- #1
- 4,796
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If f(x) is symetrical, then
[tex]f^{(n)}(x) =0, \ \ \ n=1,3,5,...[/tex]
What would be a proof of that?
[tex]f^{(n)}(x) =0, \ \ \ n=1,3,5,...[/tex]
What would be a proof of that?
Derivative of a symetrical function
- Context: Graduate
- Thread starter quasar987
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- Derivative Function
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Discussion Overview
The discussion revolves around the properties of symmetrical (even) functions and their derivatives, particularly focusing on the behavior of odd derivatives and the implications of symmetry in the context of Taylor series expansions. The scope includes theoretical exploration and mathematical reasoning.
Discussion Character
- Exploratory, Technical explanation, Debate/contested, Mathematical reasoning
Main Points Raised
- One participant asserts that if f(x) is symmetrical, then all odd derivatives f^{(n)}(x) for n=1,3,5,... are zero, seeking a proof for this claim.
- Another participant challenges this by suggesting that if f(x) is symmetrical and C^{\infty}, then the odd terms in the Taylor expansion vanish, also requesting a proof.
- A different participant proposes that if the coefficients of the odd terms are not all zero, one can analyze f(-x) and use the equality of coefficients in Taylor series to argue against the initial claim.
- One participant points out the terminology confusion, clarifying that "even" is the correct term for functions symmetric about the y-axis, while "symmetric" is used more generally.
- Another participant notes that any odd derivative of an even function is an odd function, and mentions that for odd functions, f(0) equals zero.
- Discussion includes a mention of odd functions being symmetric in a rotational sense, with a participant reflecting on the meaning of rotational symmetry and providing an example of the graph of an odd function.
Areas of Agreement / Disagreement
Participants express differing views on the definitions and implications of symmetry in functions, particularly regarding the terminology of "symmetric" versus "even." There is no consensus on the initial claims about the derivatives of symmetrical functions, and multiple competing views remain regarding the nature of odd and even functions.
Contextual Notes
There are unresolved assumptions about the definitions of symmetry and the conditions under which the Taylor series applies. The discussion does not clarify the implications of these definitions on the claims made.
- #2
- 4,796
- 32
ummm, no that doesn't make sense, does it. How about
If f(x) is symetrical and [itex]C^{\infty}[/itex], then the terms of odd powers in the Taylor expansion vanish.
What would be a proof of that?
If f(x) is symetrical and [itex]C^{\infty}[/itex], then the terms of odd powers in the Taylor expansion vanish.
What would be a proof of that?
Last edited:
- #3
Data
- 998
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Suppose that the coefficients of the odd terms are not all zero. Look at [itex]f(-x)[/itex] and use the fact that (convergent) Taylor series are equal in an interval iff the coefficients are equal.
Last edited:
- #4
- 4,796
- 32
Great, thank you Data.
- #5
dextercioby
Science Advisor
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So you meant "even" instead of "symmetric"...The function is "even",its graph is "symmetric" wrt the vertically chosen Oy axis...
Daniel.
Daniel.
- #6
HallsofIvy
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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
matt grime
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And odd functions are also symmetric, rotationally,
- #8
cepheid
Staff Emeritus
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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.
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