Boundries

**Roam** · May 5th 2009, 08:43 PM

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.

**redsoxfan325** · May 5th 2009, 10:23 PM

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

**Roam** · May 5th 2009, 10:32 PM

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.

**CaptainBlack** · May 5th 2009, 11:00 PM

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

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