Mixing
How to get the coefficients of a power series without multiplying
$begingroup$
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 Jan 3 at 14:46
mrtaurho
4,09121234
asked Jan 3 at 14:35
Str0ngerStr0nger
63
$endgroup$
add a comment |
$begingroup$
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 Jan 3 at 14:46
mrtaurho
4,09121234
asked Jan 3 at 14:35
Str0ngerStr0nger
63
$endgroup$
$begingroup$
You might want to try using the Cauchy product.
$endgroup$
– 0x539
Jan 3 at 14:51
$begingroup$
@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.
$endgroup$
– mrtaurho
Jan 3 at 14:54
add a comment |
$begingroup$
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 Jan 3 at 14:46
mrtaurho
4,09121234
asked Jan 3 at 14:35
Str0ngerStr0nger
63
$endgroup$
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 Jan 3 at 14:46
mrtaurho
4,09121234
asked Jan 3 at 14:35
Str0ngerStr0nger
63
edited Jan 3 at 14:46
mrtaurho
4,09121234
asked Jan 3 at 14:35
Str0ngerStr0nger
63
edited Jan 3 at 14:46
mrtaurho
4,09121234
edited Jan 3 at 14:46
mrtaurho
4,09121234
edited Jan 3 at 14:46
mrtaurho
4,09121234
4,09121234
asked Jan 3 at 14:35
Str0ngerStr0nger
63
asked Jan 3 at 14:35
Str0ngerStr0nger
63
asked Jan 3 at 14:35
Str0ngerStr0nger
63
63
$begingroup$
You might want to try using the Cauchy product.
$endgroup$
– 0x539
Jan 3 at 14:51
$begingroup$
@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.
$endgroup$
– mrtaurho
Jan 3 at 14:54
add a comment |
$begingroup$
You might want to try using the Cauchy product.
$endgroup$
– 0x539
Jan 3 at 14:51
$begingroup$
@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.
$endgroup$
– mrtaurho
Jan 3 at 14:54
$begingroup$
You might want to try using the Cauchy product.
$endgroup$
– 0x539
Jan 3 at 14:51
$begingroup$
You might want to try using the Cauchy product.
$endgroup$
– 0x539
Jan 3 at 14:51
$begingroup$
@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.
$endgroup$
– mrtaurho
Jan 3 at 14:54
$begingroup$
@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.
$endgroup$
– mrtaurho
Jan 3 at 14:54
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
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 Jan 3 at 14:55
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
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1 Answer
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$begingroup$
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 Jan 3 at 14:55
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
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 Jan 3 at 14:55
Ross MillikanRoss Millikan
293k23197371
$endgroup$
add a comment |
$begingroup$
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 Jan 3 at 14:55
Ross MillikanRoss Millikan
293k23197371
$endgroup$
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 Jan 3 at 14:55
Ross MillikanRoss Millikan
293k23197371
answered Jan 3 at 14:55
Ross MillikanRoss Millikan
293k23197371
answered Jan 3 at 14:55
Ross MillikanRoss Millikan
293k23197371
answered Jan 3 at 14:55
Ross MillikanRoss Millikan
293k23197371
293k23197371
add a comment |
add a comment |
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$begingroup$
You might want to try using the Cauchy product.
$endgroup$
– 0x539
Jan 3 at 14:51
$begingroup$
@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.
$endgroup$
– mrtaurho
Jan 3 at 14:54