# Recurrence Relation_discrete

Let $f(x)=\frac 1 {(1-x)^2}$,find the coefficients $a_0, a_1, a_2, \ldots$ in the expansion $f(x)=\sum_{k=0}^\infty a_k x^k$

Solution is using theorem

Let $f(x)=\sum_{k=0}^\infty a_k x^k,\quad g(x)=\sum_{k=0}^\infty b_k x^k$

$$\Rightarrow f(x)g(x)=\sum_{k=0}^\infty \left ( \sum_{j=0}^k a_j b_{k-j} \right) x^k$$

And the solution is given as -: $$\frac 1 {(1-x)^2}=\sum_{k=0}^\infty \left ( \sum_{j=0}^k 1 \right )x^k = \sum_{k=0}^\infty (k+1) x^{k+1}$$

Not getting how to get 1in second summation..please explain ! -

• Hint: $\frac{1}{1-x} = \sum_{k=0}^{\infty} x^k$.
– dxiv
Oct 5 '16 at 23:19
• are you sure of the transcription of the given solution ?
– user354674
Oct 6 '16 at 0:28

You know that

$$\frac1{1-x}=\sum_{k\ge 0}x^k=\sum_{k\ge 0}\color{crimson}1\cdot x^k\;,$$

so that if we let $a_k$ be the coefficient of $x^k$ in this series, then $\color{crimson}{a_k=1}$ for all $k$. Now

\begin{align*} \frac1{(1-x)^2}&=\frac1{1-x}\cdot\frac1{1-x}\\ &=\sum_{k\ge 0}\left(\sum_{j=0}^k\color{crimson}{a_ja_{k-j}}\right)x^k\\ &=\sum_{k\ge 0}\left(\sum_{j=0}^k\color{crimson}1\cdot\color{crimson}1\right)x^k\\ &=\sum_{k\ge 0}\left(\sum_{j=0}^k1\right)x^k\\ &=\sum_{k\ge 0}(k+1)x^k\;. \end{align*}

• still not getting how to get $a_{j}\,\,a_{k-j}$=1.1 !! Oct 6 '16 at 17:47
• @sourav: $a_j$ is the coefficient of $x^j$ in the power series for the first factor of $\frac1{1-x}$, and $a_{k-j}$ is the coefficient of $x^{k-j}$ in the power series for the second factor of $\frac1{1-x}$. Each of those coefficients is $1$. Oct 6 '16 at 17:56
• best answer anyone can recieve for this question ! Oct 6 '16 at 17:59

$${\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}\quad {\text{ for }}|x|<1\!$$ differntiate it to get $${\frac {1}{(1-x)^{2}}}=\sum _{n=1}^{\infty }nx^{n-1}$$

• I think you mean "differentiate", not "derive" Oct 5 '16 at 23:22
• @Mosquite thanks, I have corrected it Oct 5 '16 at 23:24
• At the end, it should be $$\sum_{n=1}^{\infty}nx^{n-1}$$ not $$\sum_{n=0}^{\infty}nx^{n-1}$$
– Hrhm
Oct 5 '16 at 23:27
• @Hrhm there is no problem if you let $n=0$ or $n=1$ Oct 5 '16 at 23:31
• @E.H.E Why are you sorry? A moment ago, I thought that my comment was wrong too. (Now I see that if we let $n=0$, then the expression is undefined for $x=0$.) 😊
– Hrhm
Oct 6 '16 at 1:59