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

• I
• WMDhamnekar
In summary, the conversation discusses the Taylor expansion of exponential, ##\exp(x)=\displaystyle\sum_{n=0}^\infty \frac{X^n}{n!}## and the process of finding proofs for both series, ##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##. The author suggests using the trick of differentiating the Taylor expansion twice and setting f(x) = e^x to prove the results. This method can
WMDhamnekar
MHB
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?

Last edited:
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?

WMDhamnekar
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!

WMDhamnekar
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
$$x \frac{df}{dx} = x \sum_{n=0}^\infty na_n x^{n-1} = \sum_{n=0}^\infty na_n x^n$$ 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.

WMDhamnekar, Euge, WWGD and 1 other person

## 1. What is the significance of the "Proof of ## \displaystyle\sum_{n=0}^\infty \frac{nX^n}{n}= Xe^X##" formula?

The formula is significant because it shows the relationship between the sum of a series of terms involving X raised to different powers (## \displaystyle\sum_{n=0}^\infty X^n##) and the derivative of the exponential function (##Xe^X##).

## 2. How is this formula derived?

The formula can be derived using the power series expansion of the exponential function (##e^X = \displaystyle\sum_{n=0}^\infty \frac{X^n}{n!}##) and the properties of derivatives.

## 3. Can this formula be used to solve specific problems?

Yes, this formula can be used to solve problems involving exponential growth or decay, such as population growth or radioactive decay.

## 4. Are there any limitations to using this formula?

This formula is only valid for values of X where the series (## \displaystyle\sum_{n=0}^\infty X^n##) converges, which is when the absolute value of X is less than 1.

## 5. Are there any real-world applications of this formula?

Yes, this formula has many real-world applications in fields such as physics, economics, and finance. For example, it can be used to model the growth of a company's stock price or the decay of a radioactive substance.

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