Why does $sum_{ngeq0}(1-x)^n=frac1x$ have such a poor radius of convergence?
I am confused as to why $$sum_{ngeq0}(1-x)^n=frac1x$$
only works for $xin (0,2)$. I get that it has a singularity at $x=0$, so that can't work, but there are no singularities for the rest of the positive real line. Why isn't there a power series representation of $frac1x$ which works for the whole positive real line? Are there any series representations of $1/x$ which work for $xin (0,infty)$? I can't find any. Thanks.
sequences-and-series convergence power-series taylor-expansion
edited Dec 1 '18 at 7:31
Asaf Karagila♦
301k32424754
asked Nov 30 '18 at 18:26
clathratus
3,126331
add a comment |
I am confused as to why $$sum_{ngeq0}(1-x)^n=frac1x$$
only works for $xin (0,2)$. I get that it has a singularity at $x=0$, so that can't work, but there are no singularities for the rest of the positive real line. Why isn't there a power series representation of $frac1x$ which works for the whole positive real line? Are there any series representations of $1/x$ which work for $xin (0,infty)$? I can't find any. Thanks.
sequences-and-series convergence power-series taylor-expansion
edited Dec 1 '18 at 7:31
Asaf Karagila♦
301k32424754
asked Nov 30 '18 at 18:26
clathratus
3,126331
The domain of convergence of a real power series is always either a finitely long interval or the whole real line.
– Andreas Blass
Nov 30 '18 at 18:27
3
$sum r^n$ converges iff $-1<r<1$. If $r=1-x$ that's the same as $0<x<2$.
– Lord Shark the Unknown
Nov 30 '18 at 18:28
$1/x$ has a pole at $x=0$.
– Lord Shark the Unknown
Nov 30 '18 at 18:28
1
More specifically: their intervals of convergence are symmetric around the point of evaluation, which is $1$ here, so the Taylor series does not converge beyond $x = 2$ (and you can check that it doesn't converge at $x = 2$ either).
– user3482749
Nov 30 '18 at 18:31
add a comment |
I am confused as to why $$sum_{ngeq0}(1-x)^n=frac1x$$
only works for $xin (0,2)$. I get that it has a singularity at $x=0$, so that can't work, but there are no singularities for the rest of the positive real line. Why isn't there a power series representation of $frac1x$ which works for the whole positive real line? Are there any series representations of $1/x$ which work for $xin (0,infty)$? I can't find any. Thanks.
sequences-and-series convergence power-series taylor-expansion
edited Dec 1 '18 at 7:31
Asaf Karagila♦
301k32424754
asked Nov 30 '18 at 18:26
clathratus
3,126331
I am confused as to why $$sum_{ngeq0}(1-x)^n=frac1x$$
only works for $xin (0,2)$. I get that it has a singularity at $x=0$, so that can't work, but there are no singularities for the rest of the positive real line. Why isn't there a power series representation of $frac1x$ which works for the whole positive real line? Are there any series representations of $1/x$ which work for $xin (0,infty)$? I can't find any. Thanks.
sequences-and-series convergence power-series taylor-expansion
sequences-and-series convergence power-series taylor-expansion
edited Dec 1 '18 at 7:31
Asaf Karagila♦
301k32424754
asked Nov 30 '18 at 18:26
clathratus
3,126331
edited Dec 1 '18 at 7:31
Asaf Karagila♦
301k32424754
asked Nov 30 '18 at 18:26
clathratus
3,126331
edited Dec 1 '18 at 7:31
Asaf Karagila♦
301k32424754
edited Dec 1 '18 at 7:31
Asaf Karagila♦
301k32424754
edited Dec 1 '18 at 7:31
Asaf Karagila♦
301k32424754
301k32424754
asked Nov 30 '18 at 18:26
clathratus
3,126331
asked Nov 30 '18 at 18:26
clathratus
3,126331
asked Nov 30 '18 at 18:26
clathratus
3,126331
3,126331
The domain of convergence of a real power series is always either a finitely long interval or the whole real line.
– Andreas Blass
Nov 30 '18 at 18:27
3
$sum r^n$ converges iff $-1<r<1$. If $r=1-x$ that's the same as $0<x<2$.
– Lord Shark the Unknown
Nov 30 '18 at 18:28
$1/x$ has a pole at $x=0$.
– Lord Shark the Unknown
Nov 30 '18 at 18:28
1
More specifically: their intervals of convergence are symmetric around the point of evaluation, which is $1$ here, so the Taylor series does not converge beyond $x = 2$ (and you can check that it doesn't converge at $x = 2$ either).
– user3482749
Nov 30 '18 at 18:31
add a comment |
The domain of convergence of a real power series is always either a finitely long interval or the whole real line.
– Andreas Blass
Nov 30 '18 at 18:27
3
$sum r^n$ converges iff $-1<r<1$. If $r=1-x$ that's the same as $0<x<2$.
– Lord Shark the Unknown
Nov 30 '18 at 18:28
$1/x$ has a pole at $x=0$.
– Lord Shark the Unknown
Nov 30 '18 at 18:28
1
More specifically: their intervals of convergence are symmetric around the point of evaluation, which is $1$ here, so the Taylor series does not converge beyond $x = 2$ (and you can check that it doesn't converge at $x = 2$ either).
– user3482749
Nov 30 '18 at 18:31
The domain of convergence of a real power series is always either a finitely long interval or the whole real line.
– Andreas Blass
Nov 30 '18 at 18:27
The domain of convergence of a real power series is always either a finitely long interval or the whole real line.
– Andreas Blass
Nov 30 '18 at 18:27
3
3
$sum r^n$ converges iff $-1<r<1$. If $r=1-x$ that's the same as $0<x<2$.
– Lord Shark the Unknown
Nov 30 '18 at 18:28
$sum r^n$ converges iff $-1<r<1$. If $r=1-x$ that's the same as $0<x<2$.
– Lord Shark the Unknown
Nov 30 '18 at 18:28
$1/x$ has a pole at $x=0$.
– Lord Shark the Unknown
Nov 30 '18 at 18:28
$1/x$ has a pole at $x=0$.
– Lord Shark the Unknown
Nov 30 '18 at 18:28
1
1
More specifically: their intervals of convergence are symmetric around the point of evaluation, which is $1$ here, so the Taylor series does not converge beyond $x = 2$ (and you can check that it doesn't converge at $x = 2$ either).
– user3482749
Nov 30 '18 at 18:31
More specifically: their intervals of convergence are symmetric around the point of evaluation, which is $1$ here, so the Taylor series does not converge beyond $x = 2$ (and you can check that it doesn't converge at $x = 2$ either).
– user3482749
Nov 30 '18 at 18:31
add a comment |
4 Answers
4
active
oldest
votes
The radius of convergence of a power series in $z-a$ is the largest $r$ such that the sum of the series is analytic in the disk ${z: |z-a| < r}$ in the complex plane. Thus a singularity at some point stops the series from converging
at all points farther away from the centre than that point, even though the function may be analytic at those other points.
On the other hand, you could take the series
$$ frac{1}{x} = -1 + sum_{n=1}^infty left(frac{1+x}{1+2x}right)^n$$
which converges for all $x > 0$ (in fact, everywhere outside a circle of radius $1/3$ centred at $-1/3$ in the complex plane).
Of course, this is not a power series in the usual sense.
answered Nov 30 '18 at 18:35
Robert Israel
318k23208457
What type of series is $$-1+sum_{ngeq0}bigg(frac{1+x}{1+2x}bigg)^n$$ and how do you find it? Is this type of series generally better at approximating rational functions?
– clathratus
Nov 30 '18 at 21:36
2
@clathratus It's just a geometric series where $y=frac{1+x}{1+2x}$. It is a special case for $a=2$ of: $$frac{1}{x}+a = frac{1+ax}{x}\ = frac{1+ax}{1+ax - (1+ax-x)}\ = frac{1}{1 - frac{1+(a-1)x}{1+ax}}\ =sum_{n=0}^infty left(frac{1+(a-1)x}{1+ax}right)^n $$ Note that the series given in the posted answer does not include the term $n=0$ as you wrote in your comment.
– adfriedman
Nov 30 '18 at 22:51
@adfriedman I see. So the $$sum_{n=0}^{infty}$$ in your comment is supposed to be different too?
– clathratus
Nov 30 '18 at 23:20
2
No, I wrote what I meant. you could rewrite mine as $$frac{1}{x} = 1-a + sum_{n=1}^infty left(frac{1+(a-1)x}{1+ax}right)^n$$ to be consistent if you'd like. (The $n=0$ term is just $1$, so take your pick on whether you want to index it or not)
– adfriedman
Nov 30 '18 at 23:29
@adfriedman Thank you for the clarification. I understand now.
– clathratus
Dec 1 '18 at 2:31
add a comment |
The basic answer to you question is that a "normal" power series, i.e. with non-negative integral
exponents, including Taylor series, does not have singularities in the finite complex plane
and consequently cannot reproduce them.
A function $ f(z)$ , which is analytic in a domain of the complex plane,
can be developed (by definition) in power series around a point in that domain, and the radius
of convergence of the series will be equal to the distance of that point from the nearest
singularity, of course excluded.
Thus, for $1/z$ which has a simple pole at $z=0$ :
- when developed at $z=1$ will have a convergence radius of $ 1$;
- to get a larger radius( $R$), you shall develop it around $z_0=R$ ;
- there is no possibility to develop it so as to encompass positive and negative real values of it.
The Laurent series for $1/z$ at $z_0=0$ is infact $1/z$, i.e. the function itself.
answered Nov 30 '18 at 21:29
G Cab
17.9k31237
add a comment |
The geometric series $sum_{ngeq0}(1-x)^n$ converges for
$$|1-x|<1 iff -1<1-x<1 iff 0<x<2$$
and we have
$$sum_{ngeq0}(1-x)^n=frac{1}{1-(1-x)}=frac1x$$
answered Nov 30 '18 at 18:29
gimusi
1
add a comment |
The set of those real numbers at which a power series $displaystylesum_{n=0}^infty a_n(x-a)^n$ belongs to one of these types:
${a}$;
$(a-r,a+r)$, for some $rin(0,+infty)$;
$(a-r,a+r]$, for some $rin(0,+infty)$;
$[a-r,a+r)$, for some $rin(0,+infty)$;
$[a-r,a+r]$, for some $rin(0,+infty)$;
$mathbb R$.
Since, $(0,+infty)$ doesn't appear here, no power series exists whose sum is $frac1x$ in $(0,+infty)$ and which diverges otherwise.
answered Nov 30 '18 at 18:32
José Carlos Santos
150k22121221
add a comment |
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4 Answers
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active
oldest
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4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
The radius of convergence of a power series in $z-a$ is the largest $r$ such that the sum of the series is analytic in the disk ${z: |z-a| < r}$ in the complex plane. Thus a singularity at some point stops the series from converging
at all points farther away from the centre than that point, even though the function may be analytic at those other points.
On the other hand, you could take the series
$$ frac{1}{x} = -1 + sum_{n=1}^infty left(frac{1+x}{1+2x}right)^n$$
which converges for all $x > 0$ (in fact, everywhere outside a circle of radius $1/3$ centred at $-1/3$ in the complex plane).
Of course, this is not a power series in the usual sense.
answered Nov 30 '18 at 18:35
Robert Israel
318k23208457
What type of series is $$-1+sum_{ngeq0}bigg(frac{1+x}{1+2x}bigg)^n$$ and how do you find it? Is this type of series generally better at approximating rational functions?
– clathratus
Nov 30 '18 at 21:36
2
@clathratus It's just a geometric series where $y=frac{1+x}{1+2x}$. It is a special case for $a=2$ of: $$frac{1}{x}+a = frac{1+ax}{x}\ = frac{1+ax}{1+ax - (1+ax-x)}\ = frac{1}{1 - frac{1+(a-1)x}{1+ax}}\ =sum_{n=0}^infty left(frac{1+(a-1)x}{1+ax}right)^n $$ Note that the series given in the posted answer does not include the term $n=0$ as you wrote in your comment.
– adfriedman
Nov 30 '18 at 22:51
@adfriedman I see. So the $$sum_{n=0}^{infty}$$ in your comment is supposed to be different too?
– clathratus
Nov 30 '18 at 23:20
2
No, I wrote what I meant. you could rewrite mine as $$frac{1}{x} = 1-a + sum_{n=1}^infty left(frac{1+(a-1)x}{1+ax}right)^n$$ to be consistent if you'd like. (The $n=0$ term is just $1$, so take your pick on whether you want to index it or not)
– adfriedman
Nov 30 '18 at 23:29
@adfriedman Thank you for the clarification. I understand now.
– clathratus
Dec 1 '18 at 2:31
add a comment |
The radius of convergence of a power series in $z-a$ is the largest $r$ such that the sum of the series is analytic in the disk ${z: |z-a| < r}$ in the complex plane. Thus a singularity at some point stops the series from converging
at all points farther away from the centre than that point, even though the function may be analytic at those other points.
On the other hand, you could take the series
$$ frac{1}{x} = -1 + sum_{n=1}^infty left(frac{1+x}{1+2x}right)^n$$
which converges for all $x > 0$ (in fact, everywhere outside a circle of radius $1/3$ centred at $-1/3$ in the complex plane).
Of course, this is not a power series in the usual sense.
answered Nov 30 '18 at 18:35
Robert Israel
318k23208457
What type of series is $$-1+sum_{ngeq0}bigg(frac{1+x}{1+2x}bigg)^n$$ and how do you find it? Is this type of series generally better at approximating rational functions?
– clathratus
Nov 30 '18 at 21:36
2
@clathratus It's just a geometric series where $y=frac{1+x}{1+2x}$. It is a special case for $a=2$ of: $$frac{1}{x}+a = frac{1+ax}{x}\ = frac{1+ax}{1+ax - (1+ax-x)}\ = frac{1}{1 - frac{1+(a-1)x}{1+ax}}\ =sum_{n=0}^infty left(frac{1+(a-1)x}{1+ax}right)^n $$ Note that the series given in the posted answer does not include the term $n=0$ as you wrote in your comment.
– adfriedman
Nov 30 '18 at 22:51
@adfriedman I see. So the $$sum_{n=0}^{infty}$$ in your comment is supposed to be different too?
– clathratus
Nov 30 '18 at 23:20
2
No, I wrote what I meant. you could rewrite mine as $$frac{1}{x} = 1-a + sum_{n=1}^infty left(frac{1+(a-1)x}{1+ax}right)^n$$ to be consistent if you'd like. (The $n=0$ term is just $1$, so take your pick on whether you want to index it or not)
– adfriedman
Nov 30 '18 at 23:29
@adfriedman Thank you for the clarification. I understand now.
– clathratus
Dec 1 '18 at 2:31
add a comment |
The radius of convergence of a power series in $z-a$ is the largest $r$ such that the sum of the series is analytic in the disk ${z: |z-a| < r}$ in the complex plane. Thus a singularity at some point stops the series from converging
at all points farther away from the centre than that point, even though the function may be analytic at those other points.
On the other hand, you could take the series
$$ frac{1}{x} = -1 + sum_{n=1}^infty left(frac{1+x}{1+2x}right)^n$$
which converges for all $x > 0$ (in fact, everywhere outside a circle of radius $1/3$ centred at $-1/3$ in the complex plane).
Of course, this is not a power series in the usual sense.
answered Nov 30 '18 at 18:35
Robert Israel
318k23208457
The radius of convergence of a power series in $z-a$ is the largest $r$ such that the sum of the series is analytic in the disk ${z: |z-a| < r}$ in the complex plane. Thus a singularity at some point stops the series from converging
at all points farther away from the centre than that point, even though the function may be analytic at those other points.
On the other hand, you could take the series
$$ frac{1}{x} = -1 + sum_{n=1}^infty left(frac{1+x}{1+2x}right)^n$$
which converges for all $x > 0$ (in fact, everywhere outside a circle of radius $1/3$ centred at $-1/3$ in the complex plane).
Of course, this is not a power series in the usual sense.
answered Nov 30 '18 at 18:35
Robert Israel
318k23208457
edited Nov 30 '18 at 18:47
answered Nov 30 '18 at 18:35
Robert Israel
318k23208457
answered Nov 30 '18 at 18:35
Robert Israel
318k23208457
answered Nov 30 '18 at 18:35
Robert Israel
318k23208457
318k23208457
What type of series is $$-1+sum_{ngeq0}bigg(frac{1+x}{1+2x}bigg)^n$$ and how do you find it? Is this type of series generally better at approximating rational functions?
– clathratus
Nov 30 '18 at 21:36
2
@clathratus It's just a geometric series where $y=frac{1+x}{1+2x}$. It is a special case for $a=2$ of: $$frac{1}{x}+a = frac{1+ax}{x}\ = frac{1+ax}{1+ax - (1+ax-x)}\ = frac{1}{1 - frac{1+(a-1)x}{1+ax}}\ =sum_{n=0}^infty left(frac{1+(a-1)x}{1+ax}right)^n $$ Note that the series given in the posted answer does not include the term $n=0$ as you wrote in your comment.
– adfriedman
Nov 30 '18 at 22:51
@adfriedman I see. So the $$sum_{n=0}^{infty}$$ in your comment is supposed to be different too?
– clathratus
Nov 30 '18 at 23:20
2
No, I wrote what I meant. you could rewrite mine as $$frac{1}{x} = 1-a + sum_{n=1}^infty left(frac{1+(a-1)x}{1+ax}right)^n$$ to be consistent if you'd like. (The $n=0$ term is just $1$, so take your pick on whether you want to index it or not)
– adfriedman
Nov 30 '18 at 23:29
@adfriedman Thank you for the clarification. I understand now.
– clathratus
Dec 1 '18 at 2:31
add a comment |
What type of series is $$-1+sum_{ngeq0}bigg(frac{1+x}{1+2x}bigg)^n$$ and how do you find it? Is this type of series generally better at approximating rational functions?
– clathratus
Nov 30 '18 at 21:36
2
@clathratus It's just a geometric series where $y=frac{1+x}{1+2x}$. It is a special case for $a=2$ of: $$frac{1}{x}+a = frac{1+ax}{x}\ = frac{1+ax}{1+ax - (1+ax-x)}\ = frac{1}{1 - frac{1+(a-1)x}{1+ax}}\ =sum_{n=0}^infty left(frac{1+(a-1)x}{1+ax}right)^n $$ Note that the series given in the posted answer does not include the term $n=0$ as you wrote in your comment.
– adfriedman
Nov 30 '18 at 22:51
@adfriedman I see. So the $$sum_{n=0}^{infty}$$ in your comment is supposed to be different too?
– clathratus
Nov 30 '18 at 23:20
2
No, I wrote what I meant. you could rewrite mine as $$frac{1}{x} = 1-a + sum_{n=1}^infty left(frac{1+(a-1)x}{1+ax}right)^n$$ to be consistent if you'd like. (The $n=0$ term is just $1$, so take your pick on whether you want to index it or not)
– adfriedman
Nov 30 '18 at 23:29
@adfriedman Thank you for the clarification. I understand now.
– clathratus
Dec 1 '18 at 2:31
What type of series is $$-1+sum_{ngeq0}bigg(frac{1+x}{1+2x}bigg)^n$$ and how do you find it? Is this type of series generally better at approximating rational functions?
– clathratus
Nov 30 '18 at 21:36
What type of series is $$-1+sum_{ngeq0}bigg(frac{1+x}{1+2x}bigg)^n$$ and how do you find it? Is this type of series generally better at approximating rational functions?
– clathratus
Nov 30 '18 at 21:36
2
2
@clathratus It's just a geometric series where $y=frac{1+x}{1+2x}$. It is a special case for $a=2$ of: $$frac{1}{x}+a = frac{1+ax}{x}\ = frac{1+ax}{1+ax - (1+ax-x)}\ = frac{1}{1 - frac{1+(a-1)x}{1+ax}}\ =sum_{n=0}^infty left(frac{1+(a-1)x}{1+ax}right)^n $$ Note that the series given in the posted answer does not include the term $n=0$ as you wrote in your comment.
– adfriedman
Nov 30 '18 at 22:51
@clathratus It's just a geometric series where $y=frac{1+x}{1+2x}$. It is a special case for $a=2$ of: $$frac{1}{x}+a = frac{1+ax}{x}\ = frac{1+ax}{1+ax - (1+ax-x)}\ = frac{1}{1 - frac{1+(a-1)x}{1+ax}}\ =sum_{n=0}^infty left(frac{1+(a-1)x}{1+ax}right)^n $$ Note that the series given in the posted answer does not include the term $n=0$ as you wrote in your comment.
– adfriedman
Nov 30 '18 at 22:51
@adfriedman I see. So the $$sum_{n=0}^{infty}$$ in your comment is supposed to be different too?
– clathratus
Nov 30 '18 at 23:20
@adfriedman I see. So the $$sum_{n=0}^{infty}$$ in your comment is supposed to be different too?
– clathratus
Nov 30 '18 at 23:20
2
2
No, I wrote what I meant. you could rewrite mine as $$frac{1}{x} = 1-a + sum_{n=1}^infty left(frac{1+(a-1)x}{1+ax}right)^n$$ to be consistent if you'd like. (The $n=0$ term is just $1$, so take your pick on whether you want to index it or not)
– adfriedman
Nov 30 '18 at 23:29
No, I wrote what I meant. you could rewrite mine as $$frac{1}{x} = 1-a + sum_{n=1}^infty left(frac{1+(a-1)x}{1+ax}right)^n$$ to be consistent if you'd like. (The $n=0$ term is just $1$, so take your pick on whether you want to index it or not)
– adfriedman
Nov 30 '18 at 23:29
@adfriedman Thank you for the clarification. I understand now.
– clathratus
Dec 1 '18 at 2:31
@adfriedman Thank you for the clarification. I understand now.
– clathratus
Dec 1 '18 at 2:31
add a comment |
The basic answer to you question is that a "normal" power series, i.e. with non-negative integral
exponents, including Taylor series, does not have singularities in the finite complex plane
and consequently cannot reproduce them.
A function $ f(z)$ , which is analytic in a domain of the complex plane,
can be developed (by definition) in power series around a point in that domain, and the radius
of convergence of the series will be equal to the distance of that point from the nearest
singularity, of course excluded.
Thus, for $1/z$ which has a simple pole at $z=0$ :
- when developed at $z=1$ will have a convergence radius of $ 1$;
- to get a larger radius( $R$), you shall develop it around $z_0=R$ ;
- there is no possibility to develop it so as to encompass positive and negative real values of it.
The Laurent series for $1/z$ at $z_0=0$ is infact $1/z$, i.e. the function itself.
answered Nov 30 '18 at 21:29
G Cab
17.9k31237
add a comment |
The basic answer to you question is that a "normal" power series, i.e. with non-negative integral
exponents, including Taylor series, does not have singularities in the finite complex plane
and consequently cannot reproduce them.
A function $ f(z)$ , which is analytic in a domain of the complex plane,
can be developed (by definition) in power series around a point in that domain, and the radius
of convergence of the series will be equal to the distance of that point from the nearest
singularity, of course excluded.
Thus, for $1/z$ which has a simple pole at $z=0$ :
- when developed at $z=1$ will have a convergence radius of $ 1$;
- to get a larger radius( $R$), you shall develop it around $z_0=R$ ;
- there is no possibility to develop it so as to encompass positive and negative real values of it.
The Laurent series for $1/z$ at $z_0=0$ is infact $1/z$, i.e. the function itself.
answered Nov 30 '18 at 21:29
G Cab
17.9k31237
add a comment |
The basic answer to you question is that a "normal" power series, i.e. with non-negative integral
exponents, including Taylor series, does not have singularities in the finite complex plane
and consequently cannot reproduce them.
A function $ f(z)$ , which is analytic in a domain of the complex plane,
can be developed (by definition) in power series around a point in that domain, and the radius
of convergence of the series will be equal to the distance of that point from the nearest
singularity, of course excluded.
Thus, for $1/z$ which has a simple pole at $z=0$ :
- when developed at $z=1$ will have a convergence radius of $ 1$;
- to get a larger radius( $R$), you shall develop it around $z_0=R$ ;
- there is no possibility to develop it so as to encompass positive and negative real values of it.
The Laurent series for $1/z$ at $z_0=0$ is infact $1/z$, i.e. the function itself.
answered Nov 30 '18 at 21:29
G Cab
17.9k31237
The basic answer to you question is that a "normal" power series, i.e. with non-negative integral
exponents, including Taylor series, does not have singularities in the finite complex plane
and consequently cannot reproduce them.
A function $ f(z)$ , which is analytic in a domain of the complex plane,
can be developed (by definition) in power series around a point in that domain, and the radius
of convergence of the series will be equal to the distance of that point from the nearest
singularity, of course excluded.
Thus, for $1/z$ which has a simple pole at $z=0$ :
- when developed at $z=1$ will have a convergence radius of $ 1$;
- to get a larger radius( $R$), you shall develop it around $z_0=R$ ;
- there is no possibility to develop it so as to encompass positive and negative real values of it.
The Laurent series for $1/z$ at $z_0=0$ is infact $1/z$, i.e. the function itself.
answered Nov 30 '18 at 21:29
G Cab
17.9k31237
edited Dec 1 '18 at 16:53
answered Nov 30 '18 at 21:29
G Cab
17.9k31237
answered Nov 30 '18 at 21:29
G Cab
17.9k31237
answered Nov 30 '18 at 21:29
G Cab
17.9k31237
17.9k31237
add a comment |
add a comment |
The geometric series $sum_{ngeq0}(1-x)^n$ converges for
$$|1-x|<1 iff -1<1-x<1 iff 0<x<2$$
and we have
$$sum_{ngeq0}(1-x)^n=frac{1}{1-(1-x)}=frac1x$$
answered Nov 30 '18 at 18:29
gimusi
1
add a comment |
The geometric series $sum_{ngeq0}(1-x)^n$ converges for
$$|1-x|<1 iff -1<1-x<1 iff 0<x<2$$
and we have
$$sum_{ngeq0}(1-x)^n=frac{1}{1-(1-x)}=frac1x$$
answered Nov 30 '18 at 18:29
gimusi
1
add a comment |
The geometric series $sum_{ngeq0}(1-x)^n$ converges for
$$|1-x|<1 iff -1<1-x<1 iff 0<x<2$$
and we have
$$sum_{ngeq0}(1-x)^n=frac{1}{1-(1-x)}=frac1x$$
answered Nov 30 '18 at 18:29
gimusi
1
The geometric series $sum_{ngeq0}(1-x)^n$ converges for
$$|1-x|<1 iff -1<1-x<1 iff 0<x<2$$
and we have
$$sum_{ngeq0}(1-x)^n=frac{1}{1-(1-x)}=frac1x$$
answered Nov 30 '18 at 18:29
gimusi
1
answered Nov 30 '18 at 18:29
gimusi
1
answered Nov 30 '18 at 18:29
gimusi
1
answered Nov 30 '18 at 18:29
gimusi
1
1
add a comment |
add a comment |
The set of those real numbers at which a power series $displaystylesum_{n=0}^infty a_n(x-a)^n$ belongs to one of these types:
${a}$;
$(a-r,a+r)$, for some $rin(0,+infty)$;
$(a-r,a+r]$, for some $rin(0,+infty)$;
$[a-r,a+r)$, for some $rin(0,+infty)$;
$[a-r,a+r]$, for some $rin(0,+infty)$;
$mathbb R$.
Since, $(0,+infty)$ doesn't appear here, no power series exists whose sum is $frac1x$ in $(0,+infty)$ and which diverges otherwise.
answered Nov 30 '18 at 18:32
José Carlos Santos
150k22121221
add a comment |
The set of those real numbers at which a power series $displaystylesum_{n=0}^infty a_n(x-a)^n$ belongs to one of these types:
${a}$;
$(a-r,a+r)$, for some $rin(0,+infty)$;
$(a-r,a+r]$, for some $rin(0,+infty)$;
$[a-r,a+r)$, for some $rin(0,+infty)$;
$[a-r,a+r]$, for some $rin(0,+infty)$;
$mathbb R$.
Since, $(0,+infty)$ doesn't appear here, no power series exists whose sum is $frac1x$ in $(0,+infty)$ and which diverges otherwise.
answered Nov 30 '18 at 18:32
José Carlos Santos
150k22121221
add a comment |
The set of those real numbers at which a power series $displaystylesum_{n=0}^infty a_n(x-a)^n$ belongs to one of these types:
${a}$;
$(a-r,a+r)$, for some $rin(0,+infty)$;
$(a-r,a+r]$, for some $rin(0,+infty)$;
$[a-r,a+r)$, for some $rin(0,+infty)$;
$[a-r,a+r]$, for some $rin(0,+infty)$;
$mathbb R$.
Since, $(0,+infty)$ doesn't appear here, no power series exists whose sum is $frac1x$ in $(0,+infty)$ and which diverges otherwise.
answered Nov 30 '18 at 18:32
José Carlos Santos
150k22121221
The set of those real numbers at which a power series $displaystylesum_{n=0}^infty a_n(x-a)^n$ belongs to one of these types:
${a}$;
$(a-r,a+r)$, for some $rin(0,+infty)$;
$(a-r,a+r]$, for some $rin(0,+infty)$;
$[a-r,a+r)$, for some $rin(0,+infty)$;
$[a-r,a+r]$, for some $rin(0,+infty)$;
$mathbb R$.
Since, $(0,+infty)$ doesn't appear here, no power series exists whose sum is $frac1x$ in $(0,+infty)$ and which diverges otherwise.
answered Nov 30 '18 at 18:32
José Carlos Santos
150k22121221
answered Nov 30 '18 at 18:32
José Carlos Santos
150k22121221
answered Nov 30 '18 at 18:32
José Carlos Santos
150k22121221
answered Nov 30 '18 at 18:32
José Carlos Santos
150k22121221
150k22121221
add a comment |
add a comment |
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The domain of convergence of a real power series is always either a finitely long interval or the whole real line.
– Andreas Blass
Nov 30 '18 at 18:27
3
$sum r^n$ converges iff $-1<r<1$. If $r=1-x$ that's the same as $0<x<2$.
– Lord Shark the Unknown
Nov 30 '18 at 18:28
$1/x$ has a pole at $x=0$.
– Lord Shark the Unknown
Nov 30 '18 at 18:28
1
More specifically: their intervals of convergence are symmetric around the point of evaluation, which is $1$ here, so the Taylor series does not converge beyond $x = 2$ (and you can check that it doesn't converge at $x = 2$ either).
– user3482749
Nov 30 '18 at 18:31