What am i missing here?

**gdmath** · December 5th 2009, 03:09 AM

(All integrations are in respect of dx)

Its known that:
$ e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!} \quad \quad (1)$

Now if we integrate both parts we have that:
$ \int e^x=\int \sum_{n=0}^{\infty} \frac{x^n}{n!} \Rightarrow$

$ e^x=\int\frac{x^0}{0!}+\int\frac{x^1}{1!}+...\Righ tarrow$

and correct me if i am wrong but:
$ \int\frac{x^n}{n!}=\frac{x^{n+1}}{(n+1)!}$

So
$ e^x=\frac{x^1}{1!}+\frac{x^2}{2!}+...\Rightarrow$
$ e^x=\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!}\quad \quad (2)$

And comparing 1 and 2
$ \sum_{n=0}^{\infty} \frac{x^{n}}{(n)!}=\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!}$

What am i missing here?

Thank you a lot

**HallsofIvy** · December 5th 2009, 04:10 AM

Originally Posted by gdmath

Its known that:
$ e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!} \quad \quad (1)$

Now if we integrate both parts we have that:
$ \int e^x=\int \sum_{n=0}^{\infty} \frac{x^n}{n!} \Rightarrow$

$ e^x=\int\frac{x^0}{0!}+\int\frac{x^1}{1!}+...\Righ tarrow$

and correct me if i am wrong but:
$ \int\frac{x^n}{n!}=\frac{x^{n+1}}{(n+1)!}$

So
$ e^x=\frac{x^1}{1!}+\frac{x^2}{2!}+...\Rightarrow$
$ e^x=\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!}\quad \quad (2)$

And comparing 1 and 2
$ \sum_{n=0}^{\infty} \frac{x^{n}}{(n)!}=\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!}$

What am i missing here?

Thank you a lot

You are missing the fact that $\int f(x) dx= F(x)+ C$ . That is, what you have shown is that
$\sum_{n=0}^{\infty} \frac{x^{n}}{(n)!}=\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!}+ C$
and that's obviously true: let k= n+1 on the right and each term becomes $\frac{x^k}{k!}$ . When n= 0, k= 1 so your equation becomes
$\sum_{n=0}^{\infty} \frac{x^{n}}{(n)!}=\sum_{k=1}^{\infty} \frac{x^{k}}{(k)!}+ C$
which is true with C= 1, the missing "first term" on the right.

**gdmath** · December 5th 2009, 04:41 AM

Thank you

Indeed this explains the result.

However i rely on the fact that $e^x$ is the "neutral" element in calculus (as unit is at multiplication / division).

So what about if we derive the initial relation???

I thing we lead to a similar inconvinient result.

$ \sum_{n=0}^{\infty} \frac{x^{n}}{(n)!}=\sum_{n=0}^{\infty} \frac{x^{n-1}}{(n-1)!}$

of course in this last case $n!$ (for negative n) can be defined properly so the last relation to be valid

**Defunkt** · December 5th 2009, 05:57 AM

Originally Posted by gdmath

Thank you

Indeed this explains the result.

However i rely on the fact that $e^x$ is the "neutral" element in calculus (as unit is at multiplication / division).

So what about if we derive the initial relation???

I thing we lead to a similar inconvinient result.

$ \sum_{n=0}^{\infty} \frac{x^{n}}{(n)!}=\sum_{n=0}^{\infty} \frac{x^{n-1}}{(n-1)!}$

of course in this last case $n!$ (for negative n) can be defined properly so the last relation to be valid

$\sum_{n=0}^{\infty}\frac{x^n}{n!} = \frac{1}{1} + \frac{x}{1} + \frac{x^2}{2} + \frac{x^3}{6} + ...$

Differentiating that, we get:

$0 + 1 + x + \frac{x^2}{2} + ... = \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x$

Why was your result wrong? The derivate of a constant, c, is 0 -- not $\frac{x^{n-1}}{(n-1)!}$ .

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