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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: 21,020 Thanks: 2256 
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: 21,020 Thanks: 2256 
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: 21,020 Thanks: 2256 
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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