Finding Values of 'a' for f ''(x) + f(x)=0

[oswald]

Let us take the first three non-zero terms of Maclaurin's expansion of
f(x)= Asen(ax) + Bcos(ax).
Determine for which values of a the equation f ' '(x) + f(x) = 0.

 

[Gib Z]

Welcome to Physicsforums Oswald.

In this case it seems better to leave f(x) in exact form. Can you find f''(x) and express it in terms of f(x) ?

 

[HallsofIvy]

Let us take the first three non-zero terms of Maclaurin's expansion of
f(x)= Asen(ax) + Bcos(ax).
Determine for which values of a the equation f ' '(x) + f(x) = 0.

That looks like a homework problem and you haven't said anything about what YOU have already done. Do you know how to find the MacLaurin's series for a function?

 

[oswald]

Maclaurin Serie solution

I have done this, but it doesn't work...
......∞
A sin(ax) = A[ ∑ (-1)^k . (ax)^2k+1 ] / (2k+1)! to find Maclaurin series of A sin(ax)
......0
......∞
B cos(ax) = B[ ∑ (-1)^k . (ax)^2k ] / (2k)! is Maclaurin Series of B cos(ax)
......0

hence:

A sin(ax) = Aax - Aa³x³/6 + Aa^5x^5/120

B cos(ax) = 1 - Ba²x²/2 + Ba^4x^4/24

f''(x) + f(x) = 0

what is f''(x)? I think its:

f''(x) = 0 + 1
f(x) = A sen(ax) + B cos(ax)

so,

0 + 1 + Asen(ax) + B cos(ax) = 0

Asen(ax) + Bcos(ax) = -1

the answer is a= 1 and a= -1, but i don't know how. Is the value of A, B and x important?

 

[oswald]

Hi,i found the solution in yahoo...

Take your function and differentiate it twice. observe that

f '' (x) = - a^2 f(x)

Substitute this into the equation. One possibility is that
A=B=0. Otherwise, cancel f(x) and you have a simple algebraic equation for a to solve.
have done:
f²(x)= -a²f(x)
so,
-a²f(x)+f(x)=0
f(x)[-a²+1]=0
a= 1 or -1!
i don't know why the teacher ask for Maclaurin series...

 

[Gib Z]

I don't know why you didn't just take my hint, which got at exactly the same thing, rather than go to Yahoo.