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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? Tags functions, odd Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post sheran Algebra 4 July 16th, 2013 07:37 AM johnny Calculus 2 May 11th, 2011 09:41 PM Mathmen Algebra 1 November 10th, 2009 06:58 AM you_of_eh Algebra 4 September 21st, 2009 10:44 AM haddad287 Calculus 0 September 29th, 2008 05:37 PM

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