Is f the zero function?
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I am told to find the error in the following argument:
let $f(z)= cdots + frac{1}{z^2} + frac{1}{z} + 1 + z + z^2 + cdots$
Note that $z + z^2 + cdots = frac{z}{1-z}$
and $1 + frac{1}{z} + frac{1}{z^2} + cdots = frac{-z}{1-z}$
hence f(z) = 0
My argument is to point out that $z + z^2 + cdots = frac{z}{1-z}$
only holds when |z| < 1 and $1 + frac{1}{z} + frac{1}{z^2} + cdots = frac{-z}{1-z}$ can only hold in the region where 1 < |z|. Is this a sufficient counterargument or is there something else I am missing because this argument seems too simple.
thank you
complex-analysis laurent-series
asked Nov 23 at 23:22
deco
666
add a comment |
up vote
2
down vote
favorite
I am told to find the error in the following argument:
let $f(z)= cdots + frac{1}{z^2} + frac{1}{z} + 1 + z + z^2 + cdots$
Note that $z + z^2 + cdots = frac{z}{1-z}$
and $1 + frac{1}{z} + frac{1}{z^2} + cdots = frac{-z}{1-z}$
hence f(z) = 0
My argument is to point out that $z + z^2 + cdots = frac{z}{1-z}$
only holds when |z| < 1 and $1 + frac{1}{z} + frac{1}{z^2} + cdots = frac{-z}{1-z}$ can only hold in the region where 1 < |z|. Is this a sufficient counterargument or is there something else I am missing because this argument seems too simple.
thank you
complex-analysis laurent-series
asked Nov 23 at 23:22
deco
666
It’s exactly what I would have said. The doubly-infinite series presented to you is nowhere convergent,
– Lubin
Nov 23 at 23:27
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am told to find the error in the following argument:
let $f(z)= cdots + frac{1}{z^2} + frac{1}{z} + 1 + z + z^2 + cdots$
Note that $z + z^2 + cdots = frac{z}{1-z}$
and $1 + frac{1}{z} + frac{1}{z^2} + cdots = frac{-z}{1-z}$
hence f(z) = 0
My argument is to point out that $z + z^2 + cdots = frac{z}{1-z}$
only holds when |z| < 1 and $1 + frac{1}{z} + frac{1}{z^2} + cdots = frac{-z}{1-z}$ can only hold in the region where 1 < |z|. Is this a sufficient counterargument or is there something else I am missing because this argument seems too simple.
thank you
complex-analysis laurent-series
asked Nov 23 at 23:22
deco
666
I am told to find the error in the following argument:
let $f(z)= cdots + frac{1}{z^2} + frac{1}{z} + 1 + z + z^2 + cdots$
Note that $z + z^2 + cdots = frac{z}{1-z}$
and $1 + frac{1}{z} + frac{1}{z^2} + cdots = frac{-z}{1-z}$
hence f(z) = 0
My argument is to point out that $z + z^2 + cdots = frac{z}{1-z}$
only holds when |z| < 1 and $1 + frac{1}{z} + frac{1}{z^2} + cdots = frac{-z}{1-z}$ can only hold in the region where 1 < |z|. Is this a sufficient counterargument or is there something else I am missing because this argument seems too simple.
thank you
complex-analysis laurent-series
complex-analysis laurent-series
asked Nov 23 at 23:22
deco
666
asked Nov 23 at 23:22
deco
666
asked Nov 23 at 23:22
deco
666
asked Nov 23 at 23:22
deco
666
asked Nov 23 at 23:22
deco
666
666
It’s exactly what I would have said. The doubly-infinite series presented to you is nowhere convergent,
– Lubin
Nov 23 at 23:27
add a comment |
It’s exactly what I would have said. The doubly-infinite series presented to you is nowhere convergent,
– Lubin
Nov 23 at 23:27
It’s exactly what I would have said. The doubly-infinite series presented to you is nowhere convergent,
– Lubin
Nov 23 at 23:27
It’s exactly what I would have said. The doubly-infinite series presented to you is nowhere convergent,
– Lubin
Nov 23 at 23:27
add a comment |
1 Answer
1
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oldest
votes
up vote
2
down vote
accepted
The argument actually is correct that $f(z)=0$ for all $z$ such that $f(z)$ is defined. But, as you say, $f(z)$ is actually not defined for any $z$ at all, since there do not exist any $z$ such that the series defining $f(z)$ converges on both sides. So, $f(z)=0$ is true for all $z$ such that $f(z)$ is defined, but this is a vacuous statement.
answered Nov 23 at 23:27
Eric Wofsey
176k12202327
There is no point where both the series converge so $f(z)$ is not defined for any $z$.
– Kavi Rama Murthy
Nov 23 at 23:50
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The argument actually is correct that $f(z)=0$ for all $z$ such that $f(z)$ is defined. But, as you say, $f(z)$ is actually not defined for any $z$ at all, since there do not exist any $z$ such that the series defining $f(z)$ converges on both sides. So, $f(z)=0$ is true for all $z$ such that $f(z)$ is defined, but this is a vacuous statement.
answered Nov 23 at 23:27
Eric Wofsey
176k12202327
There is no point where both the series converge so $f(z)$ is not defined for any $z$.
– Kavi Rama Murthy
Nov 23 at 23:50
add a comment |
up vote
2
down vote
accepted
The argument actually is correct that $f(z)=0$ for all $z$ such that $f(z)$ is defined. But, as you say, $f(z)$ is actually not defined for any $z$ at all, since there do not exist any $z$ such that the series defining $f(z)$ converges on both sides. So, $f(z)=0$ is true for all $z$ such that $f(z)$ is defined, but this is a vacuous statement.
answered Nov 23 at 23:27
Eric Wofsey
176k12202327
There is no point where both the series converge so $f(z)$ is not defined for any $z$.
– Kavi Rama Murthy
Nov 23 at 23:50
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The argument actually is correct that $f(z)=0$ for all $z$ such that $f(z)$ is defined. But, as you say, $f(z)$ is actually not defined for any $z$ at all, since there do not exist any $z$ such that the series defining $f(z)$ converges on both sides. So, $f(z)=0$ is true for all $z$ such that $f(z)$ is defined, but this is a vacuous statement.
answered Nov 23 at 23:27
Eric Wofsey
176k12202327
The argument actually is correct that $f(z)=0$ for all $z$ such that $f(z)$ is defined. But, as you say, $f(z)$ is actually not defined for any $z$ at all, since there do not exist any $z$ such that the series defining $f(z)$ converges on both sides. So, $f(z)=0$ is true for all $z$ such that $f(z)$ is defined, but this is a vacuous statement.
answered Nov 23 at 23:27
Eric Wofsey
176k12202327
answered Nov 23 at 23:27
Eric Wofsey
176k12202327
answered Nov 23 at 23:27
Eric Wofsey
176k12202327
answered Nov 23 at 23:27
Eric Wofsey
176k12202327
176k12202327
There is no point where both the series converge so $f(z)$ is not defined for any $z$.
– Kavi Rama Murthy
Nov 23 at 23:50
add a comment |
There is no point where both the series converge so $f(z)$ is not defined for any $z$.
– Kavi Rama Murthy
Nov 23 at 23:50
There is no point where both the series converge so $f(z)$ is not defined for any $z$.
– Kavi Rama Murthy
Nov 23 at 23:50
There is no point where both the series converge so $f(z)$ is not defined for any $z$.
– Kavi Rama Murthy
Nov 23 at 23:50
add a comment |
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It’s exactly what I would have said. The doubly-infinite series presented to you is nowhere convergent,
– Lubin
Nov 23 at 23:27