(New page: a) g(x)+h(x)=0 g(x) even h(x) odd g is both even and odd g(x)=g(-x)=-g(x) b) f(x)=f$_{e}$(x)+f$_{0}$(x) f(-x)=f$_{e}$(-x)+f$_{0}$(-x)=f$_{e}$(x)-f$_{0}$(x) solve for f$_{e}$(x) and f$_{0}$...) |
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a) | a) | ||
| − | g(x)+h(x)=0 | + | |
| − | g(x) even h(x) odd | + | <math>g(x)+h(x)=0</math> |
| + | |||
| + | <math>g(x)</math> even <math>h(x)</math> odd | ||
| + | |||
g is both even and odd | g is both even and odd | ||
| − | g(x)=g(-x)=-g(x) | + | |
| + | <math>g(x)=g(-x)=-g(x)</math> | ||
| + | |||
b) | b) | ||
| − | f(x)= | + | |
| − | f(-x)= | + | <math>f(x)=f_{e}(x)+f_{0}(x)</math> |
| − | solve for | + | |
| − | + | <math>f(-x)=f_{e}(-x)+f_{0}(-x)=f_{e}(x)-f_{0}(x)</math> | |
| − | + | ||
| + | solve for <math>f_{e}(x)</math> and <math>f_{0}(x)</math> | ||
| + | |||
| + | <math>f_{e}(x)= (f(x)+f(-x))/2</math> | ||
| + | |||
| + | <math>f_{0}(x)= (f(x)-f(-x))/2</math> | ||
Latest revision as of 08:43, 6 October 2008
a)
$ g(x)+h(x)=0 $
$ g(x) $ even $ h(x) $ odd
g is both even and odd
$ g(x)=g(-x)=-g(x) $
b)
$ f(x)=f_{e}(x)+f_{0}(x) $
$ f(-x)=f_{e}(-x)+f_{0}(-x)=f_{e}(x)-f_{0}(x) $
solve for $ f_{e}(x) $ and $ f_{0}(x) $
$ f_{e}(x)= (f(x)+f(-x))/2 $
$ f_{0}(x)= (f(x)-f(-x))/2 $
