prove by induction that for every natural n and every x>0

(e^x) > 1 + x/1! + (x^2)/2! + ... + (x^n)/n!

it says we can assume that we know basic properties of derivatives and integrals, but i just am not sure where to begin.

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- November 19th 2005, 07:03 PMcen0teinduction, how do i begin?
prove by induction that for every natural n and every x>0

(e^x) > 1 + x/1! + (x^2)/2! + ... + (x^n)/n!

it says we can assume that we know basic properties of derivatives and integrals, but i just am not sure where to begin. - November 20th 2005, 01:51 AMCaptainBlackQuote:

Originally Posted by**cen0te**

1. First we show that what we wish to prove for general

n actualy holds for the first n which is realevant to the

problem. In this case it's n=0, and we need to show that

for every x>0 that:

,which I presume you know is true.

2. Second we show that if what we have to prove is true

when n=k, then it is also true for n=k+1. So assume it

true for n=y, then we have

,

for all .

Now consider:

also:

Now the integrand in [1] is strictly greater than the integrand in

[2] at every point over which the integral is taken. So

3. That is we have proven that if what we have to prove for all ,

is true for n=k, then it is true for n=k+1, also we have proven that

it is true for n=0, so by the principle of induction we have proven

it true for all

RonL - November 20th 2005, 04:59 PMcen0te
Ok, I think I understand your proof. I understand induction, it is calculus that I am horribly rusty on. I have not taken any in over 5 years, and I do not even remember the notation.

Many thanks for your help. As an aside, could the statement be reworked to support x<0? - November 20th 2005, 07:04 PMJameson
You can also use Taylor's Theorem, but that too requires Calculus.

- November 20th 2005, 10:09 PMCaptainBlackQuote:

Originally Posted by**cen0te**

so is alternatly greater and less than as n

increases.

RonL - November 20th 2005, 10:16 PMCaptainBlackQuote:

Originally Posted by**cen0te**

indicate that you are expected to use calculus in the proof.

In fact I was a bit worried about this problem when I first

saw it because induction is not the natural way to prove it.

If you know the series expansion for the

result is hovering around the obvious pile.

RonL - November 21st 2005, 02:53 PMcen0te
There is a second part to the homework problem which says

"State and prove the corresponding theorem for x<0."

So I have tried, but I am not sure what is being sought. I assume the the statement needs to be reworked, but again...I am at a loss. - November 21st 2005, 06:10 PMCaptainBlackQuote:

Originally Posted by**cen0te**

and let , then , so for all (the set of natural

numbers 0,1, .. or 1,2, .. depending on your preference for how

they are defined).

or:

RonL :) - November 21st 2005, 06:20 PMcen0te
Hmm, I actually thought of that, but for some reason I thought that was basically just cheating. I had assumed that it would have something to do with how e^x with x<0 would just keep approaching 0. However, I can't come up with anything that satisfies me, so I will go with your more experienced suggestion. Thanks again.

- December 31st 2005, 10:11 AMredkimchiQuote:

Originally Posted by**cen0te**