Derivative of e^x Power Series: Own Power Series

In summary, the conversation revolves around finding the derivative of a power series and using the sum rule of differentiation to interchange differentiation and summation. The n factorial is treated as a constant and the power rule is used to differentiate x^n. The person seeking help is unsure of how to apply the sum rule and is looking for clarification.
  • #1
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Homework Statement


I need to demonstrate that [tex]\frac{\mathrm{d} }{\mathrm{d} x}\sum_{n=0}^{\infty }\frac{x^{n}}{n!}= \sum_{n=0}^{\infty }\frac{x^{n}}{n!}[/tex]

Homework Equations

The Attempt at a Solution



I just need a hint on how to start this problem, so how would you guys start this problem?
 
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  • #2


Carry out the differentiation explicitly.
 
  • #3


Thanks for the quick reply, but I don't see how to take the derivative of the n factorial. could you please provide me with an example of how to do it?.Thanks
 
  • #4


The n factorial is just a constant. The differentiation is with respect to x.
 
  • #5


Okay I just got a weird answer, which I think its wrong. [tex]\frac{\mathrm{d} }{\mathrm{d} x}=\frac{(n!)}{nx^{n-1}}[/tex] could you give some steps cause for me its weird to differentiate explicitly with n and factorial.
 
  • #6


Do you know how to differentiate x^n with n a constant? If so do you know how to differentiate constant*x^n? What if the constant equals 1/n!?
 
  • #7


okay. if the result its 1/n! how is that related to the power series?
 
  • #8


The result isn't 1/n!. I asked you three questions in post #6 and you avoided answering all three. If you want help you will need to cooperate.
 
  • #9


Oh sorry. The only thing I can say is this dx/dx= n(x^n-1)(1)/n!
 
  • #10


That is correct. Furthermore from the sum rule of differentiation you know that [itex](f(x)+g(x))'=f'(x)+g'(x)[/itex]. Therefore you can just interchange differentiation and summation. If you don't see it just write out the first few terms.
 
  • #11


Honestly, I don't see it. what should I consider f(x) and g(x) ? because I only see n(x^n-1)(1)/n! as f(x).Sorry if I cause you trouble..
 
  • #12


f and g are just two functions. You are dealing with a sum of more than two functions. Nevertheless the sum rule still applies in the same way and you can interchange differentiation and summation.
 

1. What is the power series representation of e^x?

The power series representation of e^x is 1 + x + (x^2)/2! + (x^3)/3! + (x^4)/4! + ... = Σ(x^n)/n!, where n ranges from 0 to infinity.

2. How do you find the derivative of e^x using its power series representation?

To find the derivative of e^x using its power series representation, we can use the fact that the derivative of a power series is equal to the derivative of each term in the series. Therefore, the derivative of e^x is equal to 1 + x + (2x)/2! + (3x^2)/3! + (4x^3)/4! + ... = Σ(x^(n-1))/n!, where n ranges from 1 to infinity.

3. Why is the derivative of e^x equal to e^x?

The derivative of e^x is equal to e^x because the derivative of any term in the power series representation of e^x is equal to the term itself. Therefore, the derivative of each term in the series is equal to e^x, resulting in a power series representation that is also equal to e^x.

4. How does the power series representation of e^x relate to its Maclaurin series?

The power series representation of e^x is actually the Maclaurin series for e^x, as a Maclaurin series is a special case of a power series where the center is located at x=0. Therefore, the power series representation of e^x is the Maclaurin series for e^x with a center of x=0.

5. Can the power series representation of e^x be used to approximate the value of e^x for any value of x?

Yes, the power series representation of e^x can be used to approximate the value of e^x for any value of x, as long as the series is truncated at a certain term. The more terms that are included, the more accurate the approximation will be. However, for large values of x, the power series may converge slowly and may not be an efficient method for approximating e^x.

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