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.
In duction follows a pattern: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
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?
The problem is that the instructions given with the problemOriginally 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
I don't know if this is what they are looking for but: suppose $\displaystyle x<0$,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
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.