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Prove that sufficiently large partial sums of the Taylor series expansion of $e^z$ have all roots outside of an arbitrarily large radius.
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Consider the complex-valued family of functions $$
f_n(z) = sum_k=0^n frac1k!z^k.
$$
Is it possible to use Rouché's Theorem to prove that for any $R in mathbb R$ there exists some $n$ such that all roots of $f_n$ have modulus strictly greater than $R$?
The solutions that I've seen for this exploit the uniform convergence of the Taylor series for $e^z$, but I'm curious if it's possible to use Rouché's Theorem to show that there are no roots inside the circle of radius $R$ for sufficiently large $n$.
complex-analysis taylor-expansion roots rouches-theorem
asked Jul 25 at 17:27
Peter Kagey
1,36761952
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up vote
1
down vote
favorite
Consider the complex-valued family of functions $$
f_n(z) = sum_k=0^n frac1k!z^k.
$$
Is it possible to use Rouché's Theorem to prove that for any $R in mathbb R$ there exists some $n$ such that all roots of $f_n$ have modulus strictly greater than $R$?
The solutions that I've seen for this exploit the uniform convergence of the Taylor series for $e^z$, but I'm curious if it's possible to use Rouché's Theorem to show that there are no roots inside the circle of radius $R$ for sufficiently large $n$.
complex-analysis taylor-expansion roots rouches-theorem
asked Jul 25 at 17:27
Peter Kagey
1,36761952
Are you interested in a proof that uses uniform convergence and Rouché's theorem?
– José Carlos Santos
Jul 25 at 17:30
@JoséCarlosSantos, I'm interested in a proof that uses Rouché's theorem in any useful way—it's fine if it uniform convergence too, but it doesn't need to.
– Peter Kagey
Jul 25 at 19:07
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the complex-valued family of functions $$
f_n(z) = sum_k=0^n frac1k!z^k.
$$
Is it possible to use Rouché's Theorem to prove that for any $R in mathbb R$ there exists some $n$ such that all roots of $f_n$ have modulus strictly greater than $R$?
The solutions that I've seen for this exploit the uniform convergence of the Taylor series for $e^z$, but I'm curious if it's possible to use Rouché's Theorem to show that there are no roots inside the circle of radius $R$ for sufficiently large $n$.
complex-analysis taylor-expansion roots rouches-theorem
asked Jul 25 at 17:27
Peter Kagey
1,36761952
Consider the complex-valued family of functions $$
f_n(z) = sum_k=0^n frac1k!z^k.
$$
Is it possible to use Rouché's Theorem to prove that for any $R in mathbb R$ there exists some $n$ such that all roots of $f_n$ have modulus strictly greater than $R$?
The solutions that I've seen for this exploit the uniform convergence of the Taylor series for $e^z$, but I'm curious if it's possible to use Rouché's Theorem to show that there are no roots inside the circle of radius $R$ for sufficiently large $n$.
complex-analysis taylor-expansion roots rouches-theorem
asked Jul 25 at 17:27
Peter Kagey
1,36761952
asked Jul 25 at 17:27
Peter Kagey
1,36761952
asked Jul 25 at 17:27
Peter Kagey
1,36761952
asked Jul 25 at 17:27
Peter Kagey
1,36761952
1,36761952
Are you interested in a proof that uses uniform convergence and Rouché's theorem?
– José Carlos Santos
Jul 25 at 17:30
@JoséCarlosSantos, I'm interested in a proof that uses Rouché's theorem in any useful way—it's fine if it uniform convergence too, but it doesn't need to.
– Peter Kagey
Jul 25 at 19:07
add a comment |Â
Are you interested in a proof that uses uniform convergence and Rouché's theorem?
– José Carlos Santos
Jul 25 at 17:30
@JoséCarlosSantos, I'm interested in a proof that uses Rouché's theorem in any useful way—it's fine if it uniform convergence too, but it doesn't need to.
– Peter Kagey
Jul 25 at 19:07
Are you interested in a proof that uses uniform convergence and Rouché's theorem?
– José Carlos Santos
Jul 25 at 17:30
Are you interested in a proof that uses uniform convergence and Rouché's theorem?
– José Carlos Santos
Jul 25 at 17:30
@JoséCarlosSantos, I'm interested in a proof that uses Rouché's theorem in any useful way—it's fine if it uniform convergence too, but it doesn't need to.
– Peter Kagey
Jul 25 at 19:07
@JoséCarlosSantos, I'm interested in a proof that uses Rouché's theorem in any useful way—it's fine if it uniform convergence too, but it doesn't need to.
– Peter Kagey
Jul 25 at 19:07
add a comment |Â
2 Answers
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Take $R>0$ and let $M=infleft,$. Since the series $sum_n=0^inftyfracz^nn!$ converges uniformly to $e^z$ on the circle $z$, there is a natural $N$ such that $bigl|f_n(z)-e^zbigr|<M$ for each $z$ there. In particular$$|z|=Rimpliesbigl|f_n(z)-e^zbigr|<|e^z|.$$Therefore, by Rouché's theorem, on the closed disk $D_R(0)$ the functions $f_n$ and $exp$ have the same number of zeros — that is, no zeros.
answered Jul 25 at 22:36
José Carlos Santos
113k1696174
add a comment |Â
up vote
1
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The radius of convergence of the power series for $e^x$ is $infty,$ so $(f_n)_n$ converges uniformly to $e^z $ on any compact subset of $Bbb C.$
The function $g(z)=|e^z|$ is continuous from $Bbb C$ to $Bbb R^+,$ and $d(R)=z$ is compact, so $m(R)=min e^z$ exists (and is positive).
Take $nin Bbb N$ large enough that $forall ngeq N;(sup : zin d(R)<m(R);).$ Then $forall ngeq N ;forall zin d(R);(|f_n(z)|>0).$
answered Jul 26 at 5:02
DanielWainfleet
31.5k31542
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Take $R>0$ and let $M=infleft,$. Since the series $sum_n=0^inftyfracz^nn!$ converges uniformly to $e^z$ on the circle $z$, there is a natural $N$ such that $bigl|f_n(z)-e^zbigr|<M$ for each $z$ there. In particular$$|z|=Rimpliesbigl|f_n(z)-e^zbigr|<|e^z|.$$Therefore, by Rouché's theorem, on the closed disk $D_R(0)$ the functions $f_n$ and $exp$ have the same number of zeros — that is, no zeros.
answered Jul 25 at 22:36
José Carlos Santos
113k1696174
add a comment |Â
up vote
2
down vote
Take $R>0$ and let $M=infleft,$. Since the series $sum_n=0^inftyfracz^nn!$ converges uniformly to $e^z$ on the circle $z$, there is a natural $N$ such that $bigl|f_n(z)-e^zbigr|<M$ for each $z$ there. In particular$$|z|=Rimpliesbigl|f_n(z)-e^zbigr|<|e^z|.$$Therefore, by Rouché's theorem, on the closed disk $D_R(0)$ the functions $f_n$ and $exp$ have the same number of zeros — that is, no zeros.
answered Jul 25 at 22:36
José Carlos Santos
113k1696174
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Take $R>0$ and let $M=infleft,$. Since the series $sum_n=0^inftyfracz^nn!$ converges uniformly to $e^z$ on the circle $z$, there is a natural $N$ such that $bigl|f_n(z)-e^zbigr|<M$ for each $z$ there. In particular$$|z|=Rimpliesbigl|f_n(z)-e^zbigr|<|e^z|.$$Therefore, by Rouché's theorem, on the closed disk $D_R(0)$ the functions $f_n$ and $exp$ have the same number of zeros — that is, no zeros.
answered Jul 25 at 22:36
José Carlos Santos
113k1696174
Take $R>0$ and let $M=infleft,$. Since the series $sum_n=0^inftyfracz^nn!$ converges uniformly to $e^z$ on the circle $z$, there is a natural $N$ such that $bigl|f_n(z)-e^zbigr|<M$ for each $z$ there. In particular$$|z|=Rimpliesbigl|f_n(z)-e^zbigr|<|e^z|.$$Therefore, by Rouché's theorem, on the closed disk $D_R(0)$ the functions $f_n$ and $exp$ have the same number of zeros — that is, no zeros.
answered Jul 25 at 22:36
José Carlos Santos
113k1696174
answered Jul 25 at 22:36
José Carlos Santos
113k1696174
answered Jul 25 at 22:36
José Carlos Santos
113k1696174
answered Jul 25 at 22:36
José Carlos Santos
113k1696174
113k1696174
add a comment |Â
add a comment |Â
up vote
1
down vote
The radius of convergence of the power series for $e^x$ is $infty,$ so $(f_n)_n$ converges uniformly to $e^z $ on any compact subset of $Bbb C.$
The function $g(z)=|e^z|$ is continuous from $Bbb C$ to $Bbb R^+,$ and $d(R)=z$ is compact, so $m(R)=min e^z$ exists (and is positive).
Take $nin Bbb N$ large enough that $forall ngeq N;(sup : zin d(R)<m(R);).$ Then $forall ngeq N ;forall zin d(R);(|f_n(z)|>0).$
answered Jul 26 at 5:02
DanielWainfleet
31.5k31542
add a comment |Â
up vote
1
down vote
The radius of convergence of the power series for $e^x$ is $infty,$ so $(f_n)_n$ converges uniformly to $e^z $ on any compact subset of $Bbb C.$
The function $g(z)=|e^z|$ is continuous from $Bbb C$ to $Bbb R^+,$ and $d(R)=z$ is compact, so $m(R)=min e^z$ exists (and is positive).
Take $nin Bbb N$ large enough that $forall ngeq N;(sup : zin d(R)<m(R);).$ Then $forall ngeq N ;forall zin d(R);(|f_n(z)|>0).$
answered Jul 26 at 5:02
DanielWainfleet
31.5k31542
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The radius of convergence of the power series for $e^x$ is $infty,$ so $(f_n)_n$ converges uniformly to $e^z $ on any compact subset of $Bbb C.$
The function $g(z)=|e^z|$ is continuous from $Bbb C$ to $Bbb R^+,$ and $d(R)=z$ is compact, so $m(R)=min e^z$ exists (and is positive).
Take $nin Bbb N$ large enough that $forall ngeq N;(sup : zin d(R)<m(R);).$ Then $forall ngeq N ;forall zin d(R);(|f_n(z)|>0).$
answered Jul 26 at 5:02
DanielWainfleet
31.5k31542
The radius of convergence of the power series for $e^x$ is $infty,$ so $(f_n)_n$ converges uniformly to $e^z $ on any compact subset of $Bbb C.$
The function $g(z)=|e^z|$ is continuous from $Bbb C$ to $Bbb R^+,$ and $d(R)=z$ is compact, so $m(R)=min e^z$ exists (and is positive).
Take $nin Bbb N$ large enough that $forall ngeq N;(sup : zin d(R)<m(R);).$ Then $forall ngeq N ;forall zin d(R);(|f_n(z)|>0).$
answered Jul 26 at 5:02
DanielWainfleet
31.5k31542
edited Jul 26 at 5:30
answered Jul 26 at 5:02
DanielWainfleet
31.5k31542
answered Jul 26 at 5:02
DanielWainfleet
31.5k31542
answered Jul 26 at 5:02
DanielWainfleet
31.5k31542
31.5k31542
add a comment |Â
add a comment |Â
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Are you interested in a proof that uses uniform convergence and Rouché's theorem?
– José Carlos Santos
Jul 25 at 17:30
@JoséCarlosSantos, I'm interested in a proof that uses Rouché's theorem in any useful way—it's fine if it uniform convergence too, but it doesn't need to.
– Peter Kagey
Jul 25 at 19:07