# Extremely difficult induction proof problem

• Jan 29th 2010, 01:38 PM
Runty
Extremely difficult induction proof problem
I dunno who thought up this one, but it is clearly for masochists.

Prove that for all $x > 0$ and all positive integers n

$e^x>1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+...+\frac {x^n}{n!}$

$n!=n(n-1)(n-2)...3\cdot 2\cdot 1$

This MUST be proven by induction on $n\geq 1$.

In addition, these hints were provided for the question. They must be justified.

$e^x=1+\int_0^x e^t dt > 1+\int_0^x dt = 1+x$

$e^x=1+\int_0^x e^t dt > 1+\int_0^x (1+t) dt = 1+x+\frac {x^2}{2}$

As I said, whoever came up with this one is a sadist.

I believe I have correct justifications on the two hints (unless I wrote something wrong), but I can't seem to provide the proof for a $n+1$ case (for proving by induction). Still, if anyone can help explain this one in detail, I'd be very appreciative.
• Jan 29th 2010, 03:26 PM
tonio
Quote:

Originally Posted by Runty
I dunno who thought up this one, but it is clearly for masochists.

Prove that for all $x > 0$ and all positive integers n

$e^x>1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+...+\frac {x^n}{n!}$

$n!=n(n-1)(n-2)...3\cdot 2\cdot 1$

This MUST be proven by induction on $n\geq 1$.

In addition, these hints were provided for the question. They must be justified.

$e^x=1+\int_0^x e^t dt > 1+\int_0^x dt = 1+x$

$e^x=1+\int_0^x e^t dt > 1+\int_0^x (1+t) dt = 1+x+\frac {x^2}{2}$

As I said, whoever came up with this one is a sadist.

I believe I have correct justifications on the two hints (unless I wrote something wrong), but I can't seem to provide the proof for a $n+1$ case (for proving by induction). Still, if anyone can help explain this one in detail, I'd be very appreciative.

Have you already studied Taylor series? If so then you must know that

$e^x=\sum\limits_{k=0}^\infty\frac{x^k}{k!}\geq 1+x+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!}$

Tonio
• Jan 29th 2010, 03:40 PM
Runty
Quote:

Originally Posted by tonio
Have you already studied Taylor series? If so then you must know that

$e^x=\sum\limits_{k=0}^\infty\frac{x^k}{k!}\geq 1+x+\frac{x^2}{2!}+\ldots+\frac{x^n}{n!}$

Tonio

Never heard of Taylor series. And I don't think it would be a valid answer in this circumstance.
• Jan 29th 2010, 03:56 PM
Jester
Clearly it's true for $n = 1$. Assume true for $n = k$ so

$
e^x>1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+...+\frac {x^k}{k!}>0$
since x > 0.

Then

$
\int_0^x e^t \,dt> \int_0^x\left(1+t+\frac {t^2}{2!}+\frac {t^3}{3!}+...+\frac {t^k}{k!}\right)\,dt
$

so

$
\left. e^t \right|_0^x > \left. t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots + \frac{t^{k+1}}{(k+1)!}\right|_0^x
$

so

$
e^x -1 > x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{k+1}}{(k+1)!}
$

so

$
e^x > 1 + x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{k+1}}{(k+1)!}
$

proving true for $n = k + 1$. Thus, it's true for all $n$.
• Jan 29th 2010, 11:28 PM
tonio
Quote:

Originally Posted by Danny
Clearly it's true for $n = 1$. Assume true for $n = k$ so

$
e^x>1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+...+\frac {x^k}{k!}>0$
since x > 0.

Then

$
\int_0^x e^t \,dt> \int_0^x\left(1+t+\frac {t^2}{2!}+\frac {t^3}{3!}+...+\frac {t^k}{k!}\right)\,dt
$

This step requires, I think, full justification. It isn't true in general, of course, but it is in this case because of continuity.

Tonio

so

$
\left. e^t \right|_0^x > \left. t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots + \frac{t^{k+1}}{(k+1)!}\right|_0^x
$

so

$
e^x -1 > x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{k+1}}{(k+1)!}
$

so

$
e^x > 1 + x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{k+1}}{(k+1)!}
$

proving true for $n = k + 1$. Thus, it's true for all $n$.

.
• Jan 30th 2010, 09:26 PM
Jester
Quote:

Originally Posted by tonio
.

And your point is . . . ?
• Jan 30th 2010, 10:26 PM
TWiX
OK guys
be relax :D
The life is still countinuing

and sorry for my bad english (Rofl)
• Jan 31st 2010, 01:04 AM
tonio
Quote:

Originally Posted by Danny
And your point is . . . ?

Exactly what I wrote: the OP will do wisely to justify that step.

Tonio
• Feb 1st 2010, 11:28 AM
Runty
Quote:

Originally Posted by Danny
Clearly it's true for $n = 1$. Assume true for $n = k$ so

$
e^x>1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+...+\frac {x^k}{k!}>0$
since x > 0.

Then

$
\int_0^x e^t \,dt> \int_0^x\left(1+t+\frac {t^2}{2!}+\frac {t^3}{3!}+...+\frac {t^k}{k!}\right)\,dt
$

so

$
\left. e^t \right|_0^x > \left. t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots + \frac{t^{k+1}}{(k+1)!}\right|_0^x
$

so

$
e^x -1 > x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{k+1}}{(k+1)!}
$

so

$
e^x > 1 + x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{k+1}}{(k+1)!}
$

proving true for $n = k + 1$. Thus, it's true for all $n$.

This will have to do. Thanks for the help.