# Harmonic Potential

1. Nov 14, 2009

### SANGHERA.JAS

Can we expand harmonic potential in a Taylor series. If so then piz tell me how?

2. Nov 14, 2009

### f95toli

Yes, but it would be rather pointless since the expression for the potential is already identical to the second term in the taylor expansion.

The harmonic potential is as simple as it gets, I can't think of any reason why you would want to use a series expansion of any kind; at least not if you are trying to simplify the calculation.

3. Nov 14, 2009

### SANGHERA.JAS

4. Nov 14, 2009

### f95toli

But the third term in that series IS the harmonic potential....
For an harmonic potential all the other terms are -by definition- zero.

5. Nov 14, 2009

### SANGHERA.JAS

You are somewhat right my friend. But my actual question is: From where the series itself come about?Although I understand Taylor's theorem well, but I am still not getting it?

6. Nov 14, 2009

### f95toli

Now, that is a complettely different question.
Although I am still not quite sure exactly what it is you do not understand.

Do you understand how one can use a Taylor series to approximate any differentiable function (such as an exponential) around some point? I.e do you understand the math?

Have you for example looked a the wiki for Taylor's theorem?
http://en.wikipedia.org/wiki/Taylor's_theorem

The reason why the theorem is so useful in physics is that it allows us to expand a complicated potential (a common example would be a potential that contains trigonometric functions) around some point we are interested in.
For many potentials of interest you can get a good approximation by truncating the series after the third term, meaning all you end up with is the harmonic potential which is easy to deal with.

7. Nov 14, 2009

### SANGHERA.JAS

My friend I am a second year bachelor student; so I understand mathematics and hence Taylor's theorem very well. But my question is we use Taylor's expansion in case of function which are explicitly given e.g. sine, cosine, exponential, {1/(1-x^2)} etc. But how can we use Taylor's theorem in case of functions which are not explicitly given; that is the case when we talk about potential function, In this case neither the function itself nor the exact dependence on any sort of variable is given.