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

  • I
  • Thread starter WMDhamnekar
  • Start date
  • Tags
    Proof
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
  • #1
WMDhamnekar
MHB
376
28
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:
Physics news on Phys.org
  • #2
WMDhamnekar said:
So, how can I get the proofs of both series?
Where exactly is the problem?
 
  • #3
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?
 
  • #4
WMDhamnekar said:
Derivative of ##e^X ## is ##e^X## So, what is wrong here?
That's not what you were asked to do.
 
  • #5
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?
 
  • Informative
Likes WMDhamnekar
  • #6
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##
 
  • #7
Differentiate, then multiply by x!
 
  • Like
Likes WMDhamnekar
  • #8
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.
 
  • #9
If [itex]f(x) = \sum_{n=0}^\infty a_n x^n[/itex] then
[tex]
x \frac{df}{dx} = x \sum_{n=0}^\infty na_n x^{n-1} = \sum_{n=0}^\infty na_n x^n[/tex] and [tex]
\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}[/tex] Setting [itex]f(x) = e^x[/itex] proves the results.

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

Similar threads

  • Set Theory, Logic, Probability, Statistics
Replies
8
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
844
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
739
  • Math POTW for University Students
Replies
1
Views
760
  • Set Theory, Logic, Probability, Statistics
Replies
3
Views
1K
Replies
3
Views
686
  • Calculus and Beyond Homework Help
Replies
13
Views
684
  • Set Theory, Logic, Probability, Statistics
Replies
7
Views
1K
  • Topology and Analysis
Replies
3
Views
970
Replies
14
Views
1K
Back
Top