Demonstrate that the derivative of the power series of e^x, it's its own power series


by xvtsx
Tags: demonstrate, derivative, power, series
xvtsx
xvtsx is offline
#1
May8-10, 10:24 AM
P: 15
1. The problem statement, all variables and given/known data
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]



2. Relevant equations3. The attempt at a solution

I just need a hint on how to start this problem, so how would you guys start this problem?
Phys.Org News Partner Science news on Phys.org
Lemurs match scent of a friend to sound of her voice
Repeated self-healing now possible in composite materials
'Heartbleed' fix may slow Web performance
Cyosis
Cyosis is offline
#2
May8-10, 11:06 AM
HW Helper
P: 1,495
Carry out the differentiation explicitly.
xvtsx
xvtsx is offline
#3
May8-10, 11:25 AM
P: 15
Thanks for the quick reply, but I dont see how to take the derivative of the n factorial. could you please provide me with an example of how to do it?.Thanks

Cyosis
Cyosis is offline
#4
May8-10, 11:26 AM
HW Helper
P: 1,495

Demonstrate that the derivative of the power series of e^x, it's its own power series


The n factorial is just a constant. The differentiation is with respect to x.
xvtsx
xvtsx is offline
#5
May8-10, 11:34 AM
P: 15
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.
Cyosis
Cyosis is offline
#6
May8-10, 11:36 AM
HW Helper
P: 1,495
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!?
xvtsx
xvtsx is offline
#7
May8-10, 11:55 AM
P: 15
okay. if the result its 1/n! how is that related to the power series?
Cyosis
Cyosis is offline
#8
May8-10, 11:57 AM
HW Helper
P: 1,495
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.
xvtsx
xvtsx is offline
#9
May8-10, 12:09 PM
P: 15
Oh sorry. The only thing I can say is this dx/dx= n(x^n-1)(1)/n!
Cyosis
Cyosis is offline
#10
May8-10, 12:15 PM
HW Helper
P: 1,495
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.
xvtsx
xvtsx is offline
#11
May8-10, 12:32 PM
P: 15
Honestly, I dont 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..
Cyosis
Cyosis is offline
#12
May8-10, 12:39 PM
HW Helper
P: 1,495
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.


Register to reply

Related Discussions
power series vs. taylor series Calculus & Beyond Homework 1
Complex power series to calculate Fourier series Calculus & Beyond Homework 1
Power Series/Taylor Series Calculus & Beyond Homework 6
Power series & Taylor series Calculus & Beyond Homework 4
Power series Precalculus Mathematics Homework 13