27 votes 27 votes Let $G(x) = \frac{1}{(1-x)^2} = \sum\limits_{i=0}^\infty g(i)x^i$, where $|x| < 1$. What is $g(i)$? $i$ $i+1$ $2i$ $2^i$ Combinatory gatecse-2005 normal generating-functions + – gatecse asked Sep 21, 2014 gatecse 8.1k views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments thor commented Jan 7, 2017 reply Follow Share https://gateoverflow.in/88476/generating-function 0 votes 0 votes ankitgupta.1729 commented Jan 28, 2020 reply Follow Share @Arjun sir has mentioned the correct link. We can use the Taylor series also. We can write the Taylor series about origin for the given generating function as : $f(x) = a_{0} x^{0} + a_{1} x^{1} +a_{2} x^{2}+a_3 x^3+........= \sum_{n}^{} \frac{f^{(n)}(0)}{n!}x^n$ for $n=0,1,2,3,....$ So, for any ordinary generating function, sequence is : $ a_n= \frac{f^{(n)}(0)}{n!},$ for $n=0,1,2,3,.........$ Here, $f^{(n)}$ means $n^{th}$ derivative of $f$. Now here, generating function $f(x)=\frac{1}{(1-x)^2}$ and $ g(i)= \frac{f^{(i)}(0)}{i!},$ for $i=0,1,2,3,.........$ Now, $f(x)= \frac{1}{(1-x)^2}$ $f'(x)= \frac{1\times2}{(1-x)^3}$ $f''(x)= \frac{1\times 2\times 3}{(1-x)^4}$ $f'''(x)= \frac{1\times 2\times 3 \times 4}{(1-x)^5}$ $.......$ $f^{(i)}(x)= \frac{1\times 2\times 3 \times .......\times (i+1)}{(1-x)^{i+2}}$ $\Rightarrow f^{(i)}(0)= (i+1)!$ So, $g(i) = \frac{f^{(i)}(0)}{i!} = \frac{(i+1)!}{i!} = i+1$ 3 votes 3 votes Deepak Poonia commented Jun 21, 2023 reply Follow Share $\color{red}{\text{Detailed Video Solution:}}$ GATE 2005 Generating Function, Click HERE. $\color{red}{\text{Generating Function Complete Playlist,}}$ ALL GATE Questions, Extended Binomial Theorem: https://youtube.com/playlist?list=PLIPZ2_p3RNHiu4mkROrhREYsslvp2lL5l Knowing the “Extended Binomial Theorem” makes Generating Function topic extremely easy. Watch the above playlist to learn everything, with Proof & Variations. 1 votes 1 votes Please log in or register to add a comment.
2 votes 2 votes $G(x)=\frac{1}{(1-x)^2} = (1-x)^{-2}$ Now we can expand $(1-x)^{-2}$ using negative binomial series theorem $(1-x)^{-2}$ $= 1 +2x + \left ( \frac{1}{1*2}\ 2*3\ \right )x^2+ \left ( \frac{1}{1*2*3}\ 2*3*4\ \right )x^3+ \left ( \frac{1}{1*2*3*4}\ 2*3*4*5\ \right )x^4+...$ $= 1x^0+ 2x^1 +3x^2 + 4x^3+...$ We can clearly see that $g(i)$ is $i+1$ $\therefore$ Option $B.$ is correct answer. Satbir answered Jun 25, 2019 Satbir comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Lets solve this using backtracking With practice you will learn how to get G(x) shashankrustagi answered Nov 27, 2020 shashankrustagi comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes B is the correct option. Let us put values S = 1 + 2x + 3x2 + 4x3 + .......... Sx = x + 2x2 + 3x3 + .......... S - Sx = 1 + x + x2 + x3 + .... S - Sx = 1/(1 - x) [sum of infinite GP series with ratio < 1 is a/(1-r)] S = 1/(1 - x)2 Regina Phalange answered Apr 15, 2017 Regina Phalange comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Simple Trial and Error Approach, Let x=0, Taking LHS, $\frac{1}{(1-0)^2} => 1$ Taking RHS, $\sum g(i)x^i => g(0)x^0 + g(1)x^1 + ...+ g(n)x^n => g(0) + 0 + ... + 0$ LHS=RHS, 1=g(0) g(i) = i+1, only satisfies. Therefore, B. arjuno answered Jan 9, 2020 arjuno comment Share Follow See all 0 reply Please log in or register to add a comment.