Mixing
Cauchy product for the reciprocal of the polynomial $x - 2x^2 + 3x^3 - 4x^4$
$begingroup$
I have come across some Laurent series in which the denominator of a fraction contains a power series. Looking around I came across this Calculate Laurent series for $1/ sin(z)$ which suggests that it is possible to calculate the reciprocal of a series up to a certain number of term. I have looked at the formula and because of the abstract notation, it is not clear to me how this formula is applied. So for example if I wanted to find the reciprocal of the following polynomial:
$$x - 2x^2 + 3x^3 - 4x^4$$
How would I compute this?
polynomials power-series cauchy-product
asked Jan 14 at 20:31
daljit97daljit97
178111
$endgroup$
|
show 4 more comments
$begingroup$
I have come across some Laurent series in which the denominator of a fraction contains a power series. Looking around I came across this Calculate Laurent series for $1/ sin(z)$ which suggests that it is possible to calculate the reciprocal of a series up to a certain number of term. I have looked at the formula and because of the abstract notation, it is not clear to me how this formula is applied. So for example if I wanted to find the reciprocal of the following polynomial:
$$x - 2x^2 + 3x^3 - 4x^4$$
How would I compute this?
polynomials power-series cauchy-product
asked Jan 14 at 20:31
daljit97daljit97
178111
$endgroup$
$begingroup$
Have you tried long division?
$endgroup$
– saulspatz
Jan 14 at 20:36
1
$begingroup$
Is the Cauchy product formula simply long division?
$endgroup$
– daljit97
Jan 14 at 20:40
$begingroup$
I would say yes, essentially. If we write$$1=(x-2x^2+3x^2-4x^4)({1over x}+a_0+a_1x+dots)$$ then the calculations to get the $a_k$ are the same ones we'd make doing a long division, or so it seems to me.
$endgroup$
– saulspatz
Jan 14 at 20:48
$begingroup$
So from there I just equate the coefficients right? Like equating powers of $x^1$ I would get $a_0 - 2 = 0 ==> a_0 = 2$?
$endgroup$
– daljit97
Jan 14 at 21:11
1
$begingroup$
To do long division in this context is simplicity itself. You have to write divisor and dividend in increasing degree, as you have already done with your polynomial $f(x)$. Write the algorithm down and you see immediately that the first monomial in the series for $1/f$ is certainly $x^{-1}$.
$endgroup$
– Lubin
Jan 14 at 23:48
|
show 4 more comments
$begingroup$
I have come across some Laurent series in which the denominator of a fraction contains a power series. Looking around I came across this Calculate Laurent series for $1/ sin(z)$ which suggests that it is possible to calculate the reciprocal of a series up to a certain number of term. I have looked at the formula and because of the abstract notation, it is not clear to me how this formula is applied. So for example if I wanted to find the reciprocal of the following polynomial:
$$x - 2x^2 + 3x^3 - 4x^4$$
How would I compute this?
polynomials power-series cauchy-product
asked Jan 14 at 20:31
daljit97daljit97
178111
$endgroup$
I have come across some Laurent series in which the denominator of a fraction contains a power series. Looking around I came across this Calculate Laurent series for $1/ sin(z)$ which suggests that it is possible to calculate the reciprocal of a series up to a certain number of term. I have looked at the formula and because of the abstract notation, it is not clear to me how this formula is applied. So for example if I wanted to find the reciprocal of the following polynomial:
$$x - 2x^2 + 3x^3 - 4x^4$$
How would I compute this?
polynomials power-series cauchy-product
polynomials power-series cauchy-product
asked Jan 14 at 20:31
daljit97daljit97
178111
asked Jan 14 at 20:31
daljit97daljit97
178111
asked Jan 14 at 20:31
daljit97daljit97
178111
asked Jan 14 at 20:31
daljit97daljit97
178111
asked Jan 14 at 20:31
daljit97daljit97
178111
178111
$begingroup$
Have you tried long division?
$endgroup$
– saulspatz
Jan 14 at 20:36
1
$begingroup$
Is the Cauchy product formula simply long division?
$endgroup$
– daljit97
Jan 14 at 20:40
$begingroup$
I would say yes, essentially. If we write$$1=(x-2x^2+3x^2-4x^4)({1over x}+a_0+a_1x+dots)$$ then the calculations to get the $a_k$ are the same ones we'd make doing a long division, or so it seems to me.
$endgroup$
– saulspatz
Jan 14 at 20:48
$begingroup$
So from there I just equate the coefficients right? Like equating powers of $x^1$ I would get $a_0 - 2 = 0 ==> a_0 = 2$?
$endgroup$
– daljit97
Jan 14 at 21:11
1
$begingroup$
To do long division in this context is simplicity itself. You have to write divisor and dividend in increasing degree, as you have already done with your polynomial $f(x)$. Write the algorithm down and you see immediately that the first monomial in the series for $1/f$ is certainly $x^{-1}$.
$endgroup$
– Lubin
Jan 14 at 23:48
|
show 4 more comments
$begingroup$
Have you tried long division?
$endgroup$
– saulspatz
Jan 14 at 20:36
1
$begingroup$
Is the Cauchy product formula simply long division?
$endgroup$
– daljit97
Jan 14 at 20:40
$begingroup$
I would say yes, essentially. If we write$$1=(x-2x^2+3x^2-4x^4)({1over x}+a_0+a_1x+dots)$$ then the calculations to get the $a_k$ are the same ones we'd make doing a long division, or so it seems to me.
$endgroup$
– saulspatz
Jan 14 at 20:48
$begingroup$
So from there I just equate the coefficients right? Like equating powers of $x^1$ I would get $a_0 - 2 = 0 ==> a_0 = 2$?
$endgroup$
– daljit97
Jan 14 at 21:11
1
$begingroup$
To do long division in this context is simplicity itself. You have to write divisor and dividend in increasing degree, as you have already done with your polynomial $f(x)$. Write the algorithm down and you see immediately that the first monomial in the series for $1/f$ is certainly $x^{-1}$.
$endgroup$
– Lubin
Jan 14 at 23:48
$begingroup$
Have you tried long division?
$endgroup$
– saulspatz
Jan 14 at 20:36
$begingroup$
Have you tried long division?
$endgroup$
– saulspatz
Jan 14 at 20:36
1
1
$begingroup$
Is the Cauchy product formula simply long division?
$endgroup$
– daljit97
Jan 14 at 20:40
$begingroup$
Is the Cauchy product formula simply long division?
$endgroup$
– daljit97
Jan 14 at 20:40
$begingroup$
I would say yes, essentially. If we write$$1=(x-2x^2+3x^2-4x^4)({1over x}+a_0+a_1x+dots)$$ then the calculations to get the $a_k$ are the same ones we'd make doing a long division, or so it seems to me.
$endgroup$
– saulspatz
Jan 14 at 20:48
$begingroup$
I would say yes, essentially. If we write$$1=(x-2x^2+3x^2-4x^4)({1over x}+a_0+a_1x+dots)$$ then the calculations to get the $a_k$ are the same ones we'd make doing a long division, or so it seems to me.
$endgroup$
– saulspatz
Jan 14 at 20:48
$begingroup$
So from there I just equate the coefficients right? Like equating powers of $x^1$ I would get $a_0 - 2 = 0 ==> a_0 = 2$?
$endgroup$
– daljit97
Jan 14 at 21:11
$begingroup$
So from there I just equate the coefficients right? Like equating powers of $x^1$ I would get $a_0 - 2 = 0 ==> a_0 = 2$?
$endgroup$
– daljit97
Jan 14 at 21:11
1
1
$begingroup$
To do long division in this context is simplicity itself. You have to write divisor and dividend in increasing degree, as you have already done with your polynomial $f(x)$. Write the algorithm down and you see immediately that the first monomial in the series for $1/f$ is certainly $x^{-1}$.
$endgroup$
– Lubin
Jan 14 at 23:48
$begingroup$
To do long division in this context is simplicity itself. You have to write divisor and dividend in increasing degree, as you have already done with your polynomial $f(x)$. Write the algorithm down and you see immediately that the first monomial in the series for $1/f$ is certainly $x^{-1}$.
$endgroup$
– Lubin
Jan 14 at 23:48
|
show 4 more comments
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$begingroup$
Have you tried long division?
$endgroup$
– saulspatz
Jan 14 at 20:36
1
$begingroup$
Is the Cauchy product formula simply long division?
$endgroup$
– daljit97
Jan 14 at 20:40
$begingroup$
I would say yes, essentially. If we write$$1=(x-2x^2+3x^2-4x^4)({1over x}+a_0+a_1x+dots)$$ then the calculations to get the $a_k$ are the same ones we'd make doing a long division, or so it seems to me.
$endgroup$
– saulspatz
Jan 14 at 20:48
$begingroup$
So from there I just equate the coefficients right? Like equating powers of $x^1$ I would get $a_0 - 2 = 0 ==> a_0 = 2$?
$endgroup$
– daljit97
Jan 14 at 21:11
1
$begingroup$
To do long division in this context is simplicity itself. You have to write divisor and dividend in increasing degree, as you have already done with your polynomial $f(x)$. Write the algorithm down and you see immediately that the first monomial in the series for $1/f$ is certainly $x^{-1}$.
$endgroup$
– Lubin
Jan 14 at 23:48