How to get the coefficients of a power series without multiplying












-2














Power series are considered:



$$begin{align*}
A(x) &= 1 + x^{2} + x^{4} + x^{6} + x^{8} + ...\
B(x) &= 1 + x^{3} + x^{6} + x^{9} + x^{12} + ...\
C(x) &= 1 + x^{4} + x^{8} + x^{12} + x^{16} + ...
end{align*}$$



Determine the values of $q_{1}$, $q_{2}$, $q_{3}$, $q_{4}$, $q_{5}$, $q_{6}$ in



$$A(x) cdot B(x) cdot C(x) = 1 + q_{1}x^{} + q_{2}x^{2} + q_{3}x^{3} + q_{4}x^{4} +
...$$



without doing the product of $A(x)$, $B(x)$, $C(x)$






power-series







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  • You might want to try using the Cauchy product.
    – 0x539
    yesterday












  • @0x539 This is kind of overkill in this situation. It is way easier to determine the values by inspection, i.e. which powers add up to $x$, which to $x^2$ and so on.
    – mrtaurho
    yesterday
















-2














Power series are considered:



$$begin{align*}
A(x) &= 1 + x^{2} + x^{4} + x^{6} + x^{8} + ...\
B(x) &= 1 + x^{3} + x^{6} + x^{9} + x^{12} + ...\
C(x) &= 1 + x^{4} + x^{8} + x^{12} + x^{16} + ...
end{align*}$$



Determine the values of $q_{1}$, $q_{2}$, $q_{3}$, $q_{4}$, $q_{5}$, $q_{6}$ in



$$A(x) cdot B(x) cdot C(x) = 1 + q_{1}x^{} + q_{2}x^{2} + q_{3}x^{3} + q_{4}x^{4} +
...$$



without doing the product of $A(x)$, $B(x)$, $C(x)$






power-series







share|cite|improve this question
























  • You might want to try using the Cauchy product.
    – 0x539
    yesterday












  • @0x539 This is kind of overkill in this situation. It is way easier to determine the values by inspection, i.e. which powers add up to $x$, which to $x^2$ and so on.
    – mrtaurho
    yesterday














-2












-2








-2







Power series are considered:



$$begin{align*}
A(x) &= 1 + x^{2} + x^{4} + x^{6} + x^{8} + ...\
B(x) &= 1 + x^{3} + x^{6} + x^{9} + x^{12} + ...\
C(x) &= 1 + x^{4} + x^{8} + x^{12} + x^{16} + ...
end{align*}$$



Determine the values of $q_{1}$, $q_{2}$, $q_{3}$, $q_{4}$, $q_{5}$, $q_{6}$ in



$$A(x) cdot B(x) cdot C(x) = 1 + q_{1}x^{} + q_{2}x^{2} + q_{3}x^{3} + q_{4}x^{4} +
...$$



without doing the product of $A(x)$, $B(x)$, $C(x)$






power-series







share|cite|improve this question















Power series are considered:



$$begin{align*}
A(x) &= 1 + x^{2} + x^{4} + x^{6} + x^{8} + ...\
B(x) &= 1 + x^{3} + x^{6} + x^{9} + x^{12} + ...\
C(x) &= 1 + x^{4} + x^{8} + x^{12} + x^{16} + ...
end{align*}$$



Determine the values of $q_{1}$, $q_{2}$, $q_{3}$, $q_{4}$, $q_{5}$, $q_{6}$ in



$$A(x) cdot B(x) cdot C(x) = 1 + q_{1}x^{} + q_{2}x^{2} + q_{3}x^{3} + q_{4}x^{4} +
...$$



without doing the product of $A(x)$, $B(x)$, $C(x)$






power-series




power-series






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share|cite|improve this question













share|cite|improve this question




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edited yesterday









mrtaurho

3,85121133




3,85121133










asked yesterday









Str0nger

12




12












  • You might want to try using the Cauchy product.
    – 0x539
    yesterday












  • @0x539 This is kind of overkill in this situation. It is way easier to determine the values by inspection, i.e. which powers add up to $x$, which to $x^2$ and so on.
    – mrtaurho
    yesterday


















  • You might want to try using the Cauchy product.
    – 0x539
    yesterday












  • @0x539 This is kind of overkill in this situation. It is way easier to determine the values by inspection, i.e. which powers add up to $x$, which to $x^2$ and so on.
    – mrtaurho
    yesterday
















You might want to try using the Cauchy product.
– 0x539
yesterday






You might want to try using the Cauchy product.
– 0x539
yesterday














@0x539 This is kind of overkill in this situation. It is way easier to determine the values by inspection, i.e. which powers add up to $x$, which to $x^2$ and so on.
– mrtaurho
yesterday




@0x539 This is kind of overkill in this situation. It is way easier to determine the values by inspection, i.e. which powers add up to $x$, which to $x^2$ and so on.
– mrtaurho
yesterday










1 Answer
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That depends on what you mean by not forming the product. Only needing terms up to $q_6$ I would just do a hand count of how many ways there are to add to each number $1$ through $6$ using $2,2,2,3,3,4$, which is not very many. That is equivalent to multiplying, but it looks different.



You could sum the series, saying $A(x)=frac 1{1-x^2}$, multiply those, and find the expansion in $x$, but that is more work.






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    1 Answer
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    1 Answer
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    That depends on what you mean by not forming the product. Only needing terms up to $q_6$ I would just do a hand count of how many ways there are to add to each number $1$ through $6$ using $2,2,2,3,3,4$, which is not very many. That is equivalent to multiplying, but it looks different.



    You could sum the series, saying $A(x)=frac 1{1-x^2}$, multiply those, and find the expansion in $x$, but that is more work.






    share|cite|improve this answer


























      1














      That depends on what you mean by not forming the product. Only needing terms up to $q_6$ I would just do a hand count of how many ways there are to add to each number $1$ through $6$ using $2,2,2,3,3,4$, which is not very many. That is equivalent to multiplying, but it looks different.



      You could sum the series, saying $A(x)=frac 1{1-x^2}$, multiply those, and find the expansion in $x$, but that is more work.






      share|cite|improve this answer
























        1












        1








        1






        That depends on what you mean by not forming the product. Only needing terms up to $q_6$ I would just do a hand count of how many ways there are to add to each number $1$ through $6$ using $2,2,2,3,3,4$, which is not very many. That is equivalent to multiplying, but it looks different.



        You could sum the series, saying $A(x)=frac 1{1-x^2}$, multiply those, and find the expansion in $x$, but that is more work.






        share|cite|improve this answer












        That depends on what you mean by not forming the product. Only needing terms up to $q_6$ I would just do a hand count of how many ways there are to add to each number $1$ through $6$ using $2,2,2,3,3,4$, which is not very many. That is equivalent to multiplying, but it looks different.



        You could sum the series, saying $A(x)=frac 1{1-x^2}$, multiply those, and find the expansion in $x$, but that is more work.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Ross Millikan

        292k23197371




        292k23197371






























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