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 October 18th, 2008, 05:18 AM #1 Newbie   Joined: Sep 2008 Posts: 13 Thanks: 0 Even and Odd Functions How do you write f(x)=1/(x+1) as the sum of an even and an odd function? And can all functions be written as the sum of an even and an odd function? Thnx.
 October 18th, 2008, 07:41 AM #2 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: Even and Odd Functions Mathematically, we say that a function f(x) is even if f(x)=f(-x) and is odd if f(-x)=-f(x) See if you can apply that principle to your problem.
 October 18th, 2008, 03:59 PM #3 Newbie   Joined: Sep 2008 Posts: 13 Thanks: 0 Re: Even and Odd Functions I know the definition of even and odd functions, but I can't seem to find a consistent way to write functions as the sum of an even and an odd. Is there anything that I seem to forget? or miss?
 October 18th, 2008, 04:27 PM #4 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: Even and Odd Functions Ok, just mulling around: Suppose f(x) can be written as h(x) + g(x) where h(x) is even, and g(x) is odd. Then f(x) = h(x) + g(x) = h(-x) - g(x) Then g(x) = [h(-x) - h(x)]/2 But h(x) = h(-x), so g(x) = [h(-x) - h(-x)]/2 = 0
 October 18th, 2008, 11:21 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,805 Thanks: 2150 Let's get it right. Suppose f(x) can be written as h(x) + g(x) where h(x) is even, and g(x) is odd, then f(x) = h(x) + g(x) and f(-x) = h(-x) + g(-x), where h(x) = h(-x) and g(x) = -g(-x). Hence h(x) = (h(x) + h(-x))/2 + (g(x) + g(-x))/2 = (f(x) + f(-x))/2 and g(x) = (h(x) - h(-x))/2 + (g(x) - g(-x))/2 = (f(x) - f(-x))/2.
 October 19th, 2008, 05:10 AM #6 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: Even and Odd Functions Sigh. It must be a sign. You are absolutely right about that, of course.
 October 19th, 2008, 06:57 AM #7 Newbie   Joined: Sep 2008 Posts: 13 Thanks: 0 Re: Even and Odd Functions Thanks, so you just add the two equations and you get f(x), but when I read up to h(x)= (h(x)+h(-x))/2 + (g(x)+g(-x))/2 = (f(x)+f(-x))/2, I was baffled. I know in the end, the whole process seems to work, but is there a logical explanation or approach to get to that step? or is it just the way it is?
 October 19th, 2008, 03:14 PM #8 Global Moderator   Joined: Dec 2006 Posts: 20,805 Thanks: 2150 By assumption, h(x) = h(-x), so (h(x) + h(-x))/2 = 2h(x)/2 = h(x). By assumption, g(x) = -g(-x), so (g(x) + g(-x))/2 = 0. Adding these equations gives h(x) = (h(x) + h(-x))/2 + (g(x) + g(-x))/2. Still baffled?
 October 24th, 2008, 02:22 PM #9 Newbie   Joined: Sep 2008 Posts: 13 Thanks: 0 Re: Even and Odd Functions I understand each one of the steps, but what I still don't get is how are you supposed to know that such and such equals to this, and such and such equals to that? When I first looked at it, I was thinking to myself: what a redundant way of writing a simple expression. But later I realized that it isn't redundant, and the equation is needed. So I was wondering if the two equations that you listed are similar to trig identities: are they something you NEED to know? Because I don't know know what your interpretation of the word "assumption". Thnx
 October 25th, 2008, 08:32 AM #10 Global Moderator   Joined: Dec 2006 Posts: 20,805 Thanks: 2150 The equation h(x) ? h(-x) holds because that is what is meant by "h(x) is even". Suppose h(x) is defined as (f(x) + f(-x))/2 for all values of x. It follows that h(-x) = (f(-x) + f(-(-x)))/2 = (f(-x) + f(x))/2, which is the same as h(x). Hence defining h(x) as (f(x) + f(-x))/2 makes it an even function. Does that help?

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