# Derivative of a symetrical function

1. May 5, 2005

### quasar987

If f(x) is symetrical, then

$$f^{(n)}(x) =0, \ \ \ n=1,3,5,...$$

What would be a proof of that?

2. May 5, 2005

### quasar987

ummm, no that doesn't make sense, does it. How about

If f(x) is symetrical and $C^{\infty}$, then the terms of odd powers in the Taylor expansion vanish.

What would be a proof of that?

Last edited: May 5, 2005
3. May 6, 2005

### Data

Suppose that the coefficients of the odd terms are not all zero. Look at $f(-x)$ and use the fact that (convergent) Taylor series are equal in an interval iff the coefficients are equal.

Last edited: May 6, 2005
4. May 6, 2005

### quasar987

Great, thank you Data.

5. May 6, 2005

### dextercioby

So you meant "even" instead of "symmetric"...The function is "even",its graph is "symmetric" wrt the vertically chosen Oy axis...

Daniel.

6. May 6, 2005

### HallsofIvy

One can also show that any odd derivative (1, 3, 5, etc.) of an even function is an odd function. Of course, for any odd function, f, f(0)= 0.

7. May 6, 2005

### matt grime

And odd functions are also symmetric, rotationally,

8. May 6, 2005

### cepheid

Staff Emeritus
Yeah, what does rotational symmetry, or symmetry about the origin mean again? Thinking of graphs of odd functions I can remember (e.g. y = x3), the only thing I can think of is that it means rotating the graph of the function 180o about the origin, gives an image the same as the original.