kennneth rosen chapter- counting

Question

1. Find the coefficient of $x^{10}$ in the power series.

$\left ( 1+x^{2}+x^{4}+x^{6}+x^{8}+.... \right )\left ( 1+x^{4}+x^{8}+x^{12}+.... \right )\left ( 1+x^{6}+x^{12}+x^{18}+.... \right )$

Ans. $\frac{1}{\left ( 1-x^{2} \right )\left ( 1-x^{4} \right )\left ( 1-x^{6} \right )}$........now not able to proceed.

2.Provide a closed formula for the sequence it determines x²+3x+7+(1/(1-x²))

Comments

in second question ,closed form is already given ,or question may be asked for sequence for given closed formula. i m not getting question

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It is asked that for this generating function (x²+3x+7+(1/(1-x²))) provide a closed formula for the sequence it determines.

Answer 1

$\left ( 1+x^{2}+x^{4}+x^{6}+x^{8}+.... \right )\left ( 1+x^{4}+x^{8}+x^{12}+.... \right )\left ( 1+x^{6}+x^{12}+x^{18}+.... \right )$

 

= $(\dfrac{1}{(1-x^2)})^1 .(\dfrac{1}{(1-x^4)})^1. (\dfrac{1}{(1-x^6)})^1$

 

=$\sum_{k_1 = 0}^{\infty }\binom{n+k_1-1}{k_1}(x^2)^{k_1}.\sum_{k_2 = 0}^{\infty }\binom{n+k_2-1}{k_2}(x^4)^{k_2}.\sum_{k_3 = 0}^{\infty }\binom{n+k_3-1}{k_3}(x^6)^{k_3}$                                             

 

 $\because \ (\dfrac{1}{1-x})^n \ = \ \sum_{k = 0}^{\infty }\binom{n+k-1}{k}(x)^{k}$

 

= $\sum_{k_1 = 0}^{\infty }\binom{k_1}{k_1}(x^2)^{k_1}.\sum_{k_2 = 0}^{\infty }\binom{k_2}{k_2}(x^4)^{k_2}.\sum_{k_3 = 0}^{\infty }\binom{k_3}{k_3}(x^6)^{k_3}$

 

= $\sum_{k_1 = 0}^{\infty }(x^2)^{k_1}.\sum_{k_2 = 0}^{\infty }(x^4)^{k_2}.\sum_{k_3 = 0}^{\infty }(x^6)^{k_3}$

 

$k_1 = 1 , k_2 = 2, k_3 = 0$

$= (x^2)(x^8). = 1.x^{10}$ , we get the $co-efficient \  of  \ x^{10} \  is \  1$

 

$k_1 = 0 , k_2 = 1, k_3 = 1$ , we get the $co-efficient \  of  \ x^{10} \  is \  1$

 

$k_1 = 2 , k_2 = 0, k_3 = 1$ , we get the $co-efficient \  of  \ x^{10} \  is \  1$

 

$k_1 = 5 , k_2 = 0, k_3 = 0$ , we get the $co-efficient \  of  \ x^{10} \  is \  1$

 

$k_1 = 3 , k_2 = 1, k_3 = 0$ , we get the $co-efficient \  of  \ x^{10} \  is \  1$

 

On adding we get $1 + 1 + 1 + 1 + 1 = 5$

Comments

u also written that in line number 4

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@pankaj

I write it in detail

Can u chk now the procedure right or not

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Prateek has explained it well.....u can check out.

Answer 2

I think answer will be 5,

We can solve this by brute force ,as we can see {046,280,406,640, and (10)00}

(046 =0 means $ x^{0} $ from first sequence  4 means $x^4 $ from second sequence  6 means $ x^6 $ from third sequence)

only these 5 can make only $X^{10}$ and all have coefficient as 1

{$X^{10}+X^{10}+X^{10}+X^{10}+X^{10}$   }

=Coeff.of $X^{10}$ will be 5

Comments

I applied this procedure which gives me 36 as ans

https://gateoverflow.in/65803/find-number-integral-solutions-using-generating-function

anything wrong?

Answer 3

for 2nd question as we can see,$(1/1-x^2)$ is the result of $ \sum_{0}^{n} x^{2n}$  ,where coefficient  for odd power is 0 and for even power is 1 but first three term we have to adjust .as we can see $7+3x+x^2+(1+x^2+x^4........) =8+3x+2x^2+x^4+x^6.......$

as we can see $a_0$=8,$a_1$=3,$ a_2$=2  ,for n>2,$a_n$=0 ( n is odd) $a_n$=1 (n is even) .

Comments

Correct answer, nicely explained... Thanks a lot.

Answer 4

$\frac{1}{\left ( 1-x^{2} \right )\left ( 1-x^{4} \right )\left ( 1-x^{6} \right )}$

$=\frac{1}{\left ( 1-x \right )\left ( 1+x \right )}.\frac{1}{\left ( 1-x \right )\left ( 1+x \right )\left ( 1+x^{2} \right )}.\frac{1}{\left ( 1-x \right )\left ( 1+x^{3} \right )\left ( 1+x+x^{2} \right )}$

$=\frac{1}{\left ( 1-x \right )^{3}\left ( 1+x \right )^{2}.\left ( 1+x^{2} \right ).\left ( 1+x+x^{2} \right )}$

$=\left ( 1-x \right )^{-3}.\left ( 1+x \right )^{-2}.\left ( 1+x^{2} \right )^{-1}\left ( 1+x+x^{2} \right )^{-1}$

$=\left ( 1-x \right )^{-2}.\left ( 1+x \right )^{-2}.\left ( 1+x^{2} \right )^{-1}\left ( 1-x^{3} \right )^{-1}$

Now deriving

$=\left ( 1-x \right )^{-2}.\left ( 1+x \right )^{-2}.\left ( 1+x^{2} \right )^{-1}\left ( 1-x^{3} \right )^{-1}$

we get

$=\sum_{r=0}^{\alpha }\left ( -1 \right )^{r}\binom{2+r-1}{r}.\left ( \sum_{r=0}^{\alpha }\left ( -1 \right )^{r}.x^{r} \right )^{2}.\left ( 1-x^{2}+x^{4}-x^{6}+.... \right )\left (1+x^{3}+x^{6}+x^{9}+... \right )$

$=1.1.\left ( -1 \right ).1+1.1.1.1+\binom{1+8}{8}\left ( -1 \right )^{2}.1.1+\binom{1+6}{6}\left ( -1 \right )^{4}.1.1+\binom{1+4}{4}\left ( -1 \right )^{6}.1.1+\binom{1+2}{2}\left ( -1 \right )^{8}.1.1+\binom{1+0}{0}\left ( -1 \right )^{10}.1.1$

Comments

@Debashish

plz chk this solution