Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
$e^4>e$ 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 at 18:15
    
@FH93 thanks.. I'll have another thought, don't see how to conclude from this attempt then. –  Warz Mar 10 at 18:19

1 Answer 1

up vote 7 down vote accepted

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.

share|improve this answer
    
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 at 20:25
    
Yes you're correct I forgot the factor $x^4$. –  Sami Ben Romdhane Mar 10 at 21:11
    
Thanks, but what about the $e^{\theta x}$ In my theorem we only have $e^{\theta}$? –  Warz Mar 11 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 $\xi=\theta x$. –  Sami Ben Romdhane Mar 11 at 16:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.