My Math Forum Even and Odd Function

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 August 14th, 2015, 09:15 AM #1 Member   Joined: Apr 2015 From: Indonesia Posts: 53 Thanks: 2 Even and Odd Function Can someone help me to solve this? I don't know where to start. Prove that any function can be uniquely expressed as a sum of an even and an odd function. You should logically explain two things: 1. For any function f(x), there is an even function g(x) and an odd function h(x) such that f = g + h. 2. For any function f(x), there is not more than one expression f = g + h, where g is even and h is odd. Last edited by skipjack; August 16th, 2015 at 02:23 PM.
 August 14th, 2015, 09:35 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Use the Taylor series for the function. Thanks from deadweight
 August 14th, 2015, 11:23 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 But a general function, f, may not have a Taylor's series- and even if it does, that Taylor's series may not converge to or tell you anything about that function. Given any function, f, define g(x)= (f(x)+ f(-x))2 and h(x)= (f(x)- f(-x))/2. Suppose there were two sets of functions, g and h and g' and h' such that g and g' are even and h and h' are odd, and f(x)= g(x)+ h(x)= g'(x)+ h'(x). So, for example, g(a)+ h(a)= g'(a)+ h'(a). And, since g is even and h is odd, g(-a)+ h(-a)= g(a)- h(a)= g'(a)- h'(a). Adding that to the first equation, we get 2g(a)= 2g'(a) or g(a)= g'(a) for all a. And from that it follows that h(a)= h'(a) for all a. Thanks from CRGreathouse
 August 15th, 2015, 09:30 PM #4 Member   Joined: Apr 2015 From: Indonesia Posts: 53 Thanks: 2 Then there's another way to express f(x)? Because you can subtitute h(a) as h'(a). Can you explain more? I'm so confused
August 16th, 2015, 12:11 PM   #5
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Quote:
 Originally Posted by Country Boy Given any function, f, define g(x)= (f(x)+ f(-x))/2 and h(x)= (f(x)- f(-x))/2.
I don't understand how this answers the OP's question.

Suppose $\displaystyle f$ is an even function. Then

$\displaystyle g(x) = \dfrac{f(x)+f(-x)}{2} = f(x)$ and
$\displaystyle h(x) = \dfrac{f(x)-f(-x)}{2} = 0$.

So you have expressed the even function $\displaystyle f$ as a sum of two even functions.

 August 16th, 2015, 02:28 PM #6 Global Moderator   Joined: Dec 2006 Posts: 21,114 Thanks: 2329 The function h(x) is always odd. In the above case, it's even as well. Note that the second question is poorly worded, as it isn't clear what "more than one expression" means. The intended proof(s) require assuming information not given in the question. However, the question may have been posed in a context that supplies that extra information.
August 17th, 2015, 11:27 AM   #7
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Quote:
 Originally Posted by skipjack The function h(x) is always odd. In the above case, it's even as well. Note that the second question is poorly worded, as it isn't clear what "more than one expression" means. The intended proof(s) require assuming information not given in the question. However, the question may have been posed in a context that supplies that extra information.
I guess my confusion was a bit deeper than I thought. Given Country Boy's definitions for $\displaystyle g(x)$ and $\displaystyle h(x)$, doesn't this reduce to saying you can express $\displaystyle f(x)$ as $\displaystyle f(x) + 0$ which is even plus odd if $\displaystyle f$ is even or odd plus even if $\displaystyle f$ is odd.?

 August 18th, 2015, 07:34 AM #8 Global Moderator   Joined: Dec 2006 Posts: 21,114 Thanks: 2329 It's true that if f(x) is even, f(x) = f(x) + 0 could be regarded as equivalent to f(x) = g(x) + h(x) (because g(x) = f(x) and h(x) = 0). Unrelated to the above point, there is a significant assumption that is vital for this question, but not explicitly stated.
September 5th, 2015, 01:07 PM   #9
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Quote:
 Originally Posted by mrtwhs I don't understand how this answers the OP's question. Suppose $\displaystyle f$ is an even function. Then $\displaystyle g(x) = \dfrac{f(x)+f(-x)}{2} = f(x)$ and $\displaystyle h(x) = \dfrac{f(x)-f(-x)}{2} = 0$. So you have expressed the even function $\displaystyle f$ as a sum of two even functions.
The question was how to express any function as a sum of an even and odd function. My response told exactly how to do that. For any function, f, g is an even function, h is an odd function, and there sum is equal to f.

In your example, where f is already even, yes, f(x)= f(x)+ 0, the sum of an even function, f, and an odd function, 0. If f is already odd then f(x)= 0+ f(x), the sum of an even function, 0, and an odd function, f.

The point you are missing is that the 0 function, f(x)= 0 for all x, is both even and odd.

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