Series representation for a specific range












0














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?










share|cite|improve this question
























  • It was a typo, I have fixed it!
    – Lechuga
    Dec 7 at 16:17










  • 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.
    – Yanko
    Dec 7 at 16:18










  • Would this function satisfy the requirements? I think it does since it is from negative infinity to positive infinity
    – Lechuga
    Dec 7 at 16:21












  • 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.
    – Yanko
    Dec 7 at 16:23












  • Does that mean that as long as the function is continuous we can represent it using that sum?
    – Lechuga
    Dec 7 at 16:26
















0














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?










share|cite|improve this question
























  • It was a typo, I have fixed it!
    – Lechuga
    Dec 7 at 16:17










  • 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.
    – Yanko
    Dec 7 at 16:18










  • Would this function satisfy the requirements? I think it does since it is from negative infinity to positive infinity
    – Lechuga
    Dec 7 at 16:21












  • 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.
    – Yanko
    Dec 7 at 16:23












  • Does that mean that as long as the function is continuous we can represent it using that sum?
    – Lechuga
    Dec 7 at 16:26














0












0








0







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?










share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 at 16:17

























asked Dec 7 at 16:04









Lechuga

105




105












  • It was a typo, I have fixed it!
    – Lechuga
    Dec 7 at 16:17










  • 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.
    – Yanko
    Dec 7 at 16:18










  • Would this function satisfy the requirements? I think it does since it is from negative infinity to positive infinity
    – Lechuga
    Dec 7 at 16:21












  • 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.
    – Yanko
    Dec 7 at 16:23












  • Does that mean that as long as the function is continuous we can represent it using that sum?
    – Lechuga
    Dec 7 at 16:26


















  • It was a typo, I have fixed it!
    – Lechuga
    Dec 7 at 16:17










  • 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.
    – Yanko
    Dec 7 at 16:18










  • Would this function satisfy the requirements? I think it does since it is from negative infinity to positive infinity
    – Lechuga
    Dec 7 at 16:21












  • 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.
    – Yanko
    Dec 7 at 16:23












  • Does that mean that as long as the function is continuous we can represent it using that sum?
    – Lechuga
    Dec 7 at 16:26
















It was a typo, I have fixed it!
– Lechuga
Dec 7 at 16:17




It was a typo, I have fixed it!
– Lechuga
Dec 7 at 16:17












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.
– Yanko
Dec 7 at 16:18




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.
– Yanko
Dec 7 at 16:18












Would this function satisfy the requirements? I think it does since it is from negative infinity to positive infinity
– Lechuga
Dec 7 at 16:21






Would this function satisfy the requirements? I think it does since it is from negative infinity to positive infinity
– Lechuga
Dec 7 at 16:21














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.
– Yanko
Dec 7 at 16:23






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.
– Yanko
Dec 7 at 16:23














Does that mean that as long as the function is continuous we can represent it using that sum?
– Lechuga
Dec 7 at 16:26




Does that mean that as long as the function is continuous we can represent it using that sum?
– Lechuga
Dec 7 at 16:26















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