How can you show that

 

[math]

\sum_{k=1}^{\infty} ka_kx^{k-1}+2\sum_{k=0}^{\infty} a_kx^k = 0

[/math]

Ann, it is one of this forum's rules that we won't do homework for you, and this at face value looks like a homework problem, so I won't do it for you. But, I will give you some hints -- the second sum goes from k=0 to infinity, while the first goes from 1 to infinity. I'd find a way to convert that second sum into k=1 to infinity or the first sum to k=0 to infinity so that you are dealing with the same sums. That should get you pretty close to finishing the problem.

Ann, it is one of this forum's rules that we won't do homework for you, and this at face value looks like a homework problem, so I won't do it for you. But, I will give you some hints -- the second sum goes from k=0 to infinity, while the first goes from 1 to infinity. I'd find a way to convert that second sum into k=1 to infinity or the first sum to k=0 to infinity so that you are dealing with the same sums. That should get you pretty close to finishing the problem.

 

You should probably note that this is a triviality, since the first summation has a factor of k involved: clearly it is the case that:

 

[math]\sum_{k=1}^{\infty} ka_kx^{k-1} = \sum_{k=0}^{\infty} ka_kx^{k-1}[/math]

 

I don't think the question you've asked is quite worded correctly: here is my corrected version, you need to tell me whether it's right or not:

 

For which [math]a_k[/math] does the equation [math]\sum_{k=1}^{\infty} ka_kx^{k-1}+2\sum_{k=0}^{\infty} a_kx^k = 0[/math'] hold?

 

If this is what you actually meant, then write [math]f(x) = \sum_{k=0}^{\infty} a_k x^k[/math], and try to re-write the equation in terms of f. From there it is simple.

What I've done is;

 

[math]

y(x) = \sum_{k=0}^{\infty} a_k x^k

[/math]

 

[math]

y'(x) = \sum_{k=1}^{\infty} ka_k x^{k-1}

[/math]

 

By developing both series and grouping I got;

 

[math]

(2a_0+a_1) + (2a_1+a_2)x + ...

[/math]

 

Because they must all equal 0 to satisfay the first equation;

 

[math]

a_{k-1} = \frac{-ka_k}{2}

[/math]

 

But I'm not sure, and I have no clue how to verify this kind of problem with Maple (the "solve" function does not allow "\sum" in it)

You're virtually there. When you substitute y(x) into the equation, one has that:

 

[math]y'(x) + 2y(x) = 0[/math]

 

This is a separable ODE which can easily be solved. You just need to figure out what to do with the constant that comes along with it (some kind of initial condition would be handy, but you haven't stated there is one).

I see the link with the differential equation, but according to the question, I only need to find the value of a.

 

[math]

a_{k-1} = \frac{-ka_k}{2}

[/math]

 

Isn't right ?

 

BTW thank you very much for the help

But I'm not sure, and I have no clue how to verify this kind of problem with Maple (the "solve" function does not allow "\sum" in it)

 

Really, I was thinking about getting maplesoft for students but I cant seem to find any honest or in depth report about it out on the net.

I see the link with the differential equation, but according to the question, I only need to find the value of a.

 

[math]

a_{k-1} = \frac{-ka_k}{2}

[/math]

 

Isn't right ?

 

BTW thank you very much for the help

 

Well that may or may not be right. My idea was to explicitly solve the ODE, then the resulting function y(x) should resemble an exponential function. Then find the Taylor expansion of y to find the co-efficients [math]a_k[/math].