Thread: 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

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



So

Now consider that since the truncated terms of the series (in other words the to infinitieth) terms will be greater than zero. Therefore taking them away will detract from the value of the summation

Therefore



And



Now combinging these and not knowing N's value gives

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