Mixing
Co-efficient of $x^{12}$ in the given Power Series
$begingroup$
Qyestion
Find Coefficient of $x^{12}$ in $${(x*frac{left(1-x^6right)}{(1-x)})}^3$$
My attempt (using Generator Functions)
Since we know that,
$$frac{left(1-x^6right)}{(1-x)} = left(x^5+x^4+x^3+x^2+x+1right)$$
therefore the given problem reduces to finding the co-efficient of $x^{12}$ in
$(x*left(x^5+x^4+x^3+x^2+x+1right))^3$,
taking that $x^{3}$ out, that means we need to find the co-efficient of $x^{9}$
in
$left(x^5+x^4+x^3+x^2+x+1right)^3$,
which in turn is equivalent to $sum_{m=0}^{infty}binom{m+2}{2}x^{m}$, the coefficient of
$x^{9}$ will be $binom{11}{2}. $ i.e $55$
(i haven't shifted the series to match the correct order though but it isn't necessary right as that is a variable change i.e substituing $m+2 = r$)
but the (book's) answer says the co-efficient isn't that, rather it's 25.
Where did i go wrong in applying the logic?
Thanks a lot for your time.
(I did check other amazing answers here, they all seem to suggest to do simplification, so I also did but don't know what went wrong. :( )
Here's what the Book's explanation says which i understand very well but what's wrong with the above said approach is what I am trying to understand,
combinatorics generating-functions
asked Jan 18 at 13:52
AdityaAditya
381413
$endgroup$
add a comment |
$begingroup$
Qyestion
Find Coefficient of $x^{12}$ in $${(x*frac{left(1-x^6right)}{(1-x)})}^3$$
My attempt (using Generator Functions)
Since we know that,
$$frac{left(1-x^6right)}{(1-x)} = left(x^5+x^4+x^3+x^2+x+1right)$$
therefore the given problem reduces to finding the co-efficient of $x^{12}$ in
$(x*left(x^5+x^4+x^3+x^2+x+1right))^3$,
taking that $x^{3}$ out, that means we need to find the co-efficient of $x^{9}$
in
$left(x^5+x^4+x^3+x^2+x+1right)^3$,
which in turn is equivalent to $sum_{m=0}^{infty}binom{m+2}{2}x^{m}$, the coefficient of
$x^{9}$ will be $binom{11}{2}. $ i.e $55$
(i haven't shifted the series to match the correct order though but it isn't necessary right as that is a variable change i.e substituing $m+2 = r$)
but the (book's) answer says the co-efficient isn't that, rather it's 25.
Where did i go wrong in applying the logic?
Thanks a lot for your time.
(I did check other amazing answers here, they all seem to suggest to do simplification, so I also did but don't know what went wrong. :( )
Here's what the Book's explanation says which i understand very well but what's wrong with the above said approach is what I am trying to understand,
combinatorics generating-functions
asked Jan 18 at 13:52
AdityaAditya
381413
$endgroup$
1
$begingroup$
"$left(x^5+x^4+x^3+x^2+x+1right)^3$, which in turn is equivalent to $sum_{m=0}^{infty}binom{m+2}{2}x^{m}$" Well, no, what makes you think they are equivalent?
$endgroup$
– Did
Jan 18 at 14:10
$begingroup$
Hmm, thanks @Did, I understood my mistake, I can't do that as it isn't correct since we have that exponent there and terms before were finite but with this change they become infinite summation, right?
$endgroup$
– Aditya
Jan 18 at 14:13
$begingroup$
Indeed you are dealing with a polynomial (of degree 15), hence it cannot coincide with the series.
$endgroup$
– Did
Jan 18 at 14:16
$begingroup$
Thanks a lot again :)
$endgroup$
– Aditya
Jan 18 at 14:17
add a comment |
$begingroup$
Qyestion
Find Coefficient of $x^{12}$ in $${(x*frac{left(1-x^6right)}{(1-x)})}^3$$
My attempt (using Generator Functions)
Since we know that,
$$frac{left(1-x^6right)}{(1-x)} = left(x^5+x^4+x^3+x^2+x+1right)$$
therefore the given problem reduces to finding the co-efficient of $x^{12}$ in
$(x*left(x^5+x^4+x^3+x^2+x+1right))^3$,
taking that $x^{3}$ out, that means we need to find the co-efficient of $x^{9}$
in
$left(x^5+x^4+x^3+x^2+x+1right)^3$,
which in turn is equivalent to $sum_{m=0}^{infty}binom{m+2}{2}x^{m}$, the coefficient of
$x^{9}$ will be $binom{11}{2}. $ i.e $55$
(i haven't shifted the series to match the correct order though but it isn't necessary right as that is a variable change i.e substituing $m+2 = r$)
but the (book's) answer says the co-efficient isn't that, rather it's 25.
Where did i go wrong in applying the logic?
Thanks a lot for your time.
(I did check other amazing answers here, they all seem to suggest to do simplification, so I also did but don't know what went wrong. :( )
Here's what the Book's explanation says which i understand very well but what's wrong with the above said approach is what I am trying to understand,
combinatorics generating-functions
asked Jan 18 at 13:52
AdityaAditya
381413
$endgroup$
Qyestion
Find Coefficient of $x^{12}$ in $${(x*frac{left(1-x^6right)}{(1-x)})}^3$$
My attempt (using Generator Functions)
Since we know that,
$$frac{left(1-x^6right)}{(1-x)} = left(x^5+x^4+x^3+x^2+x+1right)$$
therefore the given problem reduces to finding the co-efficient of $x^{12}$ in
$(x*left(x^5+x^4+x^3+x^2+x+1right))^3$,
taking that $x^{3}$ out, that means we need to find the co-efficient of $x^{9}$
in
$left(x^5+x^4+x^3+x^2+x+1right)^3$,
which in turn is equivalent to $sum_{m=0}^{infty}binom{m+2}{2}x^{m}$, the coefficient of
$x^{9}$ will be $binom{11}{2}. $ i.e $55$
(i haven't shifted the series to match the correct order though but it isn't necessary right as that is a variable change i.e substituing $m+2 = r$)
but the (book's) answer says the co-efficient isn't that, rather it's 25.
Where did i go wrong in applying the logic?
Thanks a lot for your time.
(I did check other amazing answers here, they all seem to suggest to do simplification, so I also did but don't know what went wrong. :( )
Here's what the Book's explanation says which i understand very well but what's wrong with the above said approach is what I am trying to understand,
combinatorics generating-functions
combinatorics generating-functions
asked Jan 18 at 13:52
AdityaAditya
381413
asked Jan 18 at 13:52
AdityaAditya
381413
edited Jan 18 at 14:07
Aditya
asked Jan 18 at 13:52
AdityaAditya
381413
asked Jan 18 at 13:52
AdityaAditya
381413
asked Jan 18 at 13:52
AdityaAditya
381413
381413
1
$begingroup$
"$left(x^5+x^4+x^3+x^2+x+1right)^3$, which in turn is equivalent to $sum_{m=0}^{infty}binom{m+2}{2}x^{m}$" Well, no, what makes you think they are equivalent?
$endgroup$
– Did
Jan 18 at 14:10
$begingroup$
Hmm, thanks @Did, I understood my mistake, I can't do that as it isn't correct since we have that exponent there and terms before were finite but with this change they become infinite summation, right?
$endgroup$
– Aditya
Jan 18 at 14:13
$begingroup$
Indeed you are dealing with a polynomial (of degree 15), hence it cannot coincide with the series.
$endgroup$
– Did
Jan 18 at 14:16
$begingroup$
Thanks a lot again :)
$endgroup$
– Aditya
Jan 18 at 14:17
add a comment |
1
$begingroup$
"$left(x^5+x^4+x^3+x^2+x+1right)^3$, which in turn is equivalent to $sum_{m=0}^{infty}binom{m+2}{2}x^{m}$" Well, no, what makes you think they are equivalent?
$endgroup$
– Did
Jan 18 at 14:10
$begingroup$
Hmm, thanks @Did, I understood my mistake, I can't do that as it isn't correct since we have that exponent there and terms before were finite but with this change they become infinite summation, right?
$endgroup$
– Aditya
Jan 18 at 14:13
$begingroup$
Indeed you are dealing with a polynomial (of degree 15), hence it cannot coincide with the series.
$endgroup$
– Did
Jan 18 at 14:16
$begingroup$
Thanks a lot again :)
$endgroup$
– Aditya
Jan 18 at 14:17
1
1
$begingroup$
"$left(x^5+x^4+x^3+x^2+x+1right)^3$, which in turn is equivalent to $sum_{m=0}^{infty}binom{m+2}{2}x^{m}$" Well, no, what makes you think they are equivalent?
$endgroup$
– Did
Jan 18 at 14:10
$begingroup$
"$left(x^5+x^4+x^3+x^2+x+1right)^3$, which in turn is equivalent to $sum_{m=0}^{infty}binom{m+2}{2}x^{m}$" Well, no, what makes you think they are equivalent?
$endgroup$
– Did
Jan 18 at 14:10
$begingroup$
Hmm, thanks @Did, I understood my mistake, I can't do that as it isn't correct since we have that exponent there and terms before were finite but with this change they become infinite summation, right?
$endgroup$
– Aditya
Jan 18 at 14:13
$begingroup$
Hmm, thanks @Did, I understood my mistake, I can't do that as it isn't correct since we have that exponent there and terms before were finite but with this change they become infinite summation, right?
$endgroup$
– Aditya
Jan 18 at 14:13
$begingroup$
Indeed you are dealing with a polynomial (of degree 15), hence it cannot coincide with the series.
$endgroup$
– Did
Jan 18 at 14:16
$begingroup$
Indeed you are dealing with a polynomial (of degree 15), hence it cannot coincide with the series.
$endgroup$
– Did
Jan 18 at 14:16
$begingroup$
Thanks a lot again :)
$endgroup$
– Aditya
Jan 18 at 14:17
$begingroup$
Thanks a lot again :)
$endgroup$
– Aditya
Jan 18 at 14:17
add a comment |
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1
$begingroup$
"$left(x^5+x^4+x^3+x^2+x+1right)^3$, which in turn is equivalent to $sum_{m=0}^{infty}binom{m+2}{2}x^{m}$" Well, no, what makes you think they are equivalent?
$endgroup$
– Did
Jan 18 at 14:10
$begingroup$
Hmm, thanks @Did, I understood my mistake, I can't do that as it isn't correct since we have that exponent there and terms before were finite but with this change they become infinite summation, right?
$endgroup$
– Aditya
Jan 18 at 14:13
$begingroup$
Indeed you are dealing with a polynomial (of degree 15), hence it cannot coincide with the series.
$endgroup$
– Did
Jan 18 at 14:16
$begingroup$
Thanks a lot again :)
$endgroup$
– Aditya
Jan 18 at 14:17