Proof of ## \displaystyle\sum_{n=0}^\infty \frac{nX^n}{n}= Xe^X##

[WMDhamnekar]

TL;DR Summary

Proofs of $1) \displaystyle\sum_{n=0}^\infty\frac{nX^n}{n!}= Xe^X, 2) \displaystyle\sum_{n=0}^\infty \frac{n^2X^n}{n!}= Xe^X + X^2e^X$

I know the Taylor expansion of exponential, $\exp(x)=\displaystyle\sum_{n=0}^\infty \frac{X^n}{n!}$

But if I calculate first and second derivatives of both sides of the above formula, L.H.S and R.H.S remain the same as before i-e $e^X$

So, how can I get the proofs of both series?

 

[PeroK]

So, how can I get the proofs of both series?

Where exactly is the problem?

 

[WMDhamnekar]

Where exactly is the problem?

I computed the first derivative of $e^X = \displaystyle\sum_{n=0}^\infty \frac{X^n}{n!}$ as follows:

$\displaystyle\sum_{n=1}^\infty \frac{nX^{n-1}}{n(n-1)!} =\displaystyle\sum_{n=1}^\infty \frac{X^{n-1}}{(n-1)!}= e^X$Derivative of $e^X$ is $e^X$ So, what is wrong here?

 

[PeroK]

Derivative of $e^X$ is $e^X$ So, what is wrong here?

That's not what you were asked to do.

 

[Office_Shredder]

I think I seev where you're trying to go with this, and you would save yourself a lot of pain if you just didn't cancel the n from the numerator and denominator and followed through on your original plan. What do you need to do to get to the desired function?

 

[WMDhamnekar]

I think I seev where you're trying to go with this, and you would save yourself a lot of pain if you just didn't cancel the n from the numerator and denominator and followed through on your original plan. What do you need to do to get to the desired function?

Author said he differentiated twice $e^X =\displaystyle\sum_{n=0}^\infty \frac{X^n}{n!} \Rightarrow(1)$

In my opinion, the first and second derivatives of (1) are not the proofs of $a) \displaystyle\sum_{n=0}^\infty \frac{nX^n}{n!}= Xe^X , b) \displaystyle\sum_{n=0}^\infty \frac{n^2X^n}{n!}= Xe^X + X^2e^X$

 

[Office_Shredder]

Differentiate, then multiply by x!

 

[WMDhamnekar]

Differentiate, then multiply by x!

I thought over the last night for many hours to find out author's intention behind his statement that he differentiated the Taylor expansion of exponential , $e^X$ twice. Then I realized what he wants to tell the readers. It is a trick to realize our goal.

 

[pasmith]

If $f(x) = \sum_{n=0}^\infty a_n x^n$ then
$$x \frac{df}{dx} = x \sum_{n=0}^\infty na_n x^{n-1} = \sum_{n=0}^\infty na_n x^n$$ and $$\begin{split} x^2 \frac{d^2f}{dx^2} + x\frac{df}{dx} &= x\frac{d}{dx}\left( x \frac{df}{dx}\right) \\ &= \sum_{n=0}^\infty n^2 a_nx^n.\end{split}$$ Setting $f(x) = e^x$ proves the results.

By induction, one has $$\sum_{n=0}^\infty n^k a_n x^n = \left( x \frac{d}{dx}\right)^k f$$ which is useful in finding $\sum n^ka_n$ if $\sum a_nx^n$ is known and has radius of convergence greater than 1.