A simple problem

Calculus Level 4

Denote $\displaystyle f(x) = \sum _{ n=1 }^{ \infty }{ { T }_{ n }{ x }^{ n } }$ where domain of $f(x)$ are all values of $x$ where the summation converges, and ${T}_{n}$ is the $n^{\text{th}}$ term of some sequence.

We define ${S}_{n} = {T}_{1}+{T}_{2}+ .... +T_{n}$

And $\displaystyle g(x) = \sum _{ n=1 }^{ \infty }{ { S }_{ n }{ x }^{ n } }$ with domain of $g(x)$ are all values where this summation converges.

Lastly, denote $h(x) = \dfrac { f(x) }{ g(x) }$ , domain of $h(x)$ is the intersection of domain of $f(x),g(x)$ and values where $g(x) \neq 0$

Then find $x+h(x)$ in its domain.

The answer is 1.

2 solutions

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Peter Macgregor
Mar 8, 2015

Using the definition of $g(x)$ and of $S_{n}$ we can write

$g(x)=(T_{1}x^1)+(T_{1}+T_{2})x^2+(T_{1}+T_{2}+T_{3})x^3+(T_{1}+T_{2}+T_{3}+T_{4})x^4+\dots$

Rearrange this expression by collecting the last terms in each bracket, then the second last terms, then then third last and so on to get

$g(x)=(T_{1}x^1+T_{2}x^2+\dots)+x(T_{1}x^1+T_{2}x^2+\dots)+x^2(T_{1}x^1+T_{2}x^2+\dots)+\dots$

$\implies g(x)=f(x)(1+x+x^2+\dots)=\dfrac{f(x)}{1-x}$

$\implies\dfrac{f(x)}{g(x)}=1-x$

and so

$x+h(x)=x+1-x=\boxed{1}$

Mvs Saketh
Mar 7, 2015

$\displaystyle \boxed {h(x) = 1-x}$

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This solution lacks a complete explanation.

Reason please..

Sachin Arora - 6 years, 3 months ago

It is as simple as that.

Ronak Agarwal - 6 years, 3 months ago

Really liked the question

Rinkumoni Khanikar - 6 years, 3 months ago

A simple problem

The answer is 1.

2 solutions

Discussions for this problem are now closed

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