# Thread:

1. ## What am i missing here?

(All integrations are in respect of dx)

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

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

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

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

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

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

What am i missing here?

Thank you a lot

2. Originally Posted by gdmath
Its known that:
$<br /> e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!} \quad \quad (1)$

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

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

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

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

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

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.

3. 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.

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

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

4. 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.

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

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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