Help understanding the Laurent series
$begingroup$
I just wanna see if I have some of the fundamentals nailed in terms of understanding.
Say we have some complex valued function defined as
$$f(z)=tfrac{1}{z(z-1)}$$
and we want to evaluate it's laurent series centred at $z_0=0$ on the annulus $0<|z|<1$.
Or take the function
$$f(z)=tfrac{1}{(z-1)(z+1)}$$
this time taking it's Laurent series centred at $z_0=1$ on the annulus $0<|z-1|<2$
I have a few questions regarding said laurent series:
i) my first question is just on the general situation weve got here . I believe in the first example I mentioned that we have a function which has two singularities, a simple pole at zero and another at 1. Then we have defined a region on which to calculate the laurent series that stops us from hitting the singularities by only calculating at a distance greater to one and less than the other . similarly we have in my example a complex valued function defined on some region but this time it's singularities are +-1 . and we want to centre our expansion at 1 so no we define our region in such a way that the distance between 1( which we'll never hit because we require the distance between it and any point to be greater than zero ) and any other point is less than 2 so we know have a circle that is centred at one and has one side touching 3 and the other -1 (it's singularity) but never touches either.Is this correct ?
ii)a. when actually calculating the series mentioned in my first series I know we evaluate in a region valid for $|z|<1$ in the following way :
$$tfrac{1}{z(z-1)}=tfrac{-1}{z}tfrac{1}{1-z}=tfrac{-1}{z} Sigma_{n=0}z^n$$ which is valid for $|z|<1/|c|$ c in this case being the coeffiecient of z in in the secon equality above.
My question however is this, evaluating for $|z|<1$ is the same as evaluating for $0<|z|<1$ in this case right ? , because the series i found is valid below 1 up until 0 which it is undefined, at right ?
complex-analysis laurent-series
asked Nov 30 '18 at 18:37
can'tcauchycan'tcauchy
999417
$endgroup$
add a comment |
$begingroup$
I just wanna see if I have some of the fundamentals nailed in terms of understanding.
Say we have some complex valued function defined as
$$f(z)=tfrac{1}{z(z-1)}$$
and we want to evaluate it's laurent series centred at $z_0=0$ on the annulus $0<|z|<1$.
Or take the function
$$f(z)=tfrac{1}{(z-1)(z+1)}$$
this time taking it's Laurent series centred at $z_0=1$ on the annulus $0<|z-1|<2$
I have a few questions regarding said laurent series:
i) my first question is just on the general situation weve got here . I believe in the first example I mentioned that we have a function which has two singularities, a simple pole at zero and another at 1. Then we have defined a region on which to calculate the laurent series that stops us from hitting the singularities by only calculating at a distance greater to one and less than the other . similarly we have in my example a complex valued function defined on some region but this time it's singularities are +-1 . and we want to centre our expansion at 1 so no we define our region in such a way that the distance between 1( which we'll never hit because we require the distance between it and any point to be greater than zero ) and any other point is less than 2 so we know have a circle that is centred at one and has one side touching 3 and the other -1 (it's singularity) but never touches either.Is this correct ?
ii)a. when actually calculating the series mentioned in my first series I know we evaluate in a region valid for $|z|<1$ in the following way :
$$tfrac{1}{z(z-1)}=tfrac{-1}{z}tfrac{1}{1-z}=tfrac{-1}{z} Sigma_{n=0}z^n$$ which is valid for $|z|<1/|c|$ c in this case being the coeffiecient of z in in the secon equality above.
My question however is this, evaluating for $|z|<1$ is the same as evaluating for $0<|z|<1$ in this case right ? , because the series i found is valid below 1 up until 0 which it is undefined, at right ?
complex-analysis laurent-series
asked Nov 30 '18 at 18:37
can'tcauchycan'tcauchy
999417
$endgroup$
$begingroup$
Your typography is a little difficult for me to read, but at least for your first question, I think you have the right idea. The region you define the Laurent series on won't contain a singularity, but the singularities can be on the boundary of the region (in fact, regions of convergence for Laurent series are either disks or annuli, depending on the singularities).
$endgroup$
– Clayton
Nov 30 '18 at 18:55
1
$begingroup$
We actually don't choose the region. The Laurent series itself does that. If you take the Laurent series around a pole, it's radius of convergence will be the distance to the next closest place the function is not analytic.Your final example is true everywhere that the expressions are defined. $$frac 1{1-z} = sum z^n$$ is true where the series converges, which includes $|z| < 1$ and excludes $|z| > 1$. But $$frac{-1}zfrac 1{1-z}=frac{-1}z sum z^n$$ also requires $frac 1z$ to exist, so excludes $z = 0$.
$endgroup$
– Paul Sinclair
Dec 1 '18 at 5:50
add a comment |
$begingroup$
I just wanna see if I have some of the fundamentals nailed in terms of understanding.
Say we have some complex valued function defined as
$$f(z)=tfrac{1}{z(z-1)}$$
and we want to evaluate it's laurent series centred at $z_0=0$ on the annulus $0<|z|<1$.
Or take the function
$$f(z)=tfrac{1}{(z-1)(z+1)}$$
this time taking it's Laurent series centred at $z_0=1$ on the annulus $0<|z-1|<2$
I have a few questions regarding said laurent series:
i) my first question is just on the general situation weve got here . I believe in the first example I mentioned that we have a function which has two singularities, a simple pole at zero and another at 1. Then we have defined a region on which to calculate the laurent series that stops us from hitting the singularities by only calculating at a distance greater to one and less than the other . similarly we have in my example a complex valued function defined on some region but this time it's singularities are +-1 . and we want to centre our expansion at 1 so no we define our region in such a way that the distance between 1( which we'll never hit because we require the distance between it and any point to be greater than zero ) and any other point is less than 2 so we know have a circle that is centred at one and has one side touching 3 and the other -1 (it's singularity) but never touches either.Is this correct ?
ii)a. when actually calculating the series mentioned in my first series I know we evaluate in a region valid for $|z|<1$ in the following way :
$$tfrac{1}{z(z-1)}=tfrac{-1}{z}tfrac{1}{1-z}=tfrac{-1}{z} Sigma_{n=0}z^n$$ which is valid for $|z|<1/|c|$ c in this case being the coeffiecient of z in in the secon equality above.
My question however is this, evaluating for $|z|<1$ is the same as evaluating for $0<|z|<1$ in this case right ? , because the series i found is valid below 1 up until 0 which it is undefined, at right ?
complex-analysis laurent-series
asked Nov 30 '18 at 18:37
can'tcauchycan'tcauchy
999417
$endgroup$
I just wanna see if I have some of the fundamentals nailed in terms of understanding.
Say we have some complex valued function defined as
$$f(z)=tfrac{1}{z(z-1)}$$
and we want to evaluate it's laurent series centred at $z_0=0$ on the annulus $0<|z|<1$.
Or take the function
$$f(z)=tfrac{1}{(z-1)(z+1)}$$
this time taking it's Laurent series centred at $z_0=1$ on the annulus $0<|z-1|<2$
I have a few questions regarding said laurent series:
i) my first question is just on the general situation weve got here . I believe in the first example I mentioned that we have a function which has two singularities, a simple pole at zero and another at 1. Then we have defined a region on which to calculate the laurent series that stops us from hitting the singularities by only calculating at a distance greater to one and less than the other . similarly we have in my example a complex valued function defined on some region but this time it's singularities are +-1 . and we want to centre our expansion at 1 so no we define our region in such a way that the distance between 1( which we'll never hit because we require the distance between it and any point to be greater than zero ) and any other point is less than 2 so we know have a circle that is centred at one and has one side touching 3 and the other -1 (it's singularity) but never touches either.Is this correct ?
ii)a. when actually calculating the series mentioned in my first series I know we evaluate in a region valid for $|z|<1$ in the following way :
$$tfrac{1}{z(z-1)}=tfrac{-1}{z}tfrac{1}{1-z}=tfrac{-1}{z} Sigma_{n=0}z^n$$ which is valid for $|z|<1/|c|$ c in this case being the coeffiecient of z in in the secon equality above.
My question however is this, evaluating for $|z|<1$ is the same as evaluating for $0<|z|<1$ in this case right ? , because the series i found is valid below 1 up until 0 which it is undefined, at right ?
complex-analysis laurent-series
complex-analysis laurent-series
asked Nov 30 '18 at 18:37
can'tcauchycan'tcauchy
999417
asked Nov 30 '18 at 18:37
can'tcauchycan'tcauchy
999417
edited Nov 30 '18 at 19:13
can'tcauchy
asked Nov 30 '18 at 18:37
can'tcauchycan'tcauchy
999417
asked Nov 30 '18 at 18:37
can'tcauchycan'tcauchy
999417
asked Nov 30 '18 at 18:37
can'tcauchycan'tcauchy
999417
999417
$begingroup$
Your typography is a little difficult for me to read, but at least for your first question, I think you have the right idea. The region you define the Laurent series on won't contain a singularity, but the singularities can be on the boundary of the region (in fact, regions of convergence for Laurent series are either disks or annuli, depending on the singularities).
$endgroup$
– Clayton
Nov 30 '18 at 18:55
1
$begingroup$
We actually don't choose the region. The Laurent series itself does that. If you take the Laurent series around a pole, it's radius of convergence will be the distance to the next closest place the function is not analytic.Your final example is true everywhere that the expressions are defined. $$frac 1{1-z} = sum z^n$$ is true where the series converges, which includes $|z| < 1$ and excludes $|z| > 1$. But $$frac{-1}zfrac 1{1-z}=frac{-1}z sum z^n$$ also requires $frac 1z$ to exist, so excludes $z = 0$.
$endgroup$
– Paul Sinclair
Dec 1 '18 at 5:50
add a comment |
$begingroup$
Your typography is a little difficult for me to read, but at least for your first question, I think you have the right idea. The region you define the Laurent series on won't contain a singularity, but the singularities can be on the boundary of the region (in fact, regions of convergence for Laurent series are either disks or annuli, depending on the singularities).
$endgroup$
– Clayton
Nov 30 '18 at 18:55
1
$begingroup$
We actually don't choose the region. The Laurent series itself does that. If you take the Laurent series around a pole, it's radius of convergence will be the distance to the next closest place the function is not analytic.Your final example is true everywhere that the expressions are defined. $$frac 1{1-z} = sum z^n$$ is true where the series converges, which includes $|z| < 1$ and excludes $|z| > 1$. But $$frac{-1}zfrac 1{1-z}=frac{-1}z sum z^n$$ also requires $frac 1z$ to exist, so excludes $z = 0$.
$endgroup$
– Paul Sinclair
Dec 1 '18 at 5:50
$begingroup$
Your typography is a little difficult for me to read, but at least for your first question, I think you have the right idea. The region you define the Laurent series on won't contain a singularity, but the singularities can be on the boundary of the region (in fact, regions of convergence for Laurent series are either disks or annuli, depending on the singularities).
$endgroup$
– Clayton
Nov 30 '18 at 18:55
$begingroup$
Your typography is a little difficult for me to read, but at least for your first question, I think you have the right idea. The region you define the Laurent series on won't contain a singularity, but the singularities can be on the boundary of the region (in fact, regions of convergence for Laurent series are either disks or annuli, depending on the singularities).
$endgroup$
– Clayton
Nov 30 '18 at 18:55
1
1
$begingroup$
We actually don't choose the region. The Laurent series itself does that. If you take the Laurent series around a pole, it's radius of convergence will be the distance to the next closest place the function is not analytic.Your final example is true everywhere that the expressions are defined. $$frac 1{1-z} = sum z^n$$ is true where the series converges, which includes $|z| < 1$ and excludes $|z| > 1$. But $$frac{-1}zfrac 1{1-z}=frac{-1}z sum z^n$$ also requires $frac 1z$ to exist, so excludes $z = 0$.
$endgroup$
– Paul Sinclair
Dec 1 '18 at 5:50
$begingroup$
We actually don't choose the region. The Laurent series itself does that. If you take the Laurent series around a pole, it's radius of convergence will be the distance to the next closest place the function is not analytic.Your final example is true everywhere that the expressions are defined. $$frac 1{1-z} = sum z^n$$ is true where the series converges, which includes $|z| < 1$ and excludes $|z| > 1$. But $$frac{-1}zfrac 1{1-z}=frac{-1}z sum z^n$$ also requires $frac 1z$ to exist, so excludes $z = 0$.
$endgroup$
– Paul Sinclair
Dec 1 '18 at 5:50
add a comment |
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$begingroup$
Your typography is a little difficult for me to read, but at least for your first question, I think you have the right idea. The region you define the Laurent series on won't contain a singularity, but the singularities can be on the boundary of the region (in fact, regions of convergence for Laurent series are either disks or annuli, depending on the singularities).
$endgroup$
– Clayton
Nov 30 '18 at 18:55
1
$begingroup$
We actually don't choose the region. The Laurent series itself does that. If you take the Laurent series around a pole, it's radius of convergence will be the distance to the next closest place the function is not analytic.Your final example is true everywhere that the expressions are defined. $$frac 1{1-z} = sum z^n$$ is true where the series converges, which includes $|z| < 1$ and excludes $|z| > 1$. But $$frac{-1}zfrac 1{1-z}=frac{-1}z sum z^n$$ also requires $frac 1z$ to exist, so excludes $z = 0$.
$endgroup$
– Paul Sinclair
Dec 1 '18 at 5:50