Posted by manny on Thursday, September 13, 2007 at 1:27pm.
A function f is odd iff
f(-x) = - f(x)
Put h(x) = f(x) + g(x) and calculate
h(-x):
h(-x) = f(-x) + g(-x) =
-f(x) - g(x) =
-[f(x) + g(x)] =
-h(x)
So, we see that h is odd because h(-x) = -h(x)
Now put h(x) = g[f(x)]
h(-x) = g[f(-x)] =
g[-f(x)] =
-g[f(x)] =
-h(x)
And we see that h is odd.
One more exercise you could do:
If f(x) is an arbitrary function show that it can be decomposed uniquely as:
f(x) = f_even(x) + f_odd(x)
where f_even and f_odd are even and odd functions, respectively. Give the expressions for these functions in terms of the function f.
got it.
Thank you so much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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