# Justification for upper bound in Taylor polynomial

1. Aug 6, 2017

### woe_to_hice

1. The problem statement, all variables and given/known data
I've been reviewing some Taylor polynomial material, and looking over the results and examples here.
https://math.dartmouth.edu/archive/m8w10/public_html/m8l02.pdf

I'm referring to Example 3 on the page 12 (page numbering at top-left of each page). The question is asking about an upper bound on the error.

2. Relevant equations

3. The attempt at a solution
I was able to get the result in the PDF, but in the discussion of the example, the authors mention that there is a logical issue with using an exact value of e^(0.2) while we're approximating f(x) = e^x . As an alternate bound the authors propose using f(5)(c) < 2 .

Can anyone tell me where the justification for this value of 2 comes from? Why is this a proposed value for an upper bound?

Last edited: Aug 6, 2017
2. Aug 6, 2017

### Staff: Mentor

To be honest? I can't see it either. One could take $0.2 < \frac{1}{2}$ and $f^{(5)}(c) \leq e^{0.2} < e^{0.5} < \sqrt{3} < 2$ or something like
$$f^{(5)}(c) = 1 + c + \frac{c^2}{2!}+ \frac{c^3}{3!} + \ldots < 1+\frac{1}{5}+\left( \frac{1}{5}\right)^2+ \left( \frac{1}{5}\right)^3+\ldots < 1+\frac{1}{2}+\left( \frac{1}{2}\right)^2 +\ldots = 2$$ but this is a lot of guesswork. Did they eventually had an upper bound in the previous section, or perhaps the $K$ in Rolle?

Another possibility is to use the fact (if given), that the exponential function is convex. In this case we have
$$f^{(5)}(c) < \frac{e^0+e^1}{2} < \frac{1}{2} + \frac{3}{2} = 2$$

3. Aug 6, 2017

### woe_to_hice

Thanks for replying. I will look at the Rolle's theorem possibility.

4. Aug 6, 2017

### Ray Vickson

For $x > 0$ we have $e^{-x} > 1-x$, so $e^x < 1/(1-x)$, giving $e^{0.2} < 1/(1 - 0.2) = 1.25 < 2$.