥ N. Thus, for k > N,

ak+1 ≤12

ak ≤122 ak−1 ≤ ·· · ≤ 1

2k−N aN =12k

(aN2N)

clearly limk→∞12k = 0 and so this shows the desired result that limn→∞

n! = 0.

Exercise 5.15.4 What is e to several decimal places?

From the above,∣∣e−∑

nk=0

1k!

∣∣ ≤ e(n+1)! . One can easily see that ln3 > 1 directly from

the definition of ln . Therefore, 3 > e. Also 310! ≤ 10−6. It follows that ∑

10k=0

1k! is within

10−6 of e. Therefore, ∑10k=0

1k! = 2.7182818 is within 10−6 of e.

5.16 Exercises1. Let f have n derivatives on an open interval containing c. Suppose you desire to

approximate f (x) with a polynomial pn (x) = ∑nk=0 ak (x− c)k such that both pn and

f have the same first n derivatives at c. Show that it must be the case that ak =f (k)(c)

k! .

2. Show that sin(x) = ∑nk=0 (−1)k x2k+1

(2k+1)! +sin(2n+2)(ξ )x2n+2

(2n+2)! for some ξ between 0 andx. Find sin(.1) to a few decimal places and estimate how close your approximationis using the remainder term.

3. Show that cos(x) = ∑nk=0 (−1)k x2k

(2k)! +cos(2n+2)(ξ )x2n+1

(2n+1)! for some ξ between 0 and x.Find cos(.1) to a few decimal places and estimate how close your approximation isusing the remainder term.

4. Explain why, for |x| ≤ 1, cos(x) = limn→∞ ∑nk=0 (−1)k x2k

(2k)! .

5. Suppose you want to find a function y such that y′ (x)+xy(x) = sin(x) and y(0) = 1.This is called an initial value problem for y. Find a polynomial of degree 3 which willapproximate the solution to this equation in the sense that the first three derivativesof both y and the polynomial coincide at x = 0, assuming there is such a solution y.Hint: Use Problem 1 and the differential equation to determine this polynomial.

6. The following is the graph of a function and there are two points indicated (x1,0)and (x2,0) , the latter coming from the intersection of the tangent line to the graph ofthe function at (x1, f (x1)) and the x axis as shown.

5.16. EXERCISES 145géSaaS Ga!the term at the end converges to 0 as n —> co. Now suppose you have arbitrary x. This worksx) jarcuit = 0. This isbecause x — e* is an increasing function. It suffices to show that for r > 0, limy-5.0 r =0.r+! (n+)!nl ) =a <3this happens for all n > N. Then, letting a, = x to save space, this has shown that there issome WN such that ap+4/az < 5 for all k > N. Thus, fork > N,This is pretty easy if |x| < 1. This is because le" —Yi-o x Clearlyout exactly the same for arbitrary x if it can be shown that lim, _;.0Why would this be so? It is because ! for all n large enough. Say1 1 1 1a+ S 5k < x21 Se ay2")> 5k-N ON = 5k (clearly limy_s.. 2 3¢ = 9 and so this shows the desired result that lim,—,.. a =0.Exercise 5.15.4 What is e to several decimal places?one can easily see that In3 > 1 directly from—Li=0 ral < ceaethe definition of In. Therefore, 3 >e. Also at < 10~°. It follows that ye 0 a is within10~° of e. Therefore, Yj2.9 4 = 2.7182818 is within 10~° of e.5.16 Exercises1. Let f have n derivatives on an open interval containing c. Suppose you desire toapproximate f(x) with a polynomial pp (x) = Leo ax (x — c)* such that both Pn andf have the same first derivatives at c. Show that it must be the case that a, = fe wo )2. Show that sin(x) = Tf_g(—1)* AA + eerx. Find sin(.1) to a few decimal places and estimate how close your approximationis using the remainder term.for some € between 0 and2k .(2n+2) 2n+13. Show that cos (x) = Y?_, (—1) ear eeFind cos (.1) to a few decimal places and estimate how close your approximation isusing the remainder term.for some & between 0 and x.k 2k(2k)! °4. Explain why, for |x| < 1, cos (x) = limy—oo P79 (—1)5. Suppose you want to find a function y such that y’ (x) +.xy (x) = sin (x) and y(0) = 1.This is called an initial value problem for y. Find a polynomial of degree 3 which willapproximate the solution to this equation in the sense that the first three derivativesof both y and the polynomial coincide at x = 0, assuming there is such a solution y.Hint: Use Problem | and the differential equation to determine this polynomial.6. The following is the graph of a function and there are two points indicated (x;,0)and (x,0), the latter coming from the intersection of the tangent line to the graph ofthe function at (x1, f (x1 )) and the x axis as shown.