Self doubt generating function

Question

Equation: $x+y=10$ and we are asked to find out the number of a non-negative integral solution of this equation.

Comments

@ Jason your method is not clear

can you explain in more detail??

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As it is given that no of non-negative i.e. 0, 1, 2, 3 .....

means each of x and y could take any value from 0 to 10. Therefore taking

$x^0 + x^1 + ...... + x^{10}$ for x and for y.

$(x^0 + x^1 + ...... + x^{10})^2$.

where power of x gives the value of variable x and y.

$({\frac{1-x^{11}}{1-x}})^2$

when you will solve this for co-efficient of $x^{10}$ you will get 11.

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@ Jason

please check my comments and give answer

Answer 1

We just need to find coefficient of $x^{10}$ in the expression $\frac{1}{(1-x)^2}$ which can be written as $\sum_{r=0}^{\infty}\binom{n+r-1}{r}x^r$

                                                                          Here $n=2$    $\Rightarrow\sum_{r=0}^{\infty}\binom{r+1}{r} x^r$

To have co-efficient of $x$ as $10$, we need r to be $10$     $=\binom{10+1}{10}=\binom{11}{10}=11$

Comments

@Sukannya

please see this and explain how to get coefficient in x^20

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

I do not get this

If I put r =10 then make x^20

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Where if you put r=10, you get x^20??

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This line I write doubt

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If possible please explain from starting because Generating function knew for me.

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See, you need the coefficient of x to be 20, which you can do as put r=0 in 1st and r=20 in second, r=1 in the first term and r=19 in the second term, r=2 in 1st one and r=18 in the second term, so on until r=20 in 1st term and r=0 in second.the co-efficent would be $(-1)^0$$\binom{22}{20}$, then If you put  r=1 in 1st and r=19 in the second $(-1)^1$$\binom{21}{20}$ and so on and at the end $(-1)^{20}$$\binom{2}{0}$ and then add them all.

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yes i try it

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@Sukannya 

lots of cases??

any simple method??

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@Sukannya

is that equation is $\frac{1}{(1-x)^2}$ or $\frac{(1-x^{11})^2}{(1-x)^2}$ ?

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Srestha Di, finding co-efficient of $ x^{10}$ in $\frac{1}{(1-x)^2}$ which can also be written as $\frac{1}{1-x}*\frac{1}{1-x}$ as $(1+x+x^2+x^3+......)*(1+x^2+x^3+.....)$, if we restrict both the product terms to 10 terms to find co-efficient of $x^{10}$ then $(1+x^2+x^3+.....x^{10})*(1+x^2+x^3+.....x^{10})= \frac{1-x^{11}}{1-x} * \frac{1-x^{11}}{1-x}=\frac{(1-x^{11})^2}{(1-x)^2}$.

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yes, but in 1st line how r u assuming $\frac{1}{\left ( 1-x \right )^{2}}$ ?

do u mean x,y are 2 function that is why u r assuming it at very first line?

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Yes, Di.... Both are independent sequences, thats why...

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@Sukannya

which method mostly refers??

because I followed the first method which is best answer

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The first one is the generalised one, you can follow it anywhere @Lakshman.

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Ok Thanks @Sukannya

Answer 2

$1 )\ $Brute Force method :

$=\color{Blue}{\text{(X,Y)}}  \ (0,10)  \ (1,9) \ (2,8) \ (3,7) \ (4,6) \ (5,5) \ (6,4) \ (7,3) \ (8,2) \ (9,1) \ (10,0) \\ =\color{Red}{11}  \text{ Pairs}$

$2 )\ $Function method :

• Looking above values both X and Y can have values in range $(0-10)$.

$\rightarrow \ [x^{10}]$ $(1+x+x^{2}+x^{3}....+x^{10})(1+x+x^{2}+x^{3}....+x^{10})$

$\rightarrow \ [x^{10}]$ $(1+x+x^{2}+x^{3}....+x^{10})^{2}$

$\rightarrow \ [x^{10}]$ $\Large\frac{(1-x^{11})^2}{(1-x)^2}$

$\rightarrow \ [x^{10}]$ $(1+x^{22}-2x^{11})(1-x)^{-2}$

$\rightarrow[x^{10}]$ $\Large{(^{10+2-1}_{ \  \ \  \  \ 10})}$

$\rightarrow$ $\Large(^{11}_{10})$

$\rightarrow\color{Blue}{11.}$

Comments

@saxena0612

Can you explain this

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

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@saxena0612

why do u take [x^10] outside in this line

$\rightarrow \ [\color{Red} {x^{10}}]$ $(1+x+x^{2}+x^{3}....+x^{10})(1+x+x^{2}+x^{3}....+x^{10})$ ??

[x^10] outside means there are only need to get coefficient of x {x^10.x}=x^11

but u again taking coefficient of [x^11]

isnot there a correction needed?

just verify it plz

Answer 3

This may help. 

Comments

i studied this from youtube channel trev tutor... u may watch the videos to get a better understanding..:)

also acc to what i learned ,when n is -ve ... we use formula ^n-r+1C_r  (-1)^r                                                        

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@gari

you followed Kenneth Rosen books

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no ..i don't ... i just learned this from a video. :)