TargetIIT

1) find the sum of the series

[im]http://codecogs.izyba.com/gif.latex?%5Cfrac%7B1%7D%7B4%21%7D+%5Cfrac%7B4%21%7D%7B8%21%7D+%5Cfrac%7B8%21%7D%7B12%21%7D................%5Cinfty[/im]


2) if a function defined such that f:R→R

[im]http://codecogs.izyba.com/gif.latex?f%28x%29-2f%28%5Cfrac%7Bx%7D%7B2%7D%29+f%28%5Cfrac%7Bx%7D%7B4%7D%29%3Dx%5E%7B2%7D[/im]

find f(x)

1st one is a doubt[1]

i think finding general term first would be helpful..... for prob.1

actually i got teh general term.....but cudnt proceed further[2]

to get guys started on 2 - your target: [im]http://codecogs.izyba.com/gif.latex?f%28x%29%20%3D%20%5Cfrac%7B16x%5E2%7D%7B9%7D%20+%20c[/im]

second one: More Hint:

take f(x)-f(x/2)=g(x)


After that follow post 4 ;)

ok.....is anyone trying it.....or i shud post teh soln of it

Yes, give soln of the 2nd sum.

thats not the ans karna

The sum of the first series is
[im]http://codecogs.izyba.com/gif.latex?%5Cdfrac%7B6%5Cln%202-%5Cpi%7D%7B24%7D[/im]

how u got that sir?

and yes thats teh ans[1]

for the second one,

you should always observe that
f(x)-f(x/2)=g(x)
and the given expressoin is
g(x)-g(x/2)=x[p]2[/p]
From which you can conclude that g(x)=x[p]2[/p]+x[p]2[/p]/4+x[p]2[/p]/14... = 3/4x[p][p]2[/p][/p]+constant
here conclude somehow that constant = 0
then the same logic will give f(x) = 3/4 (g(x)) = 9.16x[p]2[/p]+c





For the second one after what lion has done.. try some trick that was recently discussed here ;)

[im]http://codecogs.izyba.com/gif.latex?%3D%5Csum_%7Bk%3D0%7D%5E%7B%5Cinfty%20%7D%5Cleft%28%20%7B%5Cfrac%7B1%7D%7B6%284k+1%29%7D%7D-%7B%5Cfrac%7B1%7D%7B2%284k+2%29%7D%7D+%7B%5Cfrac%7B1%7D%7B2%284k+3%29%7D%7D-%7B%5Cfrac%7B1%7D%7B6%284k+4%29%7D%7D%20%5Cright%29[/im]

now consider these definite integrals

[im]http://codecogs.izyba.com/gif.latex?%5Cint_%7B0%7D%5E%7B1%7D%7Bx%5E%7B4k%7D%7Ddx%3D%5Cfrac%7B1%7D%7B4k+1%7D[/im]

[im]http://codecogs.izyba.com/gif.latex?%5Cint_%7B0%7D%5E%7B1%7D%7Bx%5E%7B4k+1%7D%7Ddx%3D%5Cfrac%7B1%7D%7B4k+2%7D[/im]

[im]http://codecogs.izyba.com/gif.latex?%5Cint_%7B0%7D%5E%7B1%7D%7Bx%5E%7B4k+2%7D%7Ddx%3D%5Cfrac%7B1%7D%7B4k+3%7D[/im]

[im]http://codecogs.izyba.com/gif.latex?%5Cint_%7B0%7D%5E%7B1%7D%7Bx%5E%7B4k+3%7D%7Ddx%3D%5Cfrac%7B1%7D%7B4k+4%7D[/im]

now it can be done ;)

other than a slight issue of convergence where the individual sum is not finite.. you are on the right track ;)

ok i got it........ but eragon if u were typing that much u shud hav posted the full soln........ [im]http://codecogs.izyba.com/gif.latex?%5CRightarrow%20S%3D%5Csum_%7Bk%3D0%7D%5E%7B%5Cinfty%20%7D%7B%5Cint_%7B0%7D%5E%7B1%7D%7B%5Cleft%28%5Cfrac%7B1%7D%7B6%7Dx%5E%7B4k%7D-%20%5Cfrac%7B1%7D%7B2%7Dx%5E%7B4k+1%7D+%5Cfrac%7B1%7D%7B2%7Dx%5E%7B4k+2%7D-%5Cfrac%7B1%7D%7B6%7Dx%5E%7B4k+3%7D%5Cright%29%7D%7Ddx%3D%20%7B%5Cint_%7B0%7D%5E%7B1%7D%5Csum_%7Bk%3D0%7D%5E%7B%5Cinfty%20%7D%7B%5Cleft%28%5Cfrac%7B1%7D%7B6%7Dx%5E%7B4k%7D-%20%5Cfrac%7B1%7D%7B2%7Dx%5E%7B4k+1%7D+%5Cfrac%7B1%7D%7B2%7Dx%5E%7B4k+2%7D-%5Cfrac%7B1%7D%7B6%7Dx%5E%7B4k+3%7D%5Cright%29%7D%7Ddx%20%3D%5Cfrac%7B1%7D%7B6%7D%7B%5Cint_%7B0%7D%5E%7B1%7D%5Csum_%7Bk%3D0%7D%5E%7B%5Cinfty%20%7Dx%5E%7B4k%7D%281-3x+3x%5E%7B2%7D-x%5E%7B3%7D%29%7Ddx%3D%5Cfrac%7B1%7D%7B6%7D%5Cint_%7B0%7D%5E%7B1%7D%7B%5Cfrac%7B%281-x%29%5E%7B3%7D%7D%7B1-x%5E%7B4%7D%7D%7Ddx%3D%5Cfrac%7B1%7D%7B6%7D%5Cint_%7B0%7D%5E%7B1%7D%7B%5Cfrac%7B%281-x%29%5E%7B2%7D%7D%7B%281+x%29%281+x%5E%7B2%7D%29%7D%7Ddx%3D%5Cfrac%7B1%7D%7B6%7D%5Cint_%7B0%7D%5E%7B1%7D%5Cleft%28%5Cfrac%7B2%7D%7B1+x%7D-%5Cfrac%7Bx+1%7D%7B1+x%5E%7B2%7D%7D%20%5Cright%29dx[/im]

now it can be integrated

easier way to do 2nd......we clearly see dat f(x) has to be polynomial wid degree to...else x^2 couldnt have been isolated...so let f(x) be ax^2 + bx + c...diff eqn twice we get....f(0) = 0 f'(0) = 0 and f''(0) = 32/9....hence...f(x) = 16x^2/9...do comment...:)