show that they have the same formal power series coefficients












1














$ 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







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  • 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
    5 mins ago


















1














$ 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
























  • 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
    5 mins ago
















1












1








1


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






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edited 1 hour ago

























asked 2 hours ago









M. A. SARKAR

2,1771619




2,1771619












  • 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
    5 mins 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
    5 mins 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
5 mins 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
5 mins ago












1 Answer
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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
    1 hour ago










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








  • 1




    @KaviRamaMurthy, your answer did not complete proof of the question
    – arifamath
    1 hour 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
    1 hour ago











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1 Answer
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4














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
    1 hour ago










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








  • 1




    @KaviRamaMurthy, your answer did not complete proof of the question
    – arifamath
    1 hour 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
    1 hour ago
















4














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
    1 hour ago










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








  • 1




    @KaviRamaMurthy, your answer did not complete proof of the question
    – arifamath
    1 hour 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
    1 hour ago














4












4








4






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 2 hours ago









Kavi Rama Murthy

50.3k31854




50.3k31854












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










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








  • 1




    @KaviRamaMurthy, your answer did not complete proof of the question
    – arifamath
    1 hour 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
    1 hour ago


















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










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








  • 1




    @KaviRamaMurthy, your answer did not complete proof of the question
    – arifamath
    1 hour 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
    1 hour ago
















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




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












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




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






1




1




@KaviRamaMurthy, your answer did not complete proof of the question
– arifamath
1 hour ago




@KaviRamaMurthy, your answer did not complete proof of the question
– arifamath
1 hour 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
1 hour 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
1 hour ago


















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