How to get the coefficients of a power series without multiplying
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
edited yesterday
mrtaurho
3,85121133
asked yesterday
Str0nger
12
add a comment |
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
edited yesterday
mrtaurho
3,85121133
asked yesterday
Str0nger
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
add a comment |
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
edited yesterday
mrtaurho
3,85121133
asked yesterday
Str0nger
12
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
edited yesterday
mrtaurho
3,85121133
asked yesterday
Str0nger
12
edited yesterday
mrtaurho
3,85121133
asked yesterday
Str0nger
12
edited yesterday
mrtaurho
3,85121133
edited yesterday
mrtaurho
3,85121133
edited yesterday
mrtaurho
3,85121133
3,85121133
asked yesterday
Str0nger
12
asked yesterday
Str0nger
12
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
add a comment |
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
add a comment |
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.
answered yesterday
Ross Millikan
292k23197371
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
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active
oldest
votes
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.
answered yesterday
Ross Millikan
292k23197371
add a comment |
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.
answered yesterday
Ross Millikan
292k23197371
add a comment |
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.
answered yesterday
Ross Millikan
292k23197371
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.
answered yesterday
Ross Millikan
292k23197371
answered yesterday
Ross Millikan
292k23197371
answered yesterday
Ross Millikan
292k23197371
answered yesterday
Ross Millikan
292k23197371
292k23197371
add a comment |
add a comment |
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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