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Math Help - Even and odd functions.

  1. #1
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    Even and odd functions.

    I am going to try to traduce the problem in english. :/

    Demonstrate that any function f(x), in interval  -l<x<l , can be represented as the sum of even and odd functions.

    Solution is: f(x)= \frac{1}{2} [f(x)+f(-x)]+ \frac{1}{2} [f(x)-f(-x)]

    Why is that solution?
    We just need to show that f(x) = f(x)?
    Last edited by Fabio010; October 22nd 2011 at 06:04 AM.
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  2. #2
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    Re: Even and odd functions.

    of course, the two sides are equal.

    the point is, if we define:

    g(x) = \frac{f(x) + f(-x)}{2},\ h(x) = \frac{f(x) - f(-x)}{2}

    then f(x) = g(x) + h(x), as you observed, but the "real interesting" part is:

    g(-x) = \frac{f(-x) + f(-(-x))}{2} = \frac{f(-x) + f(x)}{2} = g(x) so g is even, and:

    h(-x) = \frac{f(-x) - f(-(-x))}{2} = \frac{f(-x) - f(x)}{2} = -h(x) so h is odd.
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  3. #3
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    Re: Even and odd functions.

    So the functions can be expressed with the sum of even and odd functions. as problem say.

    Thanks for the explanation.
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  4. #4
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    Re: Even and odd functions.

    Quote Originally Posted by Fabio010 View Post
    I am going to try to traduce the problem in English. :/

    Demonstrate that any function f(x), in interval  -l<x<l , can be represented as the sum of even and odd functions.

    Solution is: f(x)=\frac{1}{2}[f(x)+(f-x)]+\frac{1}{2}[f(x)-(f-x)]

    Why is that solution?
    We just need to show that f(x) = f(x)?
    Comment #1:

    You have a mistake in your solution, not caused but translation into English; what is known as a "typo" -- a typographical error.

    The solution should be:
    f(x)=\frac{1}{2}[f(x)+f(-x)] \frac{1}{2}[f(x)-f(-x)]
    Comment #2.

    The purpose of the restriction regarding the interval,  -l<x<l\,, is not at all clear to me. If by any chance the functions are supposed to be even or odd with respect to the center of the interval,  -l<x<l\,, then the functions, g(x) and h(x) mentioned by Deveno, should also include a coordinate translation.

    Added in Edit:

    The above equation is missing a plus sign. It should be:
    f(x)=\frac{1}{2}[f(x)+f(-x)]+ \frac{1}{2}[f(x)-f(-x)]
    Last edited by SammyS; October 22nd 2011 at 08:55 AM.
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  5. #5
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    Re: Even and odd functions.

    Quote Originally Posted by SammyS View Post
    Comment #1:

    You have a mistake in your solution, not caused but translation into English; what is known as a "typo" -- a typographical error.

    The solution should be:
    f(x)=\frac{1}{2}[f(x)+f(-x)] \frac{1}{2}[f(x)-f(-x)]


    Comment #2.

    The purpose of the restriction regarding the interval,  -l<x<l\,, is not at all clear to me. If by any chance the functions are supposed to be even or odd with respect to the center of the interval,  -l<x<l\,, then the functions, g(x) and h(x) mentioned by Deveno, should also include a coordinate translation.
    1) the center of the interval \ (-l,l) is 0.

    2) the product you give is not equal to f. to see this, suppose we have f(x) = x (a perfectly ordinary function). then:

    \frac{1}{2}[f(x)+f(-x)]\frac{1}{2}[f(x) - f(-x)] = \frac{1}{4}(x + (-x))(x - (-x))

    = \frac{1}{4}(0)(2x) = 0 which is certainly NOT equal to f(x) = x.
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  6. #6
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    Re: Even and odd functions.

    Deveno explained me the solution correctly.
    Topic is Solved. Thanks for the help.
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  7. #7
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    Re: Even and odd functions.

    Quote Originally Posted by Deveno View Post
    1) the center of the interval \ (-l,l) is 0.

    2) the product you give is not equal to f. to see this, suppose we have f(x) = x (a perfectly ordinary function). then:

    \frac{1}{2}[f(x)+f(-x)]\frac{1}{2}[f(x) - f(-x)] = \frac{1}{4}(x + (-x))(x - (-x))

    = \frac{1}{4}(0)(2x) = 0 which is certainly NOT equal to f(x) = x.
    I had a typo, correcting Fabio's typo is all.

    Before Fabio edited his Original Post, he ha a couple of typos (See this quote in post #3). Then I inadvertently dropped the + sign between the two functions. I did a lot of cutting & pasting, because there are a lot of [​TEX​] [​/TEX​] tags in his equations.
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  8. #8
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    Re: Even and odd functions.

    Yes, i wrote (f-x) instead of f(-x)
    but i corrected before your post.
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