       Re: Assuming odd/even functions

• To: mathgroup at smc.vnet.net
• Subject: [mg98670] Re: [mg98634] Assuming odd/even functions
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Thu, 16 Apr 2009 04:12:06 -0400 (EDT)

```Use UpValues

Clear[f];

f /: f[a_] + f[-a_] = 0 ;

f[x] + f[-x]

0

f[-x] + f[x]

0

This isn't robust

f[x^2 - 3 x + 2] + f[-x^2 + 3 x - 2]

f[-x^2 + 3*x - 2] + f[x^2 - 3*x + 2]

However, you can help it along

Simplify[
f[x^2 - 3 x + 2] + f[-x^2 + 3 x - 2],
x^2 - 3 x + 2 == t]

0

f /: Integrate[f[x_], {x_, -a_, a_}] = 0;

Integrate[f[y], {y, -z, z}]

0

However,

Integrate[f[y], {y, z, -z}]

Integrate[f[y], {y, z, -z}]

f /: Integrate[f[x_], {x_, a_, -a_}] = 0;

Integrate[f[y], {y, z, -z}]

0

Again, not robust

Integrate[f[y], {y, x^2 - 3, -x^2 + 3}]

Integrate[f[y], {y, x^2 - 3,
3 - x^2}]

But as before

Simplify[
Integrate[f[y], {y, x^2 - 3, -x^2 + 3}],
x^2 - 3 == t]

0

Bob Hanlon

---- "Martin Sch=C3=B6necker" <ms_usenet at gmx.de> wrote:

=============
How can we give Mathematica the information of a function being odd or even=
, and have it applied automatically or in a Simplify[]?

Such that, if e.g. f[y] is odd, the sum f(a)+f(-a) yields zero, the
integral Integrate[f[y], {y, -a, a}] vanishes?

--Martin

```

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