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    I feel like I am missing the complete obvious here but how do I find A,B.C and D?


    1/(x^2+1)^4 = d/dx*(Ax^5+Bx^3+Cx)/(x^2+1)^3 + Dx^6/(x^2+1)^4

    I differentiated it and subbed in x = 1 but in the end all I'm left with is constants and no way of solving them......there must be a step i'm overlooking?

    THanks.
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    What is the actual question? Is it just to find A, B, C and D?
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    it might work ..but try integrating LHS instead of differentiating RHS ..seems much easier and then do all the rearranging stuff
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    (Original post by Chelle-belle)
    What is the actual question? Is it just to find A, B, C and D?
    well the first part of the question is "show that for 0<x<1 the largest value of x^6/(x^2+1)^4 is 1/16.

    This is the second part that I have given in my OP.

    btw, it's supposed to be greater than or equal to and not greater than!
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    (Original post by rbnphlp)
    it might work ..but try integrating LHS instead of differentiating RHS ..seems much easier and then do all the rearranging stuff
    How will that help?
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    (Original post by boromir9111)
    well the first part of the question is "show that for 0<x<1 the largest value of x^6/(x^2+1)^4 is 1/16.

    This is the second part that I have given in my OP.

    btw, it's supposed to be greater than or equal to and not greater than!
    ok in that case , your method is probably the safe bet ..just carefully rearrange everything and compare coefficients just like you would do in partial fractions ..(post your working if you wish)
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    Well after differentiating it and subbing in x = 1, I get

    1/16 = (4A - 6C + D)/16

    I can't see anyway of solving that?
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    (Original post by boromir9111)
    Well after differentiating it and subbing in x = 1, I get

    1/16 = (4A - 6C + D)/16

    I can't see anyway of solving that?
    ok ..Im not sure how you got that ..

    post you working ..
    before substituting in x , rearrange so that, say you get something like Ax^2+bx+C=1
    now compare coffecents , here you know Ax^2=0,bx=0 and C=1.
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    (Original post by boromir9111)
    Well after differentiating it and subbing in x = 1, I get

    1 = 4A - 6C + D

    I can't see anyway of solving that?
    Can you post the differentiated version so we don't have to do it lol
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    Lol. Seems like I have totally forgotten how to do partial fractions......thanks mate!
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    (Original post by Chelle-belle)
    Can you post the differentiated version so we don't have to do it lol
    I may be wrong on this but I get:

    ((Ax^5 + Bx^3 + Cx)*-6x)/(x^2+1)^2 + (5Ax^4 + 3Bx^2 + C)/(x^2+1)^3 +

    Dx^6/(x^2+1)^4
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    I'm assuming this is a STEP I question?
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    (Original post by boromir9111)
    Well after differentiating it and subbing in x = 1
    Not that I have a solution (sorry) but you subbed x=1 into

    \frac{-6x(Ax^5+Bx^3+Cx)}{(x^2+1)^2}+ \frac{5Ax^4+3Bx^2+C}{(x^2+1)^3}+ \frac{Dx^6}{(x^2+1)^4} = \frac{1}{16}
    ?
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    (Original post by Farhan.Hanif93)
    I'm assuming this is a STEP I question?
    Yeah....I kinda figured you would know that! I came across and it intrigued me and now it's annoying me!
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    (Original post by Chelle-belle)
    Not that I have a solution (sorry) but you subbed x=1 into

    \frac{-6x(Ax^5+Bx^3+Cx)}{(x^2+1)^2}+ \frac{5Ax^4+3Bx^2+C}{(x^2+1)^3}+ \frac{Dx^6}{(x^2+1)^4} = \frac{1}{16}
    ?
    :yes:
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    Write in latex please

    Never mind!!!
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    (Original post by boromir9111)
    :yes:
    From there, combine it into one fraction. Then equate the numerators of the LHS and RHS. Then expand and rearrange that expression to obtain a polynomial of degree 6. You should be able to find the value of C immediately from there. Then equate the coefficients of each power of x to what they are on the RHS to obtain a system of simultaneous equations (IIRC they should be zeroes). The values should follow from there easily. This is just an algebraic exercise, rather than a puzzling problem tbh. All you need to do is be careful with your algebra.
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    (Original post by Farhan.Hanif93)
    From there, combine it into one fraction. Then equate the numerators of the LHS and RHS. Then expand and rearrange that expression to obtain a polynomial of degree 6. You should be able to find the value of C immediately from there. Then equate the coefficients of each power of x to what they are on the RHS to obtain a system of simultaneous equations (IIRC they should be zeroes). The values should follow from there easily. This is just an algebraic exercise, rather than a puzzling problem tbh. All you need to do is be careful with your algebra.
    I was hoping I didn't have to do that lol.....but yeah, I get it now......thanks a lot buddy!
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    (Original post by boromir9111)
    :yes:
    Mmm could have been me (I tend to complete arithmetic stupidly by mistake) but try it again.
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    (Original post by Chelle-belle)
    Mmm could have been me (I tend to complete arithmetic stupidly by mistake) but try it again.
    Yeah, I know how to do it now....thanks anyway!
 
 
 
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