Series representation for a specific range
0
$begingroup$
I am wondering if there is a valid series representation using:
$f(z) = sum_{k=-infty}^{infty} a_k(z-z_0)^k$
for $r<|z−z_0|<R$
Why is this not possible?
sequences-and-series complex-numbers laurent-series
asked Dec 7 '18 at 16:04
LechugaLechuga
105
$endgroup$
|
show 2 more comments
0
$begingroup$
I am wondering if there is a valid series representation using:
$f(z) = sum_{k=-infty}^{infty} a_k(z-z_0)^k$
for $r<|z−z_0|<R$
Why is this not possible?
sequences-and-series complex-numbers laurent-series
asked Dec 7 '18 at 16:04
LechugaLechuga
105
$endgroup$
$begingroup$
It was a typo, I have fixed it!
$endgroup$
– Lechuga
Dec 7 '18 at 16:17
$begingroup$
A function of this form is called analytic. In particular you can see that such function has to be continuous. So any non-continuous function can't take that form.
$endgroup$
– Yanko
Dec 7 '18 at 16:18
$begingroup$
Would this function satisfy the requirements? I think it does since it is from negative infinity to positive infinity
$endgroup$
– Lechuga
Dec 7 '18 at 16:21
$begingroup$
I might misinterpret the question. I understand it as follows "why it isn't possible to express every function as $f(z)=sum_{k=-infty}^{infty} a_k(z-z_0)^k$". If this is what you meant then the answer is that the sum is always continuous.
$endgroup$
– Yanko
Dec 7 '18 at 16:23
$begingroup$
Does that mean that as long as the function is continuous we can represent it using that sum?
$endgroup$
– Lechuga
Dec 7 '18 at 16:26
|
show 2 more comments
0
0
0
$begingroup$
I am wondering if there is a valid series representation using:
$f(z) = sum_{k=-infty}^{infty} a_k(z-z_0)^k$
for $r<|z−z_0|<R$
Why is this not possible?
sequences-and-series complex-numbers laurent-series
asked Dec 7 '18 at 16:04
LechugaLechuga
105
$endgroup$
I am wondering if there is a valid series representation using:
$f(z) = sum_{k=-infty}^{infty} a_k(z-z_0)^k$
for $r<|z−z_0|<R$
Why is this not possible?
sequences-and-series complex-numbers laurent-series
sequences-and-series complex-numbers laurent-series
asked Dec 7 '18 at 16:04
LechugaLechuga
105
asked Dec 7 '18 at 16:04
LechugaLechuga
105
edited Dec 7 '18 at 16:17
Lechuga
asked Dec 7 '18 at 16:04
LechugaLechuga
105
asked Dec 7 '18 at 16:04
LechugaLechuga
105
asked Dec 7 '18 at 16:04
LechugaLechuga
105
105
$begingroup$
It was a typo, I have fixed it!
$endgroup$
– Lechuga
Dec 7 '18 at 16:17
$begingroup$
A function of this form is called analytic. In particular you can see that such function has to be continuous. So any non-continuous function can't take that form.
$endgroup$
– Yanko
Dec 7 '18 at 16:18
$begingroup$
Would this function satisfy the requirements? I think it does since it is from negative infinity to positive infinity
$endgroup$
– Lechuga
Dec 7 '18 at 16:21
$begingroup$
I might misinterpret the question. I understand it as follows "why it isn't possible to express every function as $f(z)=sum_{k=-infty}^{infty} a_k(z-z_0)^k$". If this is what you meant then the answer is that the sum is always continuous.
$endgroup$
– Yanko
Dec 7 '18 at 16:23
$begingroup$
Does that mean that as long as the function is continuous we can represent it using that sum?
$endgroup$
– Lechuga
Dec 7 '18 at 16:26
|
show 2 more comments
$begingroup$
It was a typo, I have fixed it!
$endgroup$
– Lechuga
Dec 7 '18 at 16:17
$begingroup$
A function of this form is called analytic. In particular you can see that such function has to be continuous. So any non-continuous function can't take that form.
$endgroup$
– Yanko
Dec 7 '18 at 16:18
$begingroup$
Would this function satisfy the requirements? I think it does since it is from negative infinity to positive infinity
$endgroup$
– Lechuga
Dec 7 '18 at 16:21
$begingroup$
I might misinterpret the question. I understand it as follows "why it isn't possible to express every function as $f(z)=sum_{k=-infty}^{infty} a_k(z-z_0)^k$". If this is what you meant then the answer is that the sum is always continuous.
$endgroup$
– Yanko
Dec 7 '18 at 16:23
$begingroup$
Does that mean that as long as the function is continuous we can represent it using that sum?
$endgroup$
– Lechuga
Dec 7 '18 at 16:26
$begingroup$
It was a typo, I have fixed it!
$endgroup$
– Lechuga
Dec 7 '18 at 16:17
$begingroup$
It was a typo, I have fixed it!
$endgroup$
– Lechuga
Dec 7 '18 at 16:17
$begingroup$
A function of this form is called analytic. In particular you can see that such function has to be continuous. So any non-continuous function can't take that form.
$endgroup$
– Yanko
Dec 7 '18 at 16:18
$begingroup$
A function of this form is called analytic. In particular you can see that such function has to be continuous. So any non-continuous function can't take that form.
$endgroup$
– Yanko
Dec 7 '18 at 16:18
$begingroup$
Would this function satisfy the requirements? I think it does since it is from negative infinity to positive infinity
$endgroup$
– Lechuga
Dec 7 '18 at 16:21
$begingroup$
Would this function satisfy the requirements? I think it does since it is from negative infinity to positive infinity
$endgroup$
– Lechuga
Dec 7 '18 at 16:21
$begingroup$
I might misinterpret the question. I understand it as follows "why it isn't possible to express every function as $f(z)=sum_{k=-infty}^{infty} a_k(z-z_0)^k$". If this is what you meant then the answer is that the sum is always continuous.
$endgroup$
– Yanko
Dec 7 '18 at 16:23
$begingroup$
I might misinterpret the question. I understand it as follows "why it isn't possible to express every function as $f(z)=sum_{k=-infty}^{infty} a_k(z-z_0)^k$". If this is what you meant then the answer is that the sum is always continuous.
$endgroup$
– Yanko
Dec 7 '18 at 16:23
$begingroup$
Does that mean that as long as the function is continuous we can represent it using that sum?
$endgroup$
– Lechuga
Dec 7 '18 at 16:26
$begingroup$
Does that mean that as long as the function is continuous we can represent it using that sum?
$endgroup$
– Lechuga
Dec 7 '18 at 16:26
|
show 2 more comments
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$begingroup$
It was a typo, I have fixed it!
$endgroup$
– Lechuga
Dec 7 '18 at 16:17
$begingroup$
A function of this form is called analytic. In particular you can see that such function has to be continuous. So any non-continuous function can't take that form.
$endgroup$
– Yanko
Dec 7 '18 at 16:18
$begingroup$
Would this function satisfy the requirements? I think it does since it is from negative infinity to positive infinity
$endgroup$
– Lechuga
Dec 7 '18 at 16:21
$begingroup$
I might misinterpret the question. I understand it as follows "why it isn't possible to express every function as $f(z)=sum_{k=-infty}^{infty} a_k(z-z_0)^k$". If this is what you meant then the answer is that the sum is always continuous.
$endgroup$
– Yanko
Dec 7 '18 at 16:23
$begingroup$
Does that mean that as long as the function is continuous we can represent it using that sum?
$endgroup$
– Lechuga
Dec 7 '18 at 16:26