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Write Taylor polynomial of degree 3 for $f(x)=\cos x$ at $a=0$, and prove that

$$0 \leq E_3(x) \leq \frac{x^4}{24}$$

This is what I have done so far. I struggle mainly with the right side of the inequality.

$E_3(x)=f(x)-P_3(x)$

$f(x)=\cos x$

$f'(x)=-\sin x$

$f''(x)=-\cos x$

$f^3(x)=\sin x$

$f^4(x)=\cos x$

So we have that

$P_3(x)=\cos (0)-\sin (0)\cdot x-\frac{\cos (0)}{2}\cdot x^2+\frac{\sin (0)}{3!}\cdot x^3=1-\frac{x^2}{2}$

$E_3(x)=f(x)-P_3(x)=\cos x-(1-\frac{x^2}{2})$

I see that $E_3(x)\geq 0$. But then I don't know how to prove the right side, does anyone have some tips?

David Diaz
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Mathomat55
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  • The taylor series for cosine is an infinite, alternating series. After some point, the absolute value of each successive term is smaller than the last. As such, the next term in the Taylor series itself is the largest the error could be. – David Diaz Nov 13 '19 at 12:37
  • @David Diaz Hmm, I'm not sure if I understand completely. How can you define the error with just the next term in the taylor series? I'm sorry, I'm a slow learner. – Mathomat55 Nov 13 '19 at 12:50

1 Answers1

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Not a proof, just an argument:

Let's look at the terms of $\cos x = \sum_{k=0}^\infty\frac{-1^kx^{2k}}{(2k)!}$ (The Taylor polynomial for cosine)

$$1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \frac{x^{10}}{10!} + \dots$$

Factorial functions grow faster than exponential functions. This means that after some point, each successive term in the Taylor series is smaller than the last. $$\frac{x^n}{n!} > \frac{x^{n+2}}{(n+2)!}$$

For the purposes of this argument, we consider small x. Since the series alternates, you're always adding and subtracting ever smaller terms. The partial summations will never be as small as $1-\frac{x^2}{2}$ again. They will never be as large as $1-\frac{x^2}{2}+\frac{x^4}{4!}$ again. It follows that the actual value of $\cos (x)$ lies somewhere between those two values.

David Diaz
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