Derivative of e^x Power Series: Own Power Series

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Homework Statement


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

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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Carry out the differentiation explicitly.
 


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
 


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


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


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!?
 


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


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.
 


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 (f(x)+g(x))'=f'(x)+g'(x). 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.
 
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