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

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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?
 
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WMDhamnekar said:
So, how can I get the proofs of both series?
Where exactly is the problem?
 
PeroK said:
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?
 
WMDhamnekar said:
Derivative of ##e^X ## is ##e^X## So, what is wrong here?
That's not what you were asked to do.
 
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?
 
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Office_Shredder said:
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##
 
Differentiate, then multiply by x!
 
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Office_Shredder said:
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.
 
If f(x) = \sum_{n=0}^\infty a_n x^n then
<br /> x \frac{df}{dx} = x \sum_{n=0}^\infty na_n x^{n-1} = \sum_{n=0}^\infty na_n x^n and <br /> \begin{split}<br /> x^2 \frac{d^2f}{dx^2} + x\frac{df}{dx} &amp;= x\frac{d}{dx}\left( x \frac{df}{dx}\right) \\<br /> &amp;= \sum_{n=0}^\infty n^2 a_nx^n.\end{split} Setting f(x) = e^x proves the results.

By induction, one has <br /> \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.
 
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