proof by induction

**cen0te** · November 19th 2005, 06:03 PM

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.

**CaptainBlack** · November 20th 2005, 12:51 AM

Originally Posted by cen0te

prove by induction that for every natural n and every x>0
(e^x) > 1 + x/1! + (x^2)/2! + ... + (x^n)/n!

In duction follows a pattern:

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:

$e^x\ >\ 1$ ,

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

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

for all $x\ >\ 0$ .

Now consider:

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

also:

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

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

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

Now the integrand in [1] is strictly greater than the integrand in
[2] at every point over which the integral is taken. So

$\int_0^y\ e^x\ dx\ >\ \int_0^y\ 1 + x/1! + x^2/2! + ... + x^k/k!\ dx$

which can be rewritten from what we have found above as:

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

So rearranging and replacing y by x we have:

$ 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 $n \geq\ 0$ ,
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 $n \geq\ 0$

RonL

**cen0te** · November 20th 2005, 03:59 PM

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?

**Jameson** · November 20th 2005, 06:04 PM

You can also use Taylor's Theorem, but that too requires Calculus.

$P(x)=\sum_{n=0}^{\infty}\frac{f^n(a)(x-a)!}{n!}$

**CaptainBlack** · November 20th 2005, 09:09 PM

Originally Posted by cen0te

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?

It's not true for x<0, since the series has alternating signs and
so is alternatly greater and less than $e^x$ as n
increases.

RonL

**CaptainBlack** · November 20th 2005, 09:16 PM

Originally Posted by cen0te

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.

The problem is that the instructions given with the problem
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 $e^x$ the
result is hovering around the obvious pile.

RonL

**cen0te** · November 21st 2005, 01:53 PM

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.

**CaptainBlack** · November 21st 2005, 05:10 PM

Originally Posted by cen0te

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.

I don't know if this is what they are looking for but: suppose $x<0$ ,
and let $y=-x$ , then $y>0$ , so for all $n\ \epsilon\ \mathbb{N}$ (the set of natural
numbers 0,1, .. or 1,2, .. depending on your preference for how
they are defined).

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

now substitute $-x$ for $y$ to get:

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

or:

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

RonL

**cen0te** · November 21st 2005, 05:20 PM

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.

**redkimchi** · December 31st 2005, 09:11 AM

Originally Posted by cen0te

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.

For all x < 0 and all positive integers n,
$ (-1)^n e^x\ <\ (-1)^n(1 + x/1! + x^2/2! + ... + x^k/k!) $

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