# Math Help - Boundries

1. ## Boundries

I'm required to show that:

$|e^x - (1+ \frac{x}{1!} + \frac{x^2}{2!})| \leq e/6$

for all $x \in [0,1]$

I'm not sure how to show that the equation is true at the boundries [0,1]. Does anyone know how exactly we can show that $e^x - (1+ \frac{x}{1!} + \frac{x^2}{2!})$ is monotonic in x?

Any help is very much appreciated.

2. I think you should just be able to use the formula for the error bound of a Taylor polynomial.

3. Originally Posted by redsoxfan325
I think you should just be able to use the formula for the error bound of a Taylor polynomial.
Could you please explain? I don't know how to do it because we haven't studied the error bound yet.

4. Originally Posted by Roam
I'm required to show that:

$|e^x - (1+ \frac{x}{1!} + \frac{x^2}{2!})| \leq e/6$

for all $x \in [0,1]$

I'm not sure how to show that the equation is true at the boundries [0,1]. Does anyone know how exactly we can show that $e^x - (1+ \frac{x}{1!} + \frac{x^2}{2!})$ is monotonic in x?

Any help is very much appreciated.
For $x \in [0,1]$

$
\left|e^x - \left(1+ \frac{x}{1!} + \frac{x^2}{2!}\right)\right| =e^x - \left(1+ \frac{x}{1!} + \frac{x^2}{2!}\right)=\sum_{n=3}^{\infty} \frac{x^n}{n!}
$

so:

$
\left|e^x - \left(1+ \frac{x}{1!} + \frac{x^2}{2!}\right)\right| = x^3 \sum_{n=0}^{\infty} \frac{x^n}{(n+3)!}
$

but fo $n\ge 0$:

$\frac{1}{(n+3)!} < \frac{1}{6 n!}$

hence:

$
\left|e^x - \left(1+ \frac{x}{1!} + \frac{x^2}{2!}\right)\right| \le \frac{x^3}{6} \sum_{n=0}^{\infty} \frac{x^n}{n!}=\frac{x^3 e^x}{6}
$

and the right hand side has it's maximum in $[0,1]$ when $x=1$ so:

$
\left|e^x - \left(1+ \frac{x}{1!} + \frac{x^2}{2!}\right)\right| \le \frac{e}{6}
$

CB