When can a rational function be represented by a Taylor series?
I am trying to prove that
$$
int_C frac{P(z)}{Q(z)} dz = 0
$$
If polynomial order of $Q$ is 2 or more than that of $P$, using the theorem stating that if a function has a finite number of singular points all interior to a contour $C$, then
$$
int_C f(z) dz = 2pi itext{Res}bigg[frac{1}{z^2} f(frac{1}{z})bigg]
$$
I received the hint that, under the conditions described above, the rational function in the integrand can be written as a McLauran Series with no negative powers of z. This would imply that the residue is zero...
My problem is, I can't seem to wrap my head around how the rational function given to me can be written into the form of a power series with only positive exponents...
So, as the title says, When can a rational function be represented as a power series? I'm not looking for a full proof, but a few details would be nice, just so I feel more comfortable running with the hint.
complex-analysis residue-calculus
asked Nov 29 '18 at 20:46
rocksNwavesrocksNwaves
369112
add a comment |
I am trying to prove that
$$
int_C frac{P(z)}{Q(z)} dz = 0
$$
If polynomial order of $Q$ is 2 or more than that of $P$, using the theorem stating that if a function has a finite number of singular points all interior to a contour $C$, then
$$
int_C f(z) dz = 2pi itext{Res}bigg[frac{1}{z^2} f(frac{1}{z})bigg]
$$
I received the hint that, under the conditions described above, the rational function in the integrand can be written as a McLauran Series with no negative powers of z. This would imply that the residue is zero...
My problem is, I can't seem to wrap my head around how the rational function given to me can be written into the form of a power series with only positive exponents...
So, as the title says, When can a rational function be represented as a power series? I'm not looking for a full proof, but a few details would be nice, just so I feel more comfortable running with the hint.
complex-analysis residue-calculus
asked Nov 29 '18 at 20:46
rocksNwavesrocksNwaves
369112
1
Not really. The idea of the residue at infinity is that $$int_{|z|=R} f(z) dz = -int_{|s|=1/R} f(1/s) d(1/s) = int_{|s|=1/R} frac{f(1/s)}{s^2} ds$$ For $f$ a rational function, let $r = |rho|$ the largest pole then the integral doesn't depend on $R> r$. Since $frac{f(1/s)}{s^2} = sum_{n=-N}^{-1} c_n s^{n}+ h(s)$ with $h$ a polynomial then you can integrate each term and obtain the result as $2ipi c_{-1}$.
– reuns
Nov 29 '18 at 21:11
What is $N$ ? When is it $0$ ?
– reuns
Nov 29 '18 at 21:16
add a comment |
I am trying to prove that
$$
int_C frac{P(z)}{Q(z)} dz = 0
$$
If polynomial order of $Q$ is 2 or more than that of $P$, using the theorem stating that if a function has a finite number of singular points all interior to a contour $C$, then
$$
int_C f(z) dz = 2pi itext{Res}bigg[frac{1}{z^2} f(frac{1}{z})bigg]
$$
I received the hint that, under the conditions described above, the rational function in the integrand can be written as a McLauran Series with no negative powers of z. This would imply that the residue is zero...
My problem is, I can't seem to wrap my head around how the rational function given to me can be written into the form of a power series with only positive exponents...
So, as the title says, When can a rational function be represented as a power series? I'm not looking for a full proof, but a few details would be nice, just so I feel more comfortable running with the hint.
complex-analysis residue-calculus
asked Nov 29 '18 at 20:46
rocksNwavesrocksNwaves
369112
I am trying to prove that
$$
int_C frac{P(z)}{Q(z)} dz = 0
$$
If polynomial order of $Q$ is 2 or more than that of $P$, using the theorem stating that if a function has a finite number of singular points all interior to a contour $C$, then
$$
int_C f(z) dz = 2pi itext{Res}bigg[frac{1}{z^2} f(frac{1}{z})bigg]
$$
I received the hint that, under the conditions described above, the rational function in the integrand can be written as a McLauran Series with no negative powers of z. This would imply that the residue is zero...
My problem is, I can't seem to wrap my head around how the rational function given to me can be written into the form of a power series with only positive exponents...
So, as the title says, When can a rational function be represented as a power series? I'm not looking for a full proof, but a few details would be nice, just so I feel more comfortable running with the hint.
complex-analysis residue-calculus
complex-analysis residue-calculus
asked Nov 29 '18 at 20:46
rocksNwavesrocksNwaves
369112
asked Nov 29 '18 at 20:46
rocksNwavesrocksNwaves
369112
edited Nov 30 '18 at 19:03
rocksNwaves
asked Nov 29 '18 at 20:46
rocksNwavesrocksNwaves
369112
asked Nov 29 '18 at 20:46
rocksNwavesrocksNwaves
369112
asked Nov 29 '18 at 20:46
rocksNwavesrocksNwaves
369112
369112
1
Not really. The idea of the residue at infinity is that $$int_{|z|=R} f(z) dz = -int_{|s|=1/R} f(1/s) d(1/s) = int_{|s|=1/R} frac{f(1/s)}{s^2} ds$$ For $f$ a rational function, let $r = |rho|$ the largest pole then the integral doesn't depend on $R> r$. Since $frac{f(1/s)}{s^2} = sum_{n=-N}^{-1} c_n s^{n}+ h(s)$ with $h$ a polynomial then you can integrate each term and obtain the result as $2ipi c_{-1}$.
– reuns
Nov 29 '18 at 21:11
What is $N$ ? When is it $0$ ?
– reuns
Nov 29 '18 at 21:16
add a comment |
1
Not really. The idea of the residue at infinity is that $$int_{|z|=R} f(z) dz = -int_{|s|=1/R} f(1/s) d(1/s) = int_{|s|=1/R} frac{f(1/s)}{s^2} ds$$ For $f$ a rational function, let $r = |rho|$ the largest pole then the integral doesn't depend on $R> r$. Since $frac{f(1/s)}{s^2} = sum_{n=-N}^{-1} c_n s^{n}+ h(s)$ with $h$ a polynomial then you can integrate each term and obtain the result as $2ipi c_{-1}$.
– reuns
Nov 29 '18 at 21:11
What is $N$ ? When is it $0$ ?
– reuns
Nov 29 '18 at 21:16
1
1
Not really. The idea of the residue at infinity is that $$int_{|z|=R} f(z) dz = -int_{|s|=1/R} f(1/s) d(1/s) = int_{|s|=1/R} frac{f(1/s)}{s^2} ds$$ For $f$ a rational function, let $r = |rho|$ the largest pole then the integral doesn't depend on $R> r$. Since $frac{f(1/s)}{s^2} = sum_{n=-N}^{-1} c_n s^{n}+ h(s)$ with $h$ a polynomial then you can integrate each term and obtain the result as $2ipi c_{-1}$.
– reuns
Nov 29 '18 at 21:11
Not really. The idea of the residue at infinity is that $$int_{|z|=R} f(z) dz = -int_{|s|=1/R} f(1/s) d(1/s) = int_{|s|=1/R} frac{f(1/s)}{s^2} ds$$ For $f$ a rational function, let $r = |rho|$ the largest pole then the integral doesn't depend on $R> r$. Since $frac{f(1/s)}{s^2} = sum_{n=-N}^{-1} c_n s^{n}+ h(s)$ with $h$ a polynomial then you can integrate each term and obtain the result as $2ipi c_{-1}$.
– reuns
Nov 29 '18 at 21:11
What is $N$ ? When is it $0$ ?
– reuns
Nov 29 '18 at 21:16
What is $N$ ? When is it $0$ ?
– reuns
Nov 29 '18 at 21:16
add a comment |
1 Answer
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The answer is that whenever you center the expansion at a point where the function is analytic, it can be represented as a Taylor series. In this specific case, we will see that the function in question is analytic at $z=0$ and therefore has a Maclaurin expansion.
To see why we can write $frac{1}{z^2}frac{P(frac{1}{z})}{Q(frac{1}{z})}$ as a Taylor series centered at $z=0$, consider distributing the $frac{1}{z^2}$ onto the rational function as shown below:
$$
g(z)=frac{1}{z^2}frac{P(frac{1}{z})}{Q(frac{1}{z})} = frac{a_0 + a_1frac{1}{z} + dots + a_nfrac{1}{z^n}}{b_0z^2 + b_1z + b_2 + dots+b_mfrac{1}{z^{m-2}}} times frac{z^{m-2}}{z^{m-2}} = frac{a_0z^{m-2} + a_1z^{m-3} + dots + a_nz^{m-n-2}}{b_0z^m + b_1z^{m-1} + dots+b_m}
$$
Now, take note of the far right hand side of the above equation and consider what happens when $z=0$. Because we know $b_m$ is non zero, the denominator is non-zero at $z=0$. This means that $g(z)$ is analytic at $z=0$
We know then that if a function is analytic within some neighborhood of $z=0$ that it can be written as a Maclaurin power series:
$$
g(z) = sum_{n=0}^infty d_n z^n
$$
Because there are no negative powers of $z$ in this expansion, its residue is zero at $z=0$.
Using the above information and the fact that $frac{P(z)}{Q(z)}$ is analytic except for a finite number $n$ of singular points (the zeros of $Q$) contained in a large enough contour $C$, then:
$$
int_C f(z)dz = 2pi i sum_{j=1}^nunderset{z = z_j}{text{Res}}[f(x)] = 2pi i { underset{z = 0}{text{Res}}bigg[frac{1}{z^2}fbigg(frac{1}{z}bigg)bigg]} = 0
$$
answered Nov 30 '18 at 18:50
rocksNwavesrocksNwaves
369112
add a comment |
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1 Answer
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The answer is that whenever you center the expansion at a point where the function is analytic, it can be represented as a Taylor series. In this specific case, we will see that the function in question is analytic at $z=0$ and therefore has a Maclaurin expansion.
To see why we can write $frac{1}{z^2}frac{P(frac{1}{z})}{Q(frac{1}{z})}$ as a Taylor series centered at $z=0$, consider distributing the $frac{1}{z^2}$ onto the rational function as shown below:
$$
g(z)=frac{1}{z^2}frac{P(frac{1}{z})}{Q(frac{1}{z})} = frac{a_0 + a_1frac{1}{z} + dots + a_nfrac{1}{z^n}}{b_0z^2 + b_1z + b_2 + dots+b_mfrac{1}{z^{m-2}}} times frac{z^{m-2}}{z^{m-2}} = frac{a_0z^{m-2} + a_1z^{m-3} + dots + a_nz^{m-n-2}}{b_0z^m + b_1z^{m-1} + dots+b_m}
$$
Now, take note of the far right hand side of the above equation and consider what happens when $z=0$. Because we know $b_m$ is non zero, the denominator is non-zero at $z=0$. This means that $g(z)$ is analytic at $z=0$
We know then that if a function is analytic within some neighborhood of $z=0$ that it can be written as a Maclaurin power series:
$$
g(z) = sum_{n=0}^infty d_n z^n
$$
Because there are no negative powers of $z$ in this expansion, its residue is zero at $z=0$.
Using the above information and the fact that $frac{P(z)}{Q(z)}$ is analytic except for a finite number $n$ of singular points (the zeros of $Q$) contained in a large enough contour $C$, then:
$$
int_C f(z)dz = 2pi i sum_{j=1}^nunderset{z = z_j}{text{Res}}[f(x)] = 2pi i { underset{z = 0}{text{Res}}bigg[frac{1}{z^2}fbigg(frac{1}{z}bigg)bigg]} = 0
$$
answered Nov 30 '18 at 18:50
rocksNwavesrocksNwaves
369112
add a comment |
The answer is that whenever you center the expansion at a point where the function is analytic, it can be represented as a Taylor series. In this specific case, we will see that the function in question is analytic at $z=0$ and therefore has a Maclaurin expansion.
To see why we can write $frac{1}{z^2}frac{P(frac{1}{z})}{Q(frac{1}{z})}$ as a Taylor series centered at $z=0$, consider distributing the $frac{1}{z^2}$ onto the rational function as shown below:
$$
g(z)=frac{1}{z^2}frac{P(frac{1}{z})}{Q(frac{1}{z})} = frac{a_0 + a_1frac{1}{z} + dots + a_nfrac{1}{z^n}}{b_0z^2 + b_1z + b_2 + dots+b_mfrac{1}{z^{m-2}}} times frac{z^{m-2}}{z^{m-2}} = frac{a_0z^{m-2} + a_1z^{m-3} + dots + a_nz^{m-n-2}}{b_0z^m + b_1z^{m-1} + dots+b_m}
$$
Now, take note of the far right hand side of the above equation and consider what happens when $z=0$. Because we know $b_m$ is non zero, the denominator is non-zero at $z=0$. This means that $g(z)$ is analytic at $z=0$
We know then that if a function is analytic within some neighborhood of $z=0$ that it can be written as a Maclaurin power series:
$$
g(z) = sum_{n=0}^infty d_n z^n
$$
Because there are no negative powers of $z$ in this expansion, its residue is zero at $z=0$.
Using the above information and the fact that $frac{P(z)}{Q(z)}$ is analytic except for a finite number $n$ of singular points (the zeros of $Q$) contained in a large enough contour $C$, then:
$$
int_C f(z)dz = 2pi i sum_{j=1}^nunderset{z = z_j}{text{Res}}[f(x)] = 2pi i { underset{z = 0}{text{Res}}bigg[frac{1}{z^2}fbigg(frac{1}{z}bigg)bigg]} = 0
$$
answered Nov 30 '18 at 18:50
rocksNwavesrocksNwaves
369112
add a comment |
The answer is that whenever you center the expansion at a point where the function is analytic, it can be represented as a Taylor series. In this specific case, we will see that the function in question is analytic at $z=0$ and therefore has a Maclaurin expansion.
To see why we can write $frac{1}{z^2}frac{P(frac{1}{z})}{Q(frac{1}{z})}$ as a Taylor series centered at $z=0$, consider distributing the $frac{1}{z^2}$ onto the rational function as shown below:
$$
g(z)=frac{1}{z^2}frac{P(frac{1}{z})}{Q(frac{1}{z})} = frac{a_0 + a_1frac{1}{z} + dots + a_nfrac{1}{z^n}}{b_0z^2 + b_1z + b_2 + dots+b_mfrac{1}{z^{m-2}}} times frac{z^{m-2}}{z^{m-2}} = frac{a_0z^{m-2} + a_1z^{m-3} + dots + a_nz^{m-n-2}}{b_0z^m + b_1z^{m-1} + dots+b_m}
$$
Now, take note of the far right hand side of the above equation and consider what happens when $z=0$. Because we know $b_m$ is non zero, the denominator is non-zero at $z=0$. This means that $g(z)$ is analytic at $z=0$
We know then that if a function is analytic within some neighborhood of $z=0$ that it can be written as a Maclaurin power series:
$$
g(z) = sum_{n=0}^infty d_n z^n
$$
Because there are no negative powers of $z$ in this expansion, its residue is zero at $z=0$.
Using the above information and the fact that $frac{P(z)}{Q(z)}$ is analytic except for a finite number $n$ of singular points (the zeros of $Q$) contained in a large enough contour $C$, then:
$$
int_C f(z)dz = 2pi i sum_{j=1}^nunderset{z = z_j}{text{Res}}[f(x)] = 2pi i { underset{z = 0}{text{Res}}bigg[frac{1}{z^2}fbigg(frac{1}{z}bigg)bigg]} = 0
$$
answered Nov 30 '18 at 18:50
rocksNwavesrocksNwaves
369112
The answer is that whenever you center the expansion at a point where the function is analytic, it can be represented as a Taylor series. In this specific case, we will see that the function in question is analytic at $z=0$ and therefore has a Maclaurin expansion.
To see why we can write $frac{1}{z^2}frac{P(frac{1}{z})}{Q(frac{1}{z})}$ as a Taylor series centered at $z=0$, consider distributing the $frac{1}{z^2}$ onto the rational function as shown below:
$$
g(z)=frac{1}{z^2}frac{P(frac{1}{z})}{Q(frac{1}{z})} = frac{a_0 + a_1frac{1}{z} + dots + a_nfrac{1}{z^n}}{b_0z^2 + b_1z + b_2 + dots+b_mfrac{1}{z^{m-2}}} times frac{z^{m-2}}{z^{m-2}} = frac{a_0z^{m-2} + a_1z^{m-3} + dots + a_nz^{m-n-2}}{b_0z^m + b_1z^{m-1} + dots+b_m}
$$
Now, take note of the far right hand side of the above equation and consider what happens when $z=0$. Because we know $b_m$ is non zero, the denominator is non-zero at $z=0$. This means that $g(z)$ is analytic at $z=0$
We know then that if a function is analytic within some neighborhood of $z=0$ that it can be written as a Maclaurin power series:
$$
g(z) = sum_{n=0}^infty d_n z^n
$$
Because there are no negative powers of $z$ in this expansion, its residue is zero at $z=0$.
Using the above information and the fact that $frac{P(z)}{Q(z)}$ is analytic except for a finite number $n$ of singular points (the zeros of $Q$) contained in a large enough contour $C$, then:
$$
int_C f(z)dz = 2pi i sum_{j=1}^nunderset{z = z_j}{text{Res}}[f(x)] = 2pi i { underset{z = 0}{text{Res}}bigg[frac{1}{z^2}fbigg(frac{1}{z}bigg)bigg]} = 0
$$
answered Nov 30 '18 at 18:50
rocksNwavesrocksNwaves
369112
edited Dec 3 '18 at 20:01
answered Nov 30 '18 at 18:50
rocksNwavesrocksNwaves
369112
answered Nov 30 '18 at 18:50
rocksNwavesrocksNwaves
369112
answered Nov 30 '18 at 18:50
rocksNwavesrocksNwaves
369112
369112
add a comment |
add a comment |
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Not really. The idea of the residue at infinity is that $$int_{|z|=R} f(z) dz = -int_{|s|=1/R} f(1/s) d(1/s) = int_{|s|=1/R} frac{f(1/s)}{s^2} ds$$ For $f$ a rational function, let $r = |rho|$ the largest pole then the integral doesn't depend on $R> r$. Since $frac{f(1/s)}{s^2} = sum_{n=-N}^{-1} c_n s^{n}+ h(s)$ with $h$ a polynomial then you can integrate each term and obtain the result as $2ipi c_{-1}$.
– reuns
Nov 29 '18 at 21:11
What is $N$ ? When is it $0$ ?
– reuns
Nov 29 '18 at 21:16