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Math Help - proving the function is even

  1. #1
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    proving the function is even

    Let f(x) be a function. prove that :

    (a) g(x) =1/2[f(x)+f(-x)] is an even function

    (b) h(x) = 1/2[f(x)-f(-x)] is an odd function

    (c) hence show that every function can be written as a sum of an even function and odd function...

    my approaches towards the question :

    i know that even function will yield f(-x) = f(x)

    and odd function will yield f(-x) = -f(x)

    so for
    (a) g(-x) = 1/2[f(-x) + f(x)] = 0

    (b) h(-x) = 1/2[f(-x)-f(x)] = 1/2[-2f(x)] = -f(x)

    (c) i dunno...

    plz help me .... thanx
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  2. #2
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    Quote Originally Posted by bobey View Post
    Let f(x) be a function. prove that :

    (a) g(x) =1/2[f(x)+f(-x)] is an even function

    (b) h(x) = 1/2[f(x)-f(-x)] is an odd function

    (c) hence show that every function can be written as a sum of an even function and odd function...

    my approaches towards the question :

    i know that even function will yield f(-x) = f(x)

    and odd function will yield f(-x) = -f(x)

    so for
    (a) g(-x) = 1/2[f(-x) + f(x)] = 0
    Where in the world did "= 0" come from?
    g(x)= (1/2)[f(x)+ f(-x)] so g(-x)= (1/2)[f(-x)+ f(x)] and, since addition is commutative, that is the same as g(-x)= (1/2)[f(x)+ f(-x)]. What does that tell you?

    (b) h(-x) = 1/2[f(-x)-f(x)] = 1/2[-2f(x)] = -f(x)
    Now, how did f(-x)- f(x) become "-2f(x)"?? It is NOT given that f(-x)= -f(x). That is, you are NOT told that f(x) is itself an odd function.

    h(x)= (1/2)[f(x)- f(-x)] so h(-x)= (1/2)[f(-x)- f(x)]= (1/2)[-(f(x)- f(-x))]= -(1/2)[f(x)- f(-x)].

    (c) i dunno...
    What is g(x)+ h(x)?

    plz help me .... thanx
    These are, by the way, refered to as the even and odd "parts" of f(x).

    For example, if f(x)= e^x then g(x)= (1/2)[e^x- e^{-x}]= cosh(x) and h(x)= (1/2)[e^x- e^{-x}]= sinh(x). cosh(x) and sinh(x) are the "even and odd parts" of e^x
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Where in the world did "= 0" come from?
    g(x)= (1/2)[f(x)+ f(-x)] so g(-x)= (1/2)[f(-x)+ f(x)] and, since addition is commutative, that is the same as g(-x)= (1/2)[f(x)+ f(-x)]. What does that tell you?


    Now, how did f(-x)- f(x) become "-2f(x)"?? It is NOT given that f(-x)= -f(x). That is, you are NOT told that f(x) is itself an odd function.

    h(x)= (1/2)[f(x)- f(-x)] so h(-x)= (1/2)[f(-x)- f(x)]= (1/2)[-(f(x)- f(-x))]= -(1/2)[f(x)- f(-x)].


    What is g(x)+ h(x)?



    These are, by the way, refered to as the even and odd "parts" of f(x).

    For example, if f(x)= e^x then g(x)= (1/2)[e^x- e^{-x}]= cosh(x) and h(x)= (1/2)[e^x- e^{-x}]= sinh(x). cosh(x) and sinh(x) are the "even and odd parts" of e^x
    from the book i referred to, it states that

    sinh(x)= (1/2)[e^x- e^{-x}] and cosh(x)= (1/2)[e^x+e^{-x}], why u said both of function g(x)=h(x) = (1/2)[e^x- e^{-x}]..

    thus f(x) = g(x) + h(x) ==> [tex](1/2)[e^x- e^{-x}] +(1/2)[e^x+e^{-x}] = e^x/math]... is it true????
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