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Let $G(x) = \frac{1}{(1-x)^2} = \sum\limits_{i=0}^\infty g(i)x^i$, where $|x| < 1$. What is $g(i)$?

  1. $i$
  2. $i+1$
  3. $2i$
  4. $2^i$
in Combinatory by Boss (17.5k points) | 1.9k views
+3
Sumbody answer this ???b is the answer by putting x=0 but how to solve such types
+2

Why don't use the same method?

https://en.wikipedia.org/wiki/Taylor_series

0
Yes but how to use compare them there are g(i) and Gx i know its easy but couldnt get it now
0
0

 @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, $g(i) = \frac{f^{(i)}(0)}{i!} = \frac{(i+1)!}{i!} = i+1$

7 Answers

+54 votes
Best answer

$\frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + x^5 + \dots + \infty$

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

by Boss (28.8k points)
edited by
+1
nice approach @mcjoshi sir
+1
Does someone has a list of all such important series. Knowing this series is essential to solve this question
+5

Some important $GF:$

0
good approach.....
+6 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
by Active (1.8k points)
+5 votes

Option B) is the answer 

by Loyal (8k points)
+3 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, $g(i) = \frac{(i+1)!}{i!} = i+1$
by Active (1.3k points)
0
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.
+1

@rawan 

Check this out. 

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 

by Loyal (10k points)
0 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.

by Boss (24.2k points)
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.

 

by (299 points)
Answer:

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