1i think finding general term first would be helpful..... for prob.1
1) find the sum of the series
\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}................\infty
2) if a function defined such that f:R→R
f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^{2}
find f(x)
1st one is a doubt[1]
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15 Answers
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341to get guys started on 2 - your target: f(x) = \frac{16x^2}{9} + c
1second one: More Hint:
take f(x)-f(x/2)=g(x)
After that follow post 4 ;)
1well i got teh general term
\sum_{k=0}^{\infty }{\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}}
now wat to do after that
i thought of sperating it into four fractions by partial fractions.....but cudnt get
62for 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)=x2
From which you can conclude that g(x)=x2+x2/4+x2/14... = 3/4x2+constant
here conclude somehow that constant = 0
then the same logic will give f(x) = 3/4 (g(x)) = 9.16x2+c
For the second one after what lion has done.. try some trick that was recently discussed here ;)
21=\sum_{k=0}^{\infty }\left( {\frac{1}{6(4k+1)}}-{\frac{1}{2(4k+2)}}+{\frac{1}{2(4k+3)}}-{\frac{1}{6(4k+4)}} \right)
now consider these definite integrals
\int_{0}^{1}{x^{4k}}dx=\frac{1}{4k+1}
\int_{0}^{1}{x^{4k+1}}dx=\frac{1}{4k+2}
\int_{0}^{1}{x^{4k+2}}dx=\frac{1}{4k+3}
\int_{0}^{1}{x^{4k+3}}dx=\frac{1}{4k+4}
now it can be done ;)
62other than a slight issue of convergence where the individual sum is not finite.. you are on the right track ;)
1ok i got it........ but eragon if u were typing that much u shud hav posted the full soln........ \Rightarrow S=\sum_{k=0}^{\infty }{\int_{0}^{1}{\left(\frac{1}{6}x^{4k}- \frac{1}{2}x^{4k+1}+\frac{1}{2}x^{4k+2}-\frac{1}{6}x^{4k+3}\right)}}dx= {\int_{0}^{1}\sum_{k=0}^{\infty }{\left(\frac{1}{6}x^{4k}- \frac{1}{2}x^{4k+1}+\frac{1}{2}x^{4k+2}-\frac{1}{6}x^{4k+3}\right)}}dx =\frac{1}{6}{\int_{0}^{1}\sum_{k=0}^{\infty }x^{4k}(1-3x+3x^{2}-x^{3})}dx=\frac{1}{6}\int_{0}^{1}{\frac{(1-x)^{3}}{1-x^{4}}}dx=\frac{1}{6}\int_{0}^{1}{\frac{(1-x)^{2}}{(1+x)(1+x^{2})}}dx=\frac{1}{6}\int_{0}^{1}\left(\frac{2}{1+x}-\frac{x+1}{1+x^{2}} \right)dx
now it can be integrated
1easier 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...:)