# prove $| \exp(x) - 1 - x - \frac{x^2}{2!} - \frac{x^3}{3!}| < \frac{e}{24}$

prove $\displaystyle\left\lvert \exp(x) - 1 - x - \frac{x^2}{2!} - \frac{x^3}{3!}\right\rvert| < \frac{e}{24}$ $\forall x \in [-1,1]$

my attempt,

we defined $\exp(x) = \displaystyle \sum_{n=0}^{\infty} \dfrac{x^n}{n!}$ using this:

$| \exp(x) - 1 - x - \frac{x^2}{2!} - \frac{x^3}{3!}| = \left | \displaystyle \sum_{n=4}^\infty \dfrac{x^n}{n!} \right | \leq \displaystyle\sum_{n=4}^\infty \left |\dfrac{x^n}{n!} \right |$ since $|x| \leq 1$ and $e = 2.718...$ $\displaystyle\sum_{n=4}^\infty \left |\dfrac{x^n}{n!} \right | \leq \displaystyle\sum_{n=4}^\infty \left |\dfrac{e^n}{n!} \right | < e^4/24 < e/24$

Is this a valid proof? The reason I ask is because we have just started Taylor Series in my analysis class and I don't see how I'm exactly using this at all in the proof

edit using maclaurin:

$\exp(x) = 1 + x + x^2/2! + x^3/3! + e^\theta x^4/4!$ for some $\theta \in (0,1)$, since $|x| \leq 1$

$$\dfrac{e^\theta}{4!}|x^4| \leq e^\theta/24 < e/24$$

is this correct?

• $e^4>e$ so the final inequality you claim is false. Furthermore, $e^4/24$ is just one term of the series of positive terms you claimed it to be larger than so that inequality is also false. – FH93 Mar 10 '14 at 18:15
• @FH93 thanks.. I'll have another thought, don't see how to conclude from this attempt then. – Warz Mar 10 '14 at 18:19

Using Maclaurin formula: there's $\theta\in(0,1)$ such that $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{e^{\theta x}}{4!}x^4$$ and since $x\in[-1,1]$ then we have $$e^{\theta x}x^4\le e^{|\theta x|}\le e^1=e$$ and we deduce the desired result.
• How did you get $\dfrac{e^{\theta x}}{4!}$? Using Maclaurin formula, there's $\theta \in (0,1)$ s.t. $exp(x) = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{e^\theta}{4!}x^4$ and since $x \in [-1,1]$ $\dfrac{e^\theta}{4!}x^4 \leq \dfrac{|e^\theta|}{|4!|}|x^4| \leq e^\theta/24 < e/24$ is this logic correct? – Warz Mar 10 '14 at 20:25
• Yes you're correct I forgot the factor $x^4$. – user63181 Mar 10 '14 at 21:11
• Thanks, but what about the $e^{\theta x}$ In my theorem we only have $e^{\theta}$? – Warz Mar 11 '14 at 16:07
• No it's $e^{\theta x}$ and in some version of the theorem we write: there's $\xi\in(0,x)$ and the last term of The Maclaurin formula is $\frac{e^{\xi}}{4!}x^4$ so we have $\xi=\theta x$. – user63181 Mar 11 '14 at 16:23