28 votes 28 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 8.4k views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply anshu commented Feb 6, 2015 i moved Nov 20, 2015 reply Follow Share Sumbody answer this ???b is the answer by putting x=0 but how to solve such types 3 votes 3 votes Arjun commented Feb 6, 2015 reply Follow Share Why don't use the same method? https://en.wikipedia.org/wiki/Taylor_series 3 votes 3 votes anshu commented Feb 6, 2015 reply Follow Share Yes but how to use compare them there are g(i) and Gx i know its easy but couldnt get it now 0 votes 0 votes 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.
Best answer 80 votes 80 votes $\frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + x^5 + \dots + \infty$ Differentiating it w.r.to $x$ $\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + \dots + \infty$ $\sum_{i=0}^{\infty} g(i)x^i = g(0) + g(1)x + g(2)x^2 + g(3)x^3 + \dots + \infty$ Comparing above two, we get $g(1) = 2, g(2) = 3 \color{red}{\Rightarrow g(i) = i+1}$ Correct Answer: B mcjoshi answered Dec 1, 2016 • edited Feb 16, 2021 by gatecse mcjoshi comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments KUSHAGRA गुप्ता commented Nov 25, 2019 reply Follow Share Some important $GF:$ 20 votes 20 votes Anju Mehral commented Jan 20, 2020 reply Follow Share good approach..... 0 votes 0 votes shashankrustagi commented Nov 27, 2020 reply Follow Share At least write copy paste from ROSEN. LOL 0 votes 0 votes Please log in or register to add a comment.
8 votes 8 votes We can use Maclaurin series $1/(1-x) = 1+x+x^{2}+x^{3}+x^{4}+...$ differentiating both side by x gives $1/(1-x)^{2} = 0+1+2x+3x^{2}+4x^{3}+...$ comparing this with given equaion $1/(1-x)^{2} = \sum g(i)x^{i} = g(0)+g(1)x+g(2)x^{2}+g(3)x^{3}+...$ $g(i)=i+1$ Option B Rakesh K answered Jan 7, 2017 Rakesh K comment Share Follow See all 0 reply Please log in or register to add a comment.
8 votes 8 votes Using the extended Binomial theorem, we can write it as: $(1-x)^{-2} = \sum_{i = 0}^{\infty}$${-n \choose k}(-x)^k$ From, this the coefficient can be written as: ${n+i-1} \choose i$, where $i = 2$ Therefore, $g(i) = \frac{(i+1)!}{i!} = i+1$ Gokulnath answered Jan 9, 2019 Gokulnath comment Share Follow See all 3 Comments See all 3 3 Comments reply rawan commented Jan 10, 2019 reply Follow Share How can you take combination of negative numbers, like you did in $-n \choose k$ and how did you arrive at $n+i-1 \choose i$? Please elaborate your answer. 0 votes 0 votes Gokulnath commented Jan 10, 2019 reply Follow Share @rawan https://www.youtube.com/watch?v=ZyUb5UxBA9Q&index=13&list=PLDDGPdw7e6Aj0amDsYInT_8p6xTSTGEi2&t=0s Check this out. 2 votes 2 votes shashankrustagi commented Nov 27, 2020 reply Follow Share Great approach. That’s what I call an answer. 0 votes 0 votes Please log in or register to add a comment.
6 votes 6 votes Option B) is the answer Vicky rix answered Sep 1, 2017 Vicky rix comment Share Follow See all 0 reply Please log in or register to add a comment.