1. Calculus - Maclaurin Polynomial

It asks me to show that the nth Maclaurin Polynomial for exp(x) is </= to exp(x) for x>0. I understand this as a theory, as you can only go so far with the Maclaurin Polynomial, whearas exp(x) goes infinitely far. I think this is what the question is getting at, and would much appreciate any help on how I would start to prove this.
Many thanks,
Eloise

2. Originally Posted by EKohler
It asks me to show that the nth Maclaurin Polynomial for exp(x) is </= to exp(x) for x>0. I understand this as a theory, as you can only go so far with the Maclaurin Polynomial, whearas exp(x) goes infinitely far. I think this is what the question is getting at, and would much appreciate any help on how I would start to prove this.
Many thanks,
Eloise
I believe that what it is getting at is

$\displaystyle \lim_{N\to\infty}\sum_{n=0}^{N}\frac{x^n}{n!}=e^x$

So $\displaystyle \sum_{n=0}^{N_1}\frac{x^n}{n!}\approx{e^x}\quad\fo rall{N_1}\in\mathbb{R}$

Now consider that since $\displaystyle x>0$ the truncated terms of the series (in other words the $\displaystyle N_{1}+1$ to infinitieth) terms will be greater than zero. Therefore taking them away will detract from the value of the summation

Therefore

$\displaystyle \lim_{N\to\infty}\sum_{n=0}^{N}\frac{x^n}{n!}=e^x$

And

$\displaystyle \sum_{n=0}^{N_1\in\mathbb{R}}\frac{x^n}{n!}<e^x\qu ad\forall{x>0}$

Now combinging these and not knowing N's value gives

$\displaystyle \boxed{\sum_{n=0}^{N}\frac{x^n}{n!}\leq{e^x}\quad\ forall{x>0}}$

3. Thanks so much. that totally explained it in a manner that made sense.