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Maths.

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.

Reply 1

Chelle-belle

What is the actual question? Is it just to find A, B, C and D?

Reply 2

rbnphlp

it might work ..but try integrating LHS instead of differentiating RHS ..seems much easier and then do all the rearranging stuff

Reply 3

boromir9111

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!

Reply 4

boromir9111

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?

Reply 5

rbnphlp

Original post by boromir9111

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)

Reply 6

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?

Reply 7

rbnphlp

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.

Reply 8

Chelle-belle

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 :colondollar:

Reply 9

boromir9111

Lol. Seems like I have totally forgotten how to do partial fractions......thanks mate!

Reply 10

boromir9111

Original post by Chelle-belle

Can you post the differentiated version so we don't have to do it lol :colondollar:

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

Reply 11

Farhan.Hanif93

I'm assuming this is a STEP I question?

Reply 12

Chelle-belle

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}

Reply 13

boromir9111

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!

Reply 14

boromir9111

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}

Reply 15

Slowbro93

Write in latex please

Never mind!!!

Reply 16

Farhan.Hanif93

Original post by boromir9111

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.