Zeros of partial sums of the exponential [duplicate]
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Complex zeros of the polynomials $sum_{k=0}^{n} z^k/k!$, inside balls
3 answers
How prove this $|z|>1$ with $1+z+frac{z^2}{2!}+cdots+frac{z^n}{n!}=0$
1 answer
I am trying to show that if $$f_n(z)=1+z+frac{z^2}{2!}+...+frac{z^n}{n!}$$
Then $f_n(z)$ don’t have zeros inside the unitary disk.
I have tryied to use Rouche’s theorem or use that in the limit the polinomial converges to the exponential, but i dont get hoy to do this.
complex-analysis power-series taylor-expansion
asked Nov 18 at 3:25
J.Rodriguez
15310
marked as duplicate by Martin R, Brahadeesh, Lord Shark the Unknown, Kelvin Lois, LutzL Nov 18 at 19:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
4
down vote
favorite
This question already has an answer here:
Complex zeros of the polynomials $sum_{k=0}^{n} z^k/k!$, inside balls
3 answers
How prove this $|z|>1$ with $1+z+frac{z^2}{2!}+cdots+frac{z^n}{n!}=0$
1 answer
I am trying to show that if $$f_n(z)=1+z+frac{z^2}{2!}+...+frac{z^n}{n!}$$
Then $f_n(z)$ don’t have zeros inside the unitary disk.
I have tryied to use Rouche’s theorem or use that in the limit the polinomial converges to the exponential, but i dont get hoy to do this.
complex-analysis power-series taylor-expansion
asked Nov 18 at 3:25
J.Rodriguez
15310
marked as duplicate by Martin R, Brahadeesh, Lord Shark the Unknown, Kelvin Lois, LutzL Nov 18 at 19:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
It converges locally uniformly to an entire function $f$. So if $f$ has no zeros it is finished. Otherwise assume $f$ has a zero of order $k$ at $z=a$, $f(z) sim C (z-a)^k$. Fix $epsilon$ very small, there is $N$ such that every $f_n,n ge N$ have $k$ zeros on $|z-a| < epsilon$ and $f_n(z) = h_n(z)prod_{l=1}^k (z-a_{l,n})=(C+O(z-a))prod_{l=1}^k (z-a_{l,n})$, $frac{f_n'}{f_n}(z) = O(z-a) + sum_{l=1}^k frac{1}{z-a_{l,n}}$. But $frac{f_n'}{f_n}(z) = 1-frac{frac{z^n}{n!}}{f_n(z)}=1-frac{a^n+O(z-a)}{n! (C+O(z-a))} frac{1}{prod_{l=1}^k (z-a_{l,n})}$, a contradiction.
– reuns
Nov 18 at 4:45
@reuns: Unless I am mistaken, your argument shows only that $f_n$ has no zeros in the unit disk for sufficiently large $n$.
– Martin R
Nov 18 at 8:31
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
This question already has an answer here:
Complex zeros of the polynomials $sum_{k=0}^{n} z^k/k!$, inside balls
3 answers
How prove this $|z|>1$ with $1+z+frac{z^2}{2!}+cdots+frac{z^n}{n!}=0$
1 answer
I am trying to show that if $$f_n(z)=1+z+frac{z^2}{2!}+...+frac{z^n}{n!}$$
Then $f_n(z)$ don’t have zeros inside the unitary disk.
I have tryied to use Rouche’s theorem or use that in the limit the polinomial converges to the exponential, but i dont get hoy to do this.
complex-analysis power-series taylor-expansion
asked Nov 18 at 3:25
J.Rodriguez
15310
This question already has an answer here:
Complex zeros of the polynomials $sum_{k=0}^{n} z^k/k!$, inside balls
3 answers
How prove this $|z|>1$ with $1+z+frac{z^2}{2!}+cdots+frac{z^n}{n!}=0$
1 answer
I am trying to show that if $$f_n(z)=1+z+frac{z^2}{2!}+...+frac{z^n}{n!}$$
Then $f_n(z)$ don’t have zeros inside the unitary disk.
I have tryied to use Rouche’s theorem or use that in the limit the polinomial converges to the exponential, but i dont get hoy to do this.
This question already has an answer here:
Complex zeros of the polynomials $sum_{k=0}^{n} z^k/k!$, inside balls
3 answers
How prove this $|z|>1$ with $1+z+frac{z^2}{2!}+cdots+frac{z^n}{n!}=0$
1 answer
complex-analysis power-series taylor-expansion
complex-analysis power-series taylor-expansion
asked Nov 18 at 3:25
J.Rodriguez
15310
asked Nov 18 at 3:25
J.Rodriguez
15310
asked Nov 18 at 3:25
J.Rodriguez
15310
asked Nov 18 at 3:25
J.Rodriguez
15310
asked Nov 18 at 3:25
J.Rodriguez
15310
15310
marked as duplicate by Martin R, Brahadeesh, Lord Shark the Unknown, Kelvin Lois, LutzL Nov 18 at 19:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, Brahadeesh, Lord Shark the Unknown, Kelvin Lois, LutzL Nov 18 at 19:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
It converges locally uniformly to an entire function $f$. So if $f$ has no zeros it is finished. Otherwise assume $f$ has a zero of order $k$ at $z=a$, $f(z) sim C (z-a)^k$. Fix $epsilon$ very small, there is $N$ such that every $f_n,n ge N$ have $k$ zeros on $|z-a| < epsilon$ and $f_n(z) = h_n(z)prod_{l=1}^k (z-a_{l,n})=(C+O(z-a))prod_{l=1}^k (z-a_{l,n})$, $frac{f_n'}{f_n}(z) = O(z-a) + sum_{l=1}^k frac{1}{z-a_{l,n}}$. But $frac{f_n'}{f_n}(z) = 1-frac{frac{z^n}{n!}}{f_n(z)}=1-frac{a^n+O(z-a)}{n! (C+O(z-a))} frac{1}{prod_{l=1}^k (z-a_{l,n})}$, a contradiction.
– reuns
Nov 18 at 4:45
@reuns: Unless I am mistaken, your argument shows only that $f_n$ has no zeros in the unit disk for sufficiently large $n$.
– Martin R
Nov 18 at 8:31
add a comment |
It converges locally uniformly to an entire function $f$. So if $f$ has no zeros it is finished. Otherwise assume $f$ has a zero of order $k$ at $z=a$, $f(z) sim C (z-a)^k$. Fix $epsilon$ very small, there is $N$ such that every $f_n,n ge N$ have $k$ zeros on $|z-a| < epsilon$ and $f_n(z) = h_n(z)prod_{l=1}^k (z-a_{l,n})=(C+O(z-a))prod_{l=1}^k (z-a_{l,n})$, $frac{f_n'}{f_n}(z) = O(z-a) + sum_{l=1}^k frac{1}{z-a_{l,n}}$. But $frac{f_n'}{f_n}(z) = 1-frac{frac{z^n}{n!}}{f_n(z)}=1-frac{a^n+O(z-a)}{n! (C+O(z-a))} frac{1}{prod_{l=1}^k (z-a_{l,n})}$, a contradiction.
– reuns
Nov 18 at 4:45
@reuns: Unless I am mistaken, your argument shows only that $f_n$ has no zeros in the unit disk for sufficiently large $n$.
– Martin R
Nov 18 at 8:31
It converges locally uniformly to an entire function $f$. So if $f$ has no zeros it is finished. Otherwise assume $f$ has a zero of order $k$ at $z=a$, $f(z) sim C (z-a)^k$. Fix $epsilon$ very small, there is $N$ such that every $f_n,n ge N$ have $k$ zeros on $|z-a| < epsilon$ and $f_n(z) = h_n(z)prod_{l=1}^k (z-a_{l,n})=(C+O(z-a))prod_{l=1}^k (z-a_{l,n})$, $frac{f_n'}{f_n}(z) = O(z-a) + sum_{l=1}^k frac{1}{z-a_{l,n}}$. But $frac{f_n'}{f_n}(z) = 1-frac{frac{z^n}{n!}}{f_n(z)}=1-frac{a^n+O(z-a)}{n! (C+O(z-a))} frac{1}{prod_{l=1}^k (z-a_{l,n})}$, a contradiction.
– reuns
Nov 18 at 4:45
It converges locally uniformly to an entire function $f$. So if $f$ has no zeros it is finished. Otherwise assume $f$ has a zero of order $k$ at $z=a$, $f(z) sim C (z-a)^k$. Fix $epsilon$ very small, there is $N$ such that every $f_n,n ge N$ have $k$ zeros on $|z-a| < epsilon$ and $f_n(z) = h_n(z)prod_{l=1}^k (z-a_{l,n})=(C+O(z-a))prod_{l=1}^k (z-a_{l,n})$, $frac{f_n'}{f_n}(z) = O(z-a) + sum_{l=1}^k frac{1}{z-a_{l,n}}$. But $frac{f_n'}{f_n}(z) = 1-frac{frac{z^n}{n!}}{f_n(z)}=1-frac{a^n+O(z-a)}{n! (C+O(z-a))} frac{1}{prod_{l=1}^k (z-a_{l,n})}$, a contradiction.
– reuns
Nov 18 at 4:45
@reuns: Unless I am mistaken, your argument shows only that $f_n$ has no zeros in the unit disk for sufficiently large $n$.
– Martin R
Nov 18 at 8:31
@reuns: Unless I am mistaken, your argument shows only that $f_n$ has no zeros in the unit disk for sufficiently large $n$.
– Martin R
Nov 18 at 8:31
add a comment |
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You have all the facts you need. Here is an outline of what you need to do:
Use Rouche's theorem with $f_n$ and $z^nover n!$ to show that if some $f_N$ has a zero within the unit disk, then every $f_n$ with $ngt N$ has a zero in the unit disk. Now since there are an infinite number of zeros in the unit disk, the set of points mapping to zero under our $f_n$'s has at least one accumulation point. So we have a sequence of pairs, $f_i, p_i$ with $f_i(p_i)=0$ for each i, with a limit point $f, p$ (as the $f_n$'s also converge). For the sake of contradiction, we want to show $f(p)=0$ as we already know that f is the exponential and so has no zeros.
To do this use an epsilon delta argument: let $epsilon gt 0$, then find $N_0$ such that $f_n(x)$ is within $epsilon over 3$ of f(x) for all x in the unit disk when $ngt N_0$, then find $delta$ such that when $|x-x_0|ltdelta$, then $|f(x)-f(x_0)| lt$ $ epsilon over 3$. Lastly find $N_1$ so that $ngt N_1$ implies $|p-p_n| gt delta$. Use the triangle inequality to show that when $n>max(N_0,N_1)$ we have $|f(p)-f_n(p_n)|<epsilon$. This completes the contradiction, as it implies $f(x)=e^x$ has a zero in the unit disk.
answered Nov 18 at 5:07
Mark
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1 Answer
1
active
oldest
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active
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active
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up vote
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You have all the facts you need. Here is an outline of what you need to do:
Use Rouche's theorem with $f_n$ and $z^nover n!$ to show that if some $f_N$ has a zero within the unit disk, then every $f_n$ with $ngt N$ has a zero in the unit disk. Now since there are an infinite number of zeros in the unit disk, the set of points mapping to zero under our $f_n$'s has at least one accumulation point. So we have a sequence of pairs, $f_i, p_i$ with $f_i(p_i)=0$ for each i, with a limit point $f, p$ (as the $f_n$'s also converge). For the sake of contradiction, we want to show $f(p)=0$ as we already know that f is the exponential and so has no zeros.
To do this use an epsilon delta argument: let $epsilon gt 0$, then find $N_0$ such that $f_n(x)$ is within $epsilon over 3$ of f(x) for all x in the unit disk when $ngt N_0$, then find $delta$ such that when $|x-x_0|ltdelta$, then $|f(x)-f(x_0)| lt$ $ epsilon over 3$. Lastly find $N_1$ so that $ngt N_1$ implies $|p-p_n| gt delta$. Use the triangle inequality to show that when $n>max(N_0,N_1)$ we have $|f(p)-f_n(p_n)|<epsilon$. This completes the contradiction, as it implies $f(x)=e^x$ has a zero in the unit disk.
answered Nov 18 at 5:07
Mark
3916
add a comment |
up vote
0
down vote
You have all the facts you need. Here is an outline of what you need to do:
Use Rouche's theorem with $f_n$ and $z^nover n!$ to show that if some $f_N$ has a zero within the unit disk, then every $f_n$ with $ngt N$ has a zero in the unit disk. Now since there are an infinite number of zeros in the unit disk, the set of points mapping to zero under our $f_n$'s has at least one accumulation point. So we have a sequence of pairs, $f_i, p_i$ with $f_i(p_i)=0$ for each i, with a limit point $f, p$ (as the $f_n$'s also converge). For the sake of contradiction, we want to show $f(p)=0$ as we already know that f is the exponential and so has no zeros.
To do this use an epsilon delta argument: let $epsilon gt 0$, then find $N_0$ such that $f_n(x)$ is within $epsilon over 3$ of f(x) for all x in the unit disk when $ngt N_0$, then find $delta$ such that when $|x-x_0|ltdelta$, then $|f(x)-f(x_0)| lt$ $ epsilon over 3$. Lastly find $N_1$ so that $ngt N_1$ implies $|p-p_n| gt delta$. Use the triangle inequality to show that when $n>max(N_0,N_1)$ we have $|f(p)-f_n(p_n)|<epsilon$. This completes the contradiction, as it implies $f(x)=e^x$ has a zero in the unit disk.
answered Nov 18 at 5:07
Mark
3916
add a comment |
up vote
0
down vote
up vote
0
down vote
You have all the facts you need. Here is an outline of what you need to do:
Use Rouche's theorem with $f_n$ and $z^nover n!$ to show that if some $f_N$ has a zero within the unit disk, then every $f_n$ with $ngt N$ has a zero in the unit disk. Now since there are an infinite number of zeros in the unit disk, the set of points mapping to zero under our $f_n$'s has at least one accumulation point. So we have a sequence of pairs, $f_i, p_i$ with $f_i(p_i)=0$ for each i, with a limit point $f, p$ (as the $f_n$'s also converge). For the sake of contradiction, we want to show $f(p)=0$ as we already know that f is the exponential and so has no zeros.
To do this use an epsilon delta argument: let $epsilon gt 0$, then find $N_0$ such that $f_n(x)$ is within $epsilon over 3$ of f(x) for all x in the unit disk when $ngt N_0$, then find $delta$ such that when $|x-x_0|ltdelta$, then $|f(x)-f(x_0)| lt$ $ epsilon over 3$. Lastly find $N_1$ so that $ngt N_1$ implies $|p-p_n| gt delta$. Use the triangle inequality to show that when $n>max(N_0,N_1)$ we have $|f(p)-f_n(p_n)|<epsilon$. This completes the contradiction, as it implies $f(x)=e^x$ has a zero in the unit disk.
answered Nov 18 at 5:07
Mark
3916
You have all the facts you need. Here is an outline of what you need to do:
Use Rouche's theorem with $f_n$ and $z^nover n!$ to show that if some $f_N$ has a zero within the unit disk, then every $f_n$ with $ngt N$ has a zero in the unit disk. Now since there are an infinite number of zeros in the unit disk, the set of points mapping to zero under our $f_n$'s has at least one accumulation point. So we have a sequence of pairs, $f_i, p_i$ with $f_i(p_i)=0$ for each i, with a limit point $f, p$ (as the $f_n$'s also converge). For the sake of contradiction, we want to show $f(p)=0$ as we already know that f is the exponential and so has no zeros.
To do this use an epsilon delta argument: let $epsilon gt 0$, then find $N_0$ such that $f_n(x)$ is within $epsilon over 3$ of f(x) for all x in the unit disk when $ngt N_0$, then find $delta$ such that when $|x-x_0|ltdelta$, then $|f(x)-f(x_0)| lt$ $ epsilon over 3$. Lastly find $N_1$ so that $ngt N_1$ implies $|p-p_n| gt delta$. Use the triangle inequality to show that when $n>max(N_0,N_1)$ we have $|f(p)-f_n(p_n)|<epsilon$. This completes the contradiction, as it implies $f(x)=e^x$ has a zero in the unit disk.
answered Nov 18 at 5:07
Mark
3916
answered Nov 18 at 5:07
Mark
3916
answered Nov 18 at 5:07
Mark
3916
answered Nov 18 at 5:07
Mark
3916
3916
add a comment |
add a comment |
It converges locally uniformly to an entire function $f$. So if $f$ has no zeros it is finished. Otherwise assume $f$ has a zero of order $k$ at $z=a$, $f(z) sim C (z-a)^k$. Fix $epsilon$ very small, there is $N$ such that every $f_n,n ge N$ have $k$ zeros on $|z-a| < epsilon$ and $f_n(z) = h_n(z)prod_{l=1}^k (z-a_{l,n})=(C+O(z-a))prod_{l=1}^k (z-a_{l,n})$, $frac{f_n'}{f_n}(z) = O(z-a) + sum_{l=1}^k frac{1}{z-a_{l,n}}$. But $frac{f_n'}{f_n}(z) = 1-frac{frac{z^n}{n!}}{f_n(z)}=1-frac{a^n+O(z-a)}{n! (C+O(z-a))} frac{1}{prod_{l=1}^k (z-a_{l,n})}$, a contradiction.
– reuns
Nov 18 at 4:45
@reuns: Unless I am mistaken, your argument shows only that $f_n$ has no zeros in the unit disk for sufficiently large $n$.
– Martin R
Nov 18 at 8:31