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Math Help - Finding Variations

  1. #1
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    Finding Variations

    Find V(x- x^2, [-1,2]) and V(cos(4(pie)x),[0,1]).

    V(x- x^2, [-1,2])= V(x- x^2, [-1,0])+V(x- x^2, [0,1])+V(x- x^2, [1,2])=
    0+0+2=2

    V(cos(4(pie)x),[0,1])= 1+1=2

    I was a little confused about finding the variation of a monotone function but can anyone please tell me if I found the total variations for both functions or did i go wrong at some point?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by summerset353 View Post
    Find V(x- x^2, [-1,2]) and V(cos(4(pie)x),[0,1]).

    V(x- x^2, [-1,2])= V(x- x^2, [-1,0])+V(x- x^2, [0,1])+V(x- x^2, [1,2])=
    0+0+2=2

    V(cos(4(pie)x),[0,1])= 1+1=2

    I was a little confused about finding the variation of a monotone function but can anyone please tell me if I found the total variations for both functions or did i go wrong at some point?
    Is this bounded variation?
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  3. #3
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    yes it is bounded variation
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by summerset353 View Post
    yes it is bounded variation
    Ok, how have you come to your conclusions? Show exactly how you got from step to step. We will help fill in/correct any of them.
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  5. #5
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    V(x- x^2, [-1,2])= V(x- x^2, [-1,0])+V(x- x^2, [0,1])+V(x- x^2, [1,2])=


    V(x- x^2, [-1,0])=(-1 - (-1)^2) + (0- 0^2)= (-1-1)+0= -2
    V(x- x^2, [0,1])= (0- 0^2) + (1 - 1^2) = 0 + 0 = 0
    V(x- x^2, [1,2])= (1 - (1^2) + (2- 2^2)= 0+(2-4) = -2

    -2+0+-2 = -4

    So, now I got different answers. I dont know what I am doing wrong.
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  6. #6
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    V(cos(4(pie)x),[0,1])= cos(4(pie)0) +cos(4(pie)1)= 1+1=2
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