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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- Nov 19th 2005, 06:03 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. - Nov 20th 2005, 12:51 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:

$\displaystyle e^x\ >\ 1$,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

$\displaystyle e^x\ >\ 1 + x/1! + x^2/2! + ... + x^k/k!

$,

for all $\displaystyle x\ >\ 0$.

Now consider:

$\displaystyle \int_0^y\ e^x\ dx\ =\ [e^x]_0^y\ =\ e^y-1\ \ \ \ \ ...[1]$

also:

$\displaystyle \int_0^y\ 1 + x/1! + x^2/2! + ... + x^k/k!\ dx\ =$

$\displaystyle \ [x + x^2/2! + x^3/3! + ... + x^{(k+1)}/(k+1)!]_0^y\ =$

$\displaystyle y + y^2/2! + y^3/3! + ... + y^{(k+1)}/(k+1)!\ \ \ \ \ ...[2]$

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

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

$\displaystyle \int_0^y\ e^x\ dx\ >\ \int_0^y\ 1 + x/1! + x^2/2! + ... + x^k/k!\ dx$which can be rewritten from what we have found above as:

$\displaystyleSo rearranging and replacing y by x we have:

e^y-1\ >\ y + y^2/2! + y^3/3! + ... + y^{(k+1)}/(k+1)!

$

$\displaystyle

e^x\ >\ 1\ +\ x + x^2/2! + x^3/3! + ... + x^{(k+1)}/(k+1)!

$

3. That is we have proven that if what we have to prove for all $\displaystyle n \geq\ 0$,

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 $\displaystyle n \geq\ 0$

RonL - Nov 20th 2005, 03:59 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? - Nov 20th 2005, 06:04 PMJameson
You can also use Taylor's Theorem, but that too requires Calculus.

$\displaystyle P(x)=\sum_{n=0}^{\infty}\frac{f^n(a)(x-a)!}{n!}$ - Nov 20th 2005, 09:09 PMCaptainBlackQuote:

Originally Posted by**cen0te**

so is alternatly greater and less than $\displaystyle e^x$ as n

increases.

RonL - Nov 20th 2005, 09:16 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 $\displaystyle e^x$ the

result is hovering around the obvious pile.

RonL - Nov 21st 2005, 01:53 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. - Nov 21st 2005, 05:10 PMCaptainBlackQuote:

Originally Posted by**cen0te**

and let $\displaystyle y=-x$, then $\displaystyle y>0$, so for all $\displaystyle n\ \epsilon\ \mathbb{N}$ (the set of natural

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

they are defined).

$\displaystylenow substitute $\displaystyle -x$ for $\displaystyle y$ to get:

e^y\ >\ 1 + y/1! + y^2/2! + ... + y^n/n!

$

$\displaystyle

e^{-x}\ >\ 1 - x/1! + x^2/2! - ... + (-1)^n\ x^n/n!

$

or:

$\displaystyle

e^x\ <\ \frac{1}{1 - x/1! + x^2/2! - ... + (-1)^n\ x^n/n!}

$

RonL :) - Nov 21st 2005, 05:20 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.

- Dec 31st 2005, 09:11 AMredkimchiQuote:

Originally Posted by**cen0te**

$\displaystyle

(-1)^n e^x\ <\ (-1)^n(1 + x/1! + x^2/2! + ... + x^k/k!)

$