Derivative of a symetrical function

[quasar987]

If f(x) is symetrical, then

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

What would be a proof of that?

 

[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?

 

[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.

 

[quasar987]

Great, thank you Data.

 

[dextercioby]

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

Daniel.

 

[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.

 

[matt grime]

And odd functions are also symmetric, rotationally,

 

[cepheid]

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 180^o about the origin, gives an image the same as the original.