Does their exist an analytic function $f(z)$ such that $f(n + frac{1}{n}) = 0$ for all $n in mathbb{N}$?












2












$begingroup$


My approach:



Let $a_n = n + frac{1}{n}$. Define $f(z) = (z-a_1)cdots(z-a_n)cdots=prod_{ninmathbb{N}}(z-a_n)$. I think that $f(z)$ satisfies the above property. But I have some reservation in defining the infinite product. So is my approach correct and can someone write down the rigorous answer to it.






complex-analysis







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$endgroup$








  • 1




    $begingroup$
    I guess you mean besides the function $f = 0$.
    $endgroup$
    – jjagmath
    Dec 5 '18 at 23:42










  • $begingroup$
    @jjagmath Yes $f neq 0$
    $endgroup$
    – henceproved
    Dec 5 '18 at 23:46
















2












$begingroup$


My approach:



Let $a_n = n + frac{1}{n}$. Define $f(z) = (z-a_1)cdots(z-a_n)cdots=prod_{ninmathbb{N}}(z-a_n)$. I think that $f(z)$ satisfies the above property. But I have some reservation in defining the infinite product. So is my approach correct and can someone write down the rigorous answer to it.






complex-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I guess you mean besides the function $f = 0$.
    $endgroup$
    – jjagmath
    Dec 5 '18 at 23:42










  • $begingroup$
    @jjagmath Yes $f neq 0$
    $endgroup$
    – henceproved
    Dec 5 '18 at 23:46














2












2








2





$begingroup$


My approach:



Let $a_n = n + frac{1}{n}$. Define $f(z) = (z-a_1)cdots(z-a_n)cdots=prod_{ninmathbb{N}}(z-a_n)$. I think that $f(z)$ satisfies the above property. But I have some reservation in defining the infinite product. So is my approach correct and can someone write down the rigorous answer to it.






complex-analysis







share|cite|improve this question











$endgroup$




My approach:



Let $a_n = n + frac{1}{n}$. Define $f(z) = (z-a_1)cdots(z-a_n)cdots=prod_{ninmathbb{N}}(z-a_n)$. I think that $f(z)$ satisfies the above property. But I have some reservation in defining the infinite product. So is my approach correct and can someone write down the rigorous answer to it.






complex-analysis




complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 0:04









Ethan Bolker

44.3k553119




44.3k553119










asked Dec 5 '18 at 23:29









henceprovedhenceproved

1628




1628








  • 1




    $begingroup$
    I guess you mean besides the function $f = 0$.
    $endgroup$
    – jjagmath
    Dec 5 '18 at 23:42










  • $begingroup$
    @jjagmath Yes $f neq 0$
    $endgroup$
    – henceproved
    Dec 5 '18 at 23:46














  • 1




    $begingroup$
    I guess you mean besides the function $f = 0$.
    $endgroup$
    – jjagmath
    Dec 5 '18 at 23:42










  • $begingroup$
    @jjagmath Yes $f neq 0$
    $endgroup$
    – henceproved
    Dec 5 '18 at 23:46








1




1




$begingroup$
I guess you mean besides the function $f = 0$.
$endgroup$
– jjagmath
Dec 5 '18 at 23:42




$begingroup$
I guess you mean besides the function $f = 0$.
$endgroup$
– jjagmath
Dec 5 '18 at 23:42












$begingroup$
@jjagmath Yes $f neq 0$
$endgroup$
– henceproved
Dec 5 '18 at 23:46




$begingroup$
@jjagmath Yes $f neq 0$
$endgroup$
– henceproved
Dec 5 '18 at 23:46










2 Answers
2






active

oldest

votes


















2












$begingroup$

This should follow immediately from the Weierstrass Factorization Theorem. There are some technical details you should attend to, but your idea is almost basically correct.



https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the link, i was not aware of that theorem. Anyways in the theorem it mentions that $mid z - a_n mid to 0 $ as $n to infty$. Somehow I am not seeing that happening!
    $endgroup$
    – henceproved
    Dec 5 '18 at 23:48












  • $begingroup$
    @henceproved Where did you see $|z-a_n|rightarrow 0$? The first form of the theorem only requires that $a_nrightarrow infty$.
    $endgroup$
    – RandomMathGuy
    Dec 5 '18 at 23:50










  • $begingroup$
    sorry typo it should be $mid z - a_n mid to 1$ and I saw this condition in the motivation section.
    $endgroup$
    – henceproved
    Dec 5 '18 at 23:52










  • $begingroup$
    @henceproved Ah, that condition is certainly required, and your proposed product certainly does not satisfy it. That is why I said you had some technical details to attend to. Notice that in the theorem the elementary factors are rather strange looking.
    $endgroup$
    – RandomMathGuy
    Dec 5 '18 at 23:53






  • 2




    $begingroup$
    If $$sum frac{1}{a^2_n}$$ is absolutely convergent (which it is in this case), then $$f(z) = prod_{n=1}^{infty} left(1 - frac{z}{a_n}right) e^{z/a_n}$$ is a holomorphic function with zeroes exactly at $a_n$. If you don't mind extra zeroes, then you could consider $F(z) = f(z)f(-z)$, which has zeroes at $a_n$ and $-a_n$ but the slightly nicer form: $$ prod_{n=1}^{infty} left(1 - frac{z^2}{a^2_n} right).$$
    $endgroup$
    – Lorem Ipsum
    Dec 6 '18 at 1:39



















0












$begingroup$

The function you have is actually nowhere analytic and not even well defined.



Suppose that $f:mathbb{C}rightarrow mathbb{C}$ where $f(x) = lim_{nrightarrowinfty}prod_{i=1}^{n}(x-a_n)$ is well defined, i.e $f(x)in mathbb{C}$ for each $xin mathbb{C}$.



Pick any $xinmathbb{C},nin mathbb{N}$.



begin{align}
|a_n-a_m|&=|n+frac{1}{n}-m-frac{1}{m}|\
&geq |n-m| - |frac{1}{n}-frac{1}{m}|\
&geq |n-m| - 2
end{align}



So that $|a_n - a_m|rightarrowinfty$ as $mrightarrow infty$. We also have that begin{align}
|x-a_m|&= |(x-a_n)-(a_n-a_m)|\
&geq |a_n -a_m| - |x-a_n|
end{align}



Now, as $|a_n-a_m|rightarrowinfty$, choose $M = 2+|x-a_n|$ so that there exists $Nin mathbb{N}$ such that $m> N$ implies $|a_n-a_m| geq M$ which implies $|x-a_m| geq 2$.



Note $|x-a_i|$ is nonzero for all $1leq ileq N$. Thus $prod_{1leq ileq N}{|x-a_i|}$ is equal to some $varepsilon>0$. Now there exists $kinmathbb{N}$ such that $frac{1}{2^k} leq varepsilon$. Now $prod_{i=1}^{N+m+k}|x-a_i| geq varepsilon2^{m+k}geq2^{m}$. Thus $f(x)$ must be unbounded so that $f(x)notin mathbb{C}$. Hence $f$ is not well defined.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This addresses "is my approach correct" but not "can someone write down the rigorous answer to it".
    $endgroup$
    – Did
    Dec 7 '18 at 20:07











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

This should follow immediately from the Weierstrass Factorization Theorem. There are some technical details you should attend to, but your idea is almost basically correct.



https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the link, i was not aware of that theorem. Anyways in the theorem it mentions that $mid z - a_n mid to 0 $ as $n to infty$. Somehow I am not seeing that happening!
    $endgroup$
    – henceproved
    Dec 5 '18 at 23:48












  • $begingroup$
    @henceproved Where did you see $|z-a_n|rightarrow 0$? The first form of the theorem only requires that $a_nrightarrow infty$.
    $endgroup$
    – RandomMathGuy
    Dec 5 '18 at 23:50










  • $begingroup$
    sorry typo it should be $mid z - a_n mid to 1$ and I saw this condition in the motivation section.
    $endgroup$
    – henceproved
    Dec 5 '18 at 23:52










  • $begingroup$
    @henceproved Ah, that condition is certainly required, and your proposed product certainly does not satisfy it. That is why I said you had some technical details to attend to. Notice that in the theorem the elementary factors are rather strange looking.
    $endgroup$
    – RandomMathGuy
    Dec 5 '18 at 23:53






  • 2




    $begingroup$
    If $$sum frac{1}{a^2_n}$$ is absolutely convergent (which it is in this case), then $$f(z) = prod_{n=1}^{infty} left(1 - frac{z}{a_n}right) e^{z/a_n}$$ is a holomorphic function with zeroes exactly at $a_n$. If you don't mind extra zeroes, then you could consider $F(z) = f(z)f(-z)$, which has zeroes at $a_n$ and $-a_n$ but the slightly nicer form: $$ prod_{n=1}^{infty} left(1 - frac{z^2}{a^2_n} right).$$
    $endgroup$
    – Lorem Ipsum
    Dec 6 '18 at 1:39
















2












$begingroup$

This should follow immediately from the Weierstrass Factorization Theorem. There are some technical details you should attend to, but your idea is almost basically correct.



https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the link, i was not aware of that theorem. Anyways in the theorem it mentions that $mid z - a_n mid to 0 $ as $n to infty$. Somehow I am not seeing that happening!
    $endgroup$
    – henceproved
    Dec 5 '18 at 23:48












  • $begingroup$
    @henceproved Where did you see $|z-a_n|rightarrow 0$? The first form of the theorem only requires that $a_nrightarrow infty$.
    $endgroup$
    – RandomMathGuy
    Dec 5 '18 at 23:50










  • $begingroup$
    sorry typo it should be $mid z - a_n mid to 1$ and I saw this condition in the motivation section.
    $endgroup$
    – henceproved
    Dec 5 '18 at 23:52










  • $begingroup$
    @henceproved Ah, that condition is certainly required, and your proposed product certainly does not satisfy it. That is why I said you had some technical details to attend to. Notice that in the theorem the elementary factors are rather strange looking.
    $endgroup$
    – RandomMathGuy
    Dec 5 '18 at 23:53






  • 2




    $begingroup$
    If $$sum frac{1}{a^2_n}$$ is absolutely convergent (which it is in this case), then $$f(z) = prod_{n=1}^{infty} left(1 - frac{z}{a_n}right) e^{z/a_n}$$ is a holomorphic function with zeroes exactly at $a_n$. If you don't mind extra zeroes, then you could consider $F(z) = f(z)f(-z)$, which has zeroes at $a_n$ and $-a_n$ but the slightly nicer form: $$ prod_{n=1}^{infty} left(1 - frac{z^2}{a^2_n} right).$$
    $endgroup$
    – Lorem Ipsum
    Dec 6 '18 at 1:39














2












2








2





$begingroup$

This should follow immediately from the Weierstrass Factorization Theorem. There are some technical details you should attend to, but your idea is almost basically correct.



https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem






share|cite|improve this answer









$endgroup$



This should follow immediately from the Weierstrass Factorization Theorem. There are some technical details you should attend to, but your idea is almost basically correct.



https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 23:39









RandomMathGuyRandomMathGuy

641




641












  • $begingroup$
    Thanks for the link, i was not aware of that theorem. Anyways in the theorem it mentions that $mid z - a_n mid to 0 $ as $n to infty$. Somehow I am not seeing that happening!
    $endgroup$
    – henceproved
    Dec 5 '18 at 23:48












  • $begingroup$
    @henceproved Where did you see $|z-a_n|rightarrow 0$? The first form of the theorem only requires that $a_nrightarrow infty$.
    $endgroup$
    – RandomMathGuy
    Dec 5 '18 at 23:50










  • $begingroup$
    sorry typo it should be $mid z - a_n mid to 1$ and I saw this condition in the motivation section.
    $endgroup$
    – henceproved
    Dec 5 '18 at 23:52










  • $begingroup$
    @henceproved Ah, that condition is certainly required, and your proposed product certainly does not satisfy it. That is why I said you had some technical details to attend to. Notice that in the theorem the elementary factors are rather strange looking.
    $endgroup$
    – RandomMathGuy
    Dec 5 '18 at 23:53






  • 2




    $begingroup$
    If $$sum frac{1}{a^2_n}$$ is absolutely convergent (which it is in this case), then $$f(z) = prod_{n=1}^{infty} left(1 - frac{z}{a_n}right) e^{z/a_n}$$ is a holomorphic function with zeroes exactly at $a_n$. If you don't mind extra zeroes, then you could consider $F(z) = f(z)f(-z)$, which has zeroes at $a_n$ and $-a_n$ but the slightly nicer form: $$ prod_{n=1}^{infty} left(1 - frac{z^2}{a^2_n} right).$$
    $endgroup$
    – Lorem Ipsum
    Dec 6 '18 at 1:39


















  • $begingroup$
    Thanks for the link, i was not aware of that theorem. Anyways in the theorem it mentions that $mid z - a_n mid to 0 $ as $n to infty$. Somehow I am not seeing that happening!
    $endgroup$
    – henceproved
    Dec 5 '18 at 23:48












  • $begingroup$
    @henceproved Where did you see $|z-a_n|rightarrow 0$? The first form of the theorem only requires that $a_nrightarrow infty$.
    $endgroup$
    – RandomMathGuy
    Dec 5 '18 at 23:50










  • $begingroup$
    sorry typo it should be $mid z - a_n mid to 1$ and I saw this condition in the motivation section.
    $endgroup$
    – henceproved
    Dec 5 '18 at 23:52










  • $begingroup$
    @henceproved Ah, that condition is certainly required, and your proposed product certainly does not satisfy it. That is why I said you had some technical details to attend to. Notice that in the theorem the elementary factors are rather strange looking.
    $endgroup$
    – RandomMathGuy
    Dec 5 '18 at 23:53






  • 2




    $begingroup$
    If $$sum frac{1}{a^2_n}$$ is absolutely convergent (which it is in this case), then $$f(z) = prod_{n=1}^{infty} left(1 - frac{z}{a_n}right) e^{z/a_n}$$ is a holomorphic function with zeroes exactly at $a_n$. If you don't mind extra zeroes, then you could consider $F(z) = f(z)f(-z)$, which has zeroes at $a_n$ and $-a_n$ but the slightly nicer form: $$ prod_{n=1}^{infty} left(1 - frac{z^2}{a^2_n} right).$$
    $endgroup$
    – Lorem Ipsum
    Dec 6 '18 at 1:39
















$begingroup$
Thanks for the link, i was not aware of that theorem. Anyways in the theorem it mentions that $mid z - a_n mid to 0 $ as $n to infty$. Somehow I am not seeing that happening!
$endgroup$
– henceproved
Dec 5 '18 at 23:48






$begingroup$
Thanks for the link, i was not aware of that theorem. Anyways in the theorem it mentions that $mid z - a_n mid to 0 $ as $n to infty$. Somehow I am not seeing that happening!
$endgroup$
– henceproved
Dec 5 '18 at 23:48














$begingroup$
@henceproved Where did you see $|z-a_n|rightarrow 0$? The first form of the theorem only requires that $a_nrightarrow infty$.
$endgroup$
– RandomMathGuy
Dec 5 '18 at 23:50




$begingroup$
@henceproved Where did you see $|z-a_n|rightarrow 0$? The first form of the theorem only requires that $a_nrightarrow infty$.
$endgroup$
– RandomMathGuy
Dec 5 '18 at 23:50












$begingroup$
sorry typo it should be $mid z - a_n mid to 1$ and I saw this condition in the motivation section.
$endgroup$
– henceproved
Dec 5 '18 at 23:52




$begingroup$
sorry typo it should be $mid z - a_n mid to 1$ and I saw this condition in the motivation section.
$endgroup$
– henceproved
Dec 5 '18 at 23:52












$begingroup$
@henceproved Ah, that condition is certainly required, and your proposed product certainly does not satisfy it. That is why I said you had some technical details to attend to. Notice that in the theorem the elementary factors are rather strange looking.
$endgroup$
– RandomMathGuy
Dec 5 '18 at 23:53




$begingroup$
@henceproved Ah, that condition is certainly required, and your proposed product certainly does not satisfy it. That is why I said you had some technical details to attend to. Notice that in the theorem the elementary factors are rather strange looking.
$endgroup$
– RandomMathGuy
Dec 5 '18 at 23:53




2




2




$begingroup$
If $$sum frac{1}{a^2_n}$$ is absolutely convergent (which it is in this case), then $$f(z) = prod_{n=1}^{infty} left(1 - frac{z}{a_n}right) e^{z/a_n}$$ is a holomorphic function with zeroes exactly at $a_n$. If you don't mind extra zeroes, then you could consider $F(z) = f(z)f(-z)$, which has zeroes at $a_n$ and $-a_n$ but the slightly nicer form: $$ prod_{n=1}^{infty} left(1 - frac{z^2}{a^2_n} right).$$
$endgroup$
– Lorem Ipsum
Dec 6 '18 at 1:39




$begingroup$
If $$sum frac{1}{a^2_n}$$ is absolutely convergent (which it is in this case), then $$f(z) = prod_{n=1}^{infty} left(1 - frac{z}{a_n}right) e^{z/a_n}$$ is a holomorphic function with zeroes exactly at $a_n$. If you don't mind extra zeroes, then you could consider $F(z) = f(z)f(-z)$, which has zeroes at $a_n$ and $-a_n$ but the slightly nicer form: $$ prod_{n=1}^{infty} left(1 - frac{z^2}{a^2_n} right).$$
$endgroup$
– Lorem Ipsum
Dec 6 '18 at 1:39











0












$begingroup$

The function you have is actually nowhere analytic and not even well defined.



Suppose that $f:mathbb{C}rightarrow mathbb{C}$ where $f(x) = lim_{nrightarrowinfty}prod_{i=1}^{n}(x-a_n)$ is well defined, i.e $f(x)in mathbb{C}$ for each $xin mathbb{C}$.



Pick any $xinmathbb{C},nin mathbb{N}$.



begin{align}
|a_n-a_m|&=|n+frac{1}{n}-m-frac{1}{m}|\
&geq |n-m| - |frac{1}{n}-frac{1}{m}|\
&geq |n-m| - 2
end{align}



So that $|a_n - a_m|rightarrowinfty$ as $mrightarrow infty$. We also have that begin{align}
|x-a_m|&= |(x-a_n)-(a_n-a_m)|\
&geq |a_n -a_m| - |x-a_n|
end{align}



Now, as $|a_n-a_m|rightarrowinfty$, choose $M = 2+|x-a_n|$ so that there exists $Nin mathbb{N}$ such that $m> N$ implies $|a_n-a_m| geq M$ which implies $|x-a_m| geq 2$.



Note $|x-a_i|$ is nonzero for all $1leq ileq N$. Thus $prod_{1leq ileq N}{|x-a_i|}$ is equal to some $varepsilon>0$. Now there exists $kinmathbb{N}$ such that $frac{1}{2^k} leq varepsilon$. Now $prod_{i=1}^{N+m+k}|x-a_i| geq varepsilon2^{m+k}geq2^{m}$. Thus $f(x)$ must be unbounded so that $f(x)notin mathbb{C}$. Hence $f$ is not well defined.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This addresses "is my approach correct" but not "can someone write down the rigorous answer to it".
    $endgroup$
    – Did
    Dec 7 '18 at 20:07
















0












$begingroup$

The function you have is actually nowhere analytic and not even well defined.



Suppose that $f:mathbb{C}rightarrow mathbb{C}$ where $f(x) = lim_{nrightarrowinfty}prod_{i=1}^{n}(x-a_n)$ is well defined, i.e $f(x)in mathbb{C}$ for each $xin mathbb{C}$.



Pick any $xinmathbb{C},nin mathbb{N}$.



begin{align}
|a_n-a_m|&=|n+frac{1}{n}-m-frac{1}{m}|\
&geq |n-m| - |frac{1}{n}-frac{1}{m}|\
&geq |n-m| - 2
end{align}



So that $|a_n - a_m|rightarrowinfty$ as $mrightarrow infty$. We also have that begin{align}
|x-a_m|&= |(x-a_n)-(a_n-a_m)|\
&geq |a_n -a_m| - |x-a_n|
end{align}



Now, as $|a_n-a_m|rightarrowinfty$, choose $M = 2+|x-a_n|$ so that there exists $Nin mathbb{N}$ such that $m> N$ implies $|a_n-a_m| geq M$ which implies $|x-a_m| geq 2$.



Note $|x-a_i|$ is nonzero for all $1leq ileq N$. Thus $prod_{1leq ileq N}{|x-a_i|}$ is equal to some $varepsilon>0$. Now there exists $kinmathbb{N}$ such that $frac{1}{2^k} leq varepsilon$. Now $prod_{i=1}^{N+m+k}|x-a_i| geq varepsilon2^{m+k}geq2^{m}$. Thus $f(x)$ must be unbounded so that $f(x)notin mathbb{C}$. Hence $f$ is not well defined.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This addresses "is my approach correct" but not "can someone write down the rigorous answer to it".
    $endgroup$
    – Did
    Dec 7 '18 at 20:07














0












0








0





$begingroup$

The function you have is actually nowhere analytic and not even well defined.



Suppose that $f:mathbb{C}rightarrow mathbb{C}$ where $f(x) = lim_{nrightarrowinfty}prod_{i=1}^{n}(x-a_n)$ is well defined, i.e $f(x)in mathbb{C}$ for each $xin mathbb{C}$.



Pick any $xinmathbb{C},nin mathbb{N}$.



begin{align}
|a_n-a_m|&=|n+frac{1}{n}-m-frac{1}{m}|\
&geq |n-m| - |frac{1}{n}-frac{1}{m}|\
&geq |n-m| - 2
end{align}



So that $|a_n - a_m|rightarrowinfty$ as $mrightarrow infty$. We also have that begin{align}
|x-a_m|&= |(x-a_n)-(a_n-a_m)|\
&geq |a_n -a_m| - |x-a_n|
end{align}



Now, as $|a_n-a_m|rightarrowinfty$, choose $M = 2+|x-a_n|$ so that there exists $Nin mathbb{N}$ such that $m> N$ implies $|a_n-a_m| geq M$ which implies $|x-a_m| geq 2$.



Note $|x-a_i|$ is nonzero for all $1leq ileq N$. Thus $prod_{1leq ileq N}{|x-a_i|}$ is equal to some $varepsilon>0$. Now there exists $kinmathbb{N}$ such that $frac{1}{2^k} leq varepsilon$. Now $prod_{i=1}^{N+m+k}|x-a_i| geq varepsilon2^{m+k}geq2^{m}$. Thus $f(x)$ must be unbounded so that $f(x)notin mathbb{C}$. Hence $f$ is not well defined.






share|cite|improve this answer











$endgroup$



The function you have is actually nowhere analytic and not even well defined.



Suppose that $f:mathbb{C}rightarrow mathbb{C}$ where $f(x) = lim_{nrightarrowinfty}prod_{i=1}^{n}(x-a_n)$ is well defined, i.e $f(x)in mathbb{C}$ for each $xin mathbb{C}$.



Pick any $xinmathbb{C},nin mathbb{N}$.



begin{align}
|a_n-a_m|&=|n+frac{1}{n}-m-frac{1}{m}|\
&geq |n-m| - |frac{1}{n}-frac{1}{m}|\
&geq |n-m| - 2
end{align}



So that $|a_n - a_m|rightarrowinfty$ as $mrightarrow infty$. We also have that begin{align}
|x-a_m|&= |(x-a_n)-(a_n-a_m)|\
&geq |a_n -a_m| - |x-a_n|
end{align}



Now, as $|a_n-a_m|rightarrowinfty$, choose $M = 2+|x-a_n|$ so that there exists $Nin mathbb{N}$ such that $m> N$ implies $|a_n-a_m| geq M$ which implies $|x-a_m| geq 2$.



Note $|x-a_i|$ is nonzero for all $1leq ileq N$. Thus $prod_{1leq ileq N}{|x-a_i|}$ is equal to some $varepsilon>0$. Now there exists $kinmathbb{N}$ such that $frac{1}{2^k} leq varepsilon$. Now $prod_{i=1}^{N+m+k}|x-a_i| geq varepsilon2^{m+k}geq2^{m}$. Thus $f(x)$ must be unbounded so that $f(x)notin mathbb{C}$. Hence $f$ is not well defined.







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edited Dec 6 '18 at 1:31

























answered Dec 6 '18 at 1:20









TomTom

736




736












  • $begingroup$
    This addresses "is my approach correct" but not "can someone write down the rigorous answer to it".
    $endgroup$
    – Did
    Dec 7 '18 at 20:07


















  • $begingroup$
    This addresses "is my approach correct" but not "can someone write down the rigorous answer to it".
    $endgroup$
    – Did
    Dec 7 '18 at 20:07
















$begingroup$
This addresses "is my approach correct" but not "can someone write down the rigorous answer to it".
$endgroup$
– Did
Dec 7 '18 at 20:07




$begingroup$
This addresses "is my approach correct" but not "can someone write down the rigorous answer to it".
$endgroup$
– Did
Dec 7 '18 at 20:07


















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