# Thread: Extremely difficult induction proof problem

1. ## Extremely difficult induction proof problem

I dunno who thought up this one, but it is clearly for masochists.

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

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

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

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

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

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

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

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 $\displaystyle n+1$ case (for proving by induction). Still, if anyone can help explain this one in detail, I'd be very appreciative.

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

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

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

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

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

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

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

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

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 $\displaystyle n+1$ 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

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

Tonio

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

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

Tonio
Never heard of Taylor series. And I don't think it would be a valid answer in this circumstance.

4. Clearly it's true for $\displaystyle n = 1$. Assume true for $\displaystyle n = k$ so

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

Then

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

so

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

so

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

so

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

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

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

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

Then

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

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

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

so

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

so

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

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

6. Originally Posted by tonio
.
And your point is . . . ?

7. OK guys
be relax
The life is still countinuing

and sorry for my bad english

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

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

Tonio

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

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

Then

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

so

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

so

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

so

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

proving true for $\displaystyle n = k + 1$. Thus, it's true for all $\displaystyle n$.
This will have to do. Thanks for the help.