show that they have the same formal power series coefficients












2














$ underline{text{This is about formal power series:}}$



If two power series (in one or two variables) converge on a disc and agree on infinitely many values within that disc, then show that they have the same formal power series coefficients.



Answer:



Let $f(x)=sum_{n geq 0} a_n x^n$ and $g(x)=sum_{n geq 0} b_n x^n$ be two power series converges on the disc $D(r^{-})={x in mathbb{C}: |x|<r }$.



Their corresponding formal power series are respectively $ F(X)=sum_{n geq 0} a_n X^n$ and $G(X)=sum_{n geq 0} b_n X^n $, say.



Given that the power series agree on $D(r^{-})$, so we have



$sum_{n geq 0} a_n x^n=sum_{n geq 0} b_n x^n$ for infinitely many $x in D(r^{-})$,



Does this imply $ sum_{n geq 0} a_n X^n=sum_{n geq 0} a_n X^n$?



If so then, let



$h(X)=sum_{n geq 0} a_n X^n-sum_{n geq 0} b_n X^n=0$,



$Rightarrow h(X)=0$



$Rightarrow sum_{n geq 0} (a_n-b_n) X^n=0$,



$Rightarrow a_n=b_n, forall n$



I think some mistake happened .



Please help me






power-series formal-power-series







share|cite|improve this question




















  • 1




    Given $f(x)=sum_{n geq 0} a_n x^n$, then what do you mean by its corresponding formal power series is $F(x)=sum_{n geq 0} a_n X^n$? Now, the left side is a function of $x$ and the right of $X$ If you really meant $x$ on the right side, then $f(x)=F(x)$ by definition.
    – Somos
    1 hour ago












  • @Somos, I know I have done something wrong. Can you please answer the question or theorem?
    – M. A. SARKAR
    1 hour ago










  • @Somos, I have corrected the typo. Can you please answer the question or theorem?
    – M. A. SARKAR
    48 mins ago










  • Read the answer to Mathoverflow question 134310 "construct a power series with infinitely many zeros in the complex plane, bounded coefficients???" with more explanations than the answer by Murthy.
    – Somos
    45 mins ago


















2














$ underline{text{This is about formal power series:}}$



If two power series (in one or two variables) converge on a disc and agree on infinitely many values within that disc, then show that they have the same formal power series coefficients.



Answer:



Let $f(x)=sum_{n geq 0} a_n x^n$ and $g(x)=sum_{n geq 0} b_n x^n$ be two power series converges on the disc $D(r^{-})={x in mathbb{C}: |x|<r }$.



Their corresponding formal power series are respectively $ F(X)=sum_{n geq 0} a_n X^n$ and $G(X)=sum_{n geq 0} b_n X^n $, say.



Given that the power series agree on $D(r^{-})$, so we have



$sum_{n geq 0} a_n x^n=sum_{n geq 0} b_n x^n$ for infinitely many $x in D(r^{-})$,



Does this imply $ sum_{n geq 0} a_n X^n=sum_{n geq 0} a_n X^n$?



If so then, let



$h(X)=sum_{n geq 0} a_n X^n-sum_{n geq 0} b_n X^n=0$,



$Rightarrow h(X)=0$



$Rightarrow sum_{n geq 0} (a_n-b_n) X^n=0$,



$Rightarrow a_n=b_n, forall n$



I think some mistake happened .



Please help me






power-series formal-power-series







share|cite|improve this question




















  • 1




    Given $f(x)=sum_{n geq 0} a_n x^n$, then what do you mean by its corresponding formal power series is $F(x)=sum_{n geq 0} a_n X^n$? Now, the left side is a function of $x$ and the right of $X$ If you really meant $x$ on the right side, then $f(x)=F(x)$ by definition.
    – Somos
    1 hour ago












  • @Somos, I know I have done something wrong. Can you please answer the question or theorem?
    – M. A. SARKAR
    1 hour ago










  • @Somos, I have corrected the typo. Can you please answer the question or theorem?
    – M. A. SARKAR
    48 mins ago










  • Read the answer to Mathoverflow question 134310 "construct a power series with infinitely many zeros in the complex plane, bounded coefficients???" with more explanations than the answer by Murthy.
    – Somos
    45 mins ago
















2












2








2


2





$ underline{text{This is about formal power series:}}$



If two power series (in one or two variables) converge on a disc and agree on infinitely many values within that disc, then show that they have the same formal power series coefficients.



Answer:



Let $f(x)=sum_{n geq 0} a_n x^n$ and $g(x)=sum_{n geq 0} b_n x^n$ be two power series converges on the disc $D(r^{-})={x in mathbb{C}: |x|<r }$.



Their corresponding formal power series are respectively $ F(X)=sum_{n geq 0} a_n X^n$ and $G(X)=sum_{n geq 0} b_n X^n $, say.



Given that the power series agree on $D(r^{-})$, so we have



$sum_{n geq 0} a_n x^n=sum_{n geq 0} b_n x^n$ for infinitely many $x in D(r^{-})$,



Does this imply $ sum_{n geq 0} a_n X^n=sum_{n geq 0} a_n X^n$?



If so then, let



$h(X)=sum_{n geq 0} a_n X^n-sum_{n geq 0} b_n X^n=0$,



$Rightarrow h(X)=0$



$Rightarrow sum_{n geq 0} (a_n-b_n) X^n=0$,



$Rightarrow a_n=b_n, forall n$



I think some mistake happened .



Please help me






power-series formal-power-series







share|cite|improve this question















$ underline{text{This is about formal power series:}}$



If two power series (in one or two variables) converge on a disc and agree on infinitely many values within that disc, then show that they have the same formal power series coefficients.



Answer:



Let $f(x)=sum_{n geq 0} a_n x^n$ and $g(x)=sum_{n geq 0} b_n x^n$ be two power series converges on the disc $D(r^{-})={x in mathbb{C}: |x|<r }$.



Their corresponding formal power series are respectively $ F(X)=sum_{n geq 0} a_n X^n$ and $G(X)=sum_{n geq 0} b_n X^n $, say.



Given that the power series agree on $D(r^{-})$, so we have



$sum_{n geq 0} a_n x^n=sum_{n geq 0} b_n x^n$ for infinitely many $x in D(r^{-})$,



Does this imply $ sum_{n geq 0} a_n X^n=sum_{n geq 0} a_n X^n$?



If so then, let



$h(X)=sum_{n geq 0} a_n X^n-sum_{n geq 0} b_n X^n=0$,



$Rightarrow h(X)=0$



$Rightarrow sum_{n geq 0} (a_n-b_n) X^n=0$,



$Rightarrow a_n=b_n, forall n$



I think some mistake happened .



Please help me






power-series formal-power-series




power-series formal-power-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 49 mins ago

























asked 3 hours ago









M. A. SARKAR

2,1821619




2,1821619








  • 1




    Given $f(x)=sum_{n geq 0} a_n x^n$, then what do you mean by its corresponding formal power series is $F(x)=sum_{n geq 0} a_n X^n$? Now, the left side is a function of $x$ and the right of $X$ If you really meant $x$ on the right side, then $f(x)=F(x)$ by definition.
    – Somos
    1 hour ago












  • @Somos, I know I have done something wrong. Can you please answer the question or theorem?
    – M. A. SARKAR
    1 hour ago










  • @Somos, I have corrected the typo. Can you please answer the question or theorem?
    – M. A. SARKAR
    48 mins ago










  • Read the answer to Mathoverflow question 134310 "construct a power series with infinitely many zeros in the complex plane, bounded coefficients???" with more explanations than the answer by Murthy.
    – Somos
    45 mins ago
















  • 1




    Given $f(x)=sum_{n geq 0} a_n x^n$, then what do you mean by its corresponding formal power series is $F(x)=sum_{n geq 0} a_n X^n$? Now, the left side is a function of $x$ and the right of $X$ If you really meant $x$ on the right side, then $f(x)=F(x)$ by definition.
    – Somos
    1 hour ago












  • @Somos, I know I have done something wrong. Can you please answer the question or theorem?
    – M. A. SARKAR
    1 hour ago










  • @Somos, I have corrected the typo. Can you please answer the question or theorem?
    – M. A. SARKAR
    48 mins ago










  • Read the answer to Mathoverflow question 134310 "construct a power series with infinitely many zeros in the complex plane, bounded coefficients???" with more explanations than the answer by Murthy.
    – Somos
    45 mins ago










1




1




Given $f(x)=sum_{n geq 0} a_n x^n$, then what do you mean by its corresponding formal power series is $F(x)=sum_{n geq 0} a_n X^n$? Now, the left side is a function of $x$ and the right of $X$ If you really meant $x$ on the right side, then $f(x)=F(x)$ by definition.
– Somos
1 hour ago






Given $f(x)=sum_{n geq 0} a_n x^n$, then what do you mean by its corresponding formal power series is $F(x)=sum_{n geq 0} a_n X^n$? Now, the left side is a function of $x$ and the right of $X$ If you really meant $x$ on the right side, then $f(x)=F(x)$ by definition.
– Somos
1 hour ago














@Somos, I know I have done something wrong. Can you please answer the question or theorem?
– M. A. SARKAR
1 hour ago




@Somos, I know I have done something wrong. Can you please answer the question or theorem?
– M. A. SARKAR
1 hour ago












@Somos, I have corrected the typo. Can you please answer the question or theorem?
– M. A. SARKAR
48 mins ago




@Somos, I have corrected the typo. Can you please answer the question or theorem?
– M. A. SARKAR
48 mins ago












Read the answer to Mathoverflow question 134310 "construct a power series with infinitely many zeros in the complex plane, bounded coefficients???" with more explanations than the answer by Murthy.
– Somos
45 mins ago






Read the answer to Mathoverflow question 134310 "construct a power series with infinitely many zeros in the complex plane, bounded coefficients???" with more explanations than the answer by Murthy.
– Somos
45 mins ago












1 Answer
1






active

oldest

votes


















6














Not true. Let $f(z)=sin (frac 1 {z-1})$. This function has a power series expansion $sum a_n z^{n}$ in ${z:|z| <1}$. The sum of its series is $0 (equiv sum b_nz^{n}$ where $b_n=0$ for all $n$) at the points $z=1- frac 1 {npi}, n=1,2,cdots$. But its coefficients are not all $0$.






share|cite|improve this answer





















  • Then what would be the answer regarding formal power series?
    – M. A. SARKAR
    3 hours ago










  • remember there is the concept of formal power series. please read the question again
    – M. A. SARKAR
    3 hours ago










  • You talk about formal power series when you don't know that the series converges. Here it is given the the two series converge on a disk.
    – Kavi Rama Murthy
    3 hours ago








  • 1




    @KaviRamaMurthy, your answer did not complete proof of the question
    – arifamath
    3 hours ago






  • 1




    Does it really have the formal power series you think it does? The series $(z-1)^{-1}-frac1{3!}(z-1)^{-3}+frac1{5!}(z-1)^5-cdots$ doesn't converge in the formal power series sense - there are infinitely many terms with nonzero constant term.
    – jmerry
    3 hours ago











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1 Answer
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1 Answer
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active

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6














Not true. Let $f(z)=sin (frac 1 {z-1})$. This function has a power series expansion $sum a_n z^{n}$ in ${z:|z| <1}$. The sum of its series is $0 (equiv sum b_nz^{n}$ where $b_n=0$ for all $n$) at the points $z=1- frac 1 {npi}, n=1,2,cdots$. But its coefficients are not all $0$.






share|cite|improve this answer





















  • Then what would be the answer regarding formal power series?
    – M. A. SARKAR
    3 hours ago










  • remember there is the concept of formal power series. please read the question again
    – M. A. SARKAR
    3 hours ago










  • You talk about formal power series when you don't know that the series converges. Here it is given the the two series converge on a disk.
    – Kavi Rama Murthy
    3 hours ago








  • 1




    @KaviRamaMurthy, your answer did not complete proof of the question
    – arifamath
    3 hours ago






  • 1




    Does it really have the formal power series you think it does? The series $(z-1)^{-1}-frac1{3!}(z-1)^{-3}+frac1{5!}(z-1)^5-cdots$ doesn't converge in the formal power series sense - there are infinitely many terms with nonzero constant term.
    – jmerry
    3 hours ago
















6














Not true. Let $f(z)=sin (frac 1 {z-1})$. This function has a power series expansion $sum a_n z^{n}$ in ${z:|z| <1}$. The sum of its series is $0 (equiv sum b_nz^{n}$ where $b_n=0$ for all $n$) at the points $z=1- frac 1 {npi}, n=1,2,cdots$. But its coefficients are not all $0$.






share|cite|improve this answer





















  • Then what would be the answer regarding formal power series?
    – M. A. SARKAR
    3 hours ago










  • remember there is the concept of formal power series. please read the question again
    – M. A. SARKAR
    3 hours ago










  • You talk about formal power series when you don't know that the series converges. Here it is given the the two series converge on a disk.
    – Kavi Rama Murthy
    3 hours ago








  • 1




    @KaviRamaMurthy, your answer did not complete proof of the question
    – arifamath
    3 hours ago






  • 1




    Does it really have the formal power series you think it does? The series $(z-1)^{-1}-frac1{3!}(z-1)^{-3}+frac1{5!}(z-1)^5-cdots$ doesn't converge in the formal power series sense - there are infinitely many terms with nonzero constant term.
    – jmerry
    3 hours ago














6












6








6






Not true. Let $f(z)=sin (frac 1 {z-1})$. This function has a power series expansion $sum a_n z^{n}$ in ${z:|z| <1}$. The sum of its series is $0 (equiv sum b_nz^{n}$ where $b_n=0$ for all $n$) at the points $z=1- frac 1 {npi}, n=1,2,cdots$. But its coefficients are not all $0$.






share|cite|improve this answer












Not true. Let $f(z)=sin (frac 1 {z-1})$. This function has a power series expansion $sum a_n z^{n}$ in ${z:|z| <1}$. The sum of its series is $0 (equiv sum b_nz^{n}$ where $b_n=0$ for all $n$) at the points $z=1- frac 1 {npi}, n=1,2,cdots$. But its coefficients are not all $0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









Kavi Rama Murthy

50.3k31854




50.3k31854












  • Then what would be the answer regarding formal power series?
    – M. A. SARKAR
    3 hours ago










  • remember there is the concept of formal power series. please read the question again
    – M. A. SARKAR
    3 hours ago










  • You talk about formal power series when you don't know that the series converges. Here it is given the the two series converge on a disk.
    – Kavi Rama Murthy
    3 hours ago








  • 1




    @KaviRamaMurthy, your answer did not complete proof of the question
    – arifamath
    3 hours ago






  • 1




    Does it really have the formal power series you think it does? The series $(z-1)^{-1}-frac1{3!}(z-1)^{-3}+frac1{5!}(z-1)^5-cdots$ doesn't converge in the formal power series sense - there are infinitely many terms with nonzero constant term.
    – jmerry
    3 hours ago


















  • Then what would be the answer regarding formal power series?
    – M. A. SARKAR
    3 hours ago










  • remember there is the concept of formal power series. please read the question again
    – M. A. SARKAR
    3 hours ago










  • You talk about formal power series when you don't know that the series converges. Here it is given the the two series converge on a disk.
    – Kavi Rama Murthy
    3 hours ago








  • 1




    @KaviRamaMurthy, your answer did not complete proof of the question
    – arifamath
    3 hours ago






  • 1




    Does it really have the formal power series you think it does? The series $(z-1)^{-1}-frac1{3!}(z-1)^{-3}+frac1{5!}(z-1)^5-cdots$ doesn't converge in the formal power series sense - there are infinitely many terms with nonzero constant term.
    – jmerry
    3 hours ago
















Then what would be the answer regarding formal power series?
– M. A. SARKAR
3 hours ago




Then what would be the answer regarding formal power series?
– M. A. SARKAR
3 hours ago












remember there is the concept of formal power series. please read the question again
– M. A. SARKAR
3 hours ago




remember there is the concept of formal power series. please read the question again
– M. A. SARKAR
3 hours ago












You talk about formal power series when you don't know that the series converges. Here it is given the the two series converge on a disk.
– Kavi Rama Murthy
3 hours ago






You talk about formal power series when you don't know that the series converges. Here it is given the the two series converge on a disk.
– Kavi Rama Murthy
3 hours ago






1




1




@KaviRamaMurthy, your answer did not complete proof of the question
– arifamath
3 hours ago




@KaviRamaMurthy, your answer did not complete proof of the question
– arifamath
3 hours ago




1




1




Does it really have the formal power series you think it does? The series $(z-1)^{-1}-frac1{3!}(z-1)^{-3}+frac1{5!}(z-1)^5-cdots$ doesn't converge in the formal power series sense - there are infinitely many terms with nonzero constant term.
– jmerry
3 hours ago




Does it really have the formal power series you think it does? The series $(z-1)^{-1}-frac1{3!}(z-1)^{-3}+frac1{5!}(z-1)^5-cdots$ doesn't converge in the formal power series sense - there are infinitely many terms with nonzero constant term.
– jmerry
3 hours ago


















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