# proof by induction

• Nov 19th 2005, 06:03 PM
cen0te
induction, 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 AM
CaptainBlack
Quote:

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

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

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
• Nov 20th 2005, 09:16 PM
CaptainBlack
Quote:

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.

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
• Nov 21st 2005, 01:53 PM
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.
• Nov 21st 2005, 05:10 PM
CaptainBlack
Quote:

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 :)
• Nov 21st 2005, 05:20 PM
cen0te
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 AM
redkimchi
Quote:

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!)
$