Taylor Expansion: Do Assumptions Apply?

Niles
Messages
1,834
Reaction score
0

Homework Statement


Hi

Say I want to Taylor-expand
<br /> f(\omega + m\sin(\Omega t))<br />
where ω and Ω are frequencies, m is some constant and t denotes time. Then I would get
<br /> f(\omega + m\sin(\Omega t)) = f(\omega) + (m\sin(\Omega t)\frac{dI}{d\omega} + \ldots<br />
Is it necessary to make any assumptions on the sizes of m and Ω in order to make the above expansion?


Niles.
 
Physics news on Phys.org
If you are Taylor expanding in the quantity you are adding to ω and I means f(ω) that's fine. If you are going to truncate the expansion there and wondering if it's a accurate expansion that's going to depend on the size of the quantity you are adding to ω and the behaviour of f''(ω). You'll want to look at Taylor series remainder terms if you are concerned about how good it is.
 
Thanks!Niles.
 
Niles said:
Actually, isn't this an expansion around the point ω rather than msin(Ωt) if we use the definition here http://en.wikipedia.org/wiki/Taylor_series#Definition?

Sure it is. It's 'around' ω. I say the expansion is 'in' msin(Ωt) because that's the thing that appears in all the powers. Just terminology.
 
Ah, I see. Thanks for clarifying.Niles.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top