If $f : Bbb{C} to Bbb{C}$ is entire, then $|f^{(n)}(0)| < n! n^n$ for some $n in Bbb{N}$
I'm trying to solve a problem that I can't seem to work out.
$f$ is an entire function. Prove that $|f^{(n)}(0)|< n!n^n$ for at least 1 $n$.
I've been thinking to use the Cauchy estimates somehow but there's no reason for me to believe that $f$ is bounded from above.
Any help is appreciated.
complex-analysis
asked Dec 3 '18 at 6:54
Pjames
162
add a comment |
I'm trying to solve a problem that I can't seem to work out.
$f$ is an entire function. Prove that $|f^{(n)}(0)|< n!n^n$ for at least 1 $n$.
I've been thinking to use the Cauchy estimates somehow but there's no reason for me to believe that $f$ is bounded from above.
Any help is appreciated.
complex-analysis
asked Dec 3 '18 at 6:54
Pjames
162
add a comment |
I'm trying to solve a problem that I can't seem to work out.
$f$ is an entire function. Prove that $|f^{(n)}(0)|< n!n^n$ for at least 1 $n$.
I've been thinking to use the Cauchy estimates somehow but there's no reason for me to believe that $f$ is bounded from above.
Any help is appreciated.
complex-analysis
asked Dec 3 '18 at 6:54
Pjames
162
I'm trying to solve a problem that I can't seem to work out.
$f$ is an entire function. Prove that $|f^{(n)}(0)|< n!n^n$ for at least 1 $n$.
I've been thinking to use the Cauchy estimates somehow but there's no reason for me to believe that $f$ is bounded from above.
Any help is appreciated.
complex-analysis
complex-analysis
asked Dec 3 '18 at 6:54
Pjames
162
asked Dec 3 '18 at 6:54
Pjames
162
edited Dec 3 '18 at 7:35
asked Dec 3 '18 at 6:54
Pjames
162
asked Dec 3 '18 at 6:54
Pjames
162
asked Dec 3 '18 at 6:54
Pjames
162
162
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Hint: Try proving it by contradiction. Suppose $|f^{(n)}(0)| geq n! n^n$ for all $n in Bbb{N}$. You are given that $f$ is entire. Can you say anything about the radius of convergence of $f$ around $0$?
answered Dec 3 '18 at 6:59
Brahadeesh
6,11742361
add a comment |
Let $M = sup { |f(z)| : |z| = 1}$. Since $f$ is entire, for all $n ge 0$, we have
$$f^{(n)}(0) = frac{n!}{2pi i}int_{|z|=1} frac{f(z)}{z^{n+1}}dz$$
Whenever $n > max{ 1, M }$, this leads to
$$|f^{(n)}(0)| le frac{n!}{2pi} int_{|z|=1} |f(z)| |dz| le
frac{n!}{2pi} int_{|z|=1} M |dz| = n!M
< n! n < n!n^n
$$
answered Dec 3 '18 at 7:20
achille hui
95.4k5130256
add a comment |
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2 Answers
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2 Answers
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Hint: Try proving it by contradiction. Suppose $|f^{(n)}(0)| geq n! n^n$ for all $n in Bbb{N}$. You are given that $f$ is entire. Can you say anything about the radius of convergence of $f$ around $0$?
answered Dec 3 '18 at 6:59
Brahadeesh
6,11742361
add a comment |
Hint: Try proving it by contradiction. Suppose $|f^{(n)}(0)| geq n! n^n$ for all $n in Bbb{N}$. You are given that $f$ is entire. Can you say anything about the radius of convergence of $f$ around $0$?
answered Dec 3 '18 at 6:59
Brahadeesh
6,11742361
add a comment |
Hint: Try proving it by contradiction. Suppose $|f^{(n)}(0)| geq n! n^n$ for all $n in Bbb{N}$. You are given that $f$ is entire. Can you say anything about the radius of convergence of $f$ around $0$?
answered Dec 3 '18 at 6:59
Brahadeesh
6,11742361
Hint: Try proving it by contradiction. Suppose $|f^{(n)}(0)| geq n! n^n$ for all $n in Bbb{N}$. You are given that $f$ is entire. Can you say anything about the radius of convergence of $f$ around $0$?
answered Dec 3 '18 at 6:59
Brahadeesh
6,11742361
answered Dec 3 '18 at 6:59
Brahadeesh
6,11742361
answered Dec 3 '18 at 6:59
Brahadeesh
6,11742361
answered Dec 3 '18 at 6:59
Brahadeesh
6,11742361
6,11742361
add a comment |
add a comment |
Let $M = sup { |f(z)| : |z| = 1}$. Since $f$ is entire, for all $n ge 0$, we have
$$f^{(n)}(0) = frac{n!}{2pi i}int_{|z|=1} frac{f(z)}{z^{n+1}}dz$$
Whenever $n > max{ 1, M }$, this leads to
$$|f^{(n)}(0)| le frac{n!}{2pi} int_{|z|=1} |f(z)| |dz| le
frac{n!}{2pi} int_{|z|=1} M |dz| = n!M
< n! n < n!n^n
$$
answered Dec 3 '18 at 7:20
achille hui
95.4k5130256
add a comment |
Let $M = sup { |f(z)| : |z| = 1}$. Since $f$ is entire, for all $n ge 0$, we have
$$f^{(n)}(0) = frac{n!}{2pi i}int_{|z|=1} frac{f(z)}{z^{n+1}}dz$$
Whenever $n > max{ 1, M }$, this leads to
$$|f^{(n)}(0)| le frac{n!}{2pi} int_{|z|=1} |f(z)| |dz| le
frac{n!}{2pi} int_{|z|=1} M |dz| = n!M
< n! n < n!n^n
$$
answered Dec 3 '18 at 7:20
achille hui
95.4k5130256
add a comment |
Let $M = sup { |f(z)| : |z| = 1}$. Since $f$ is entire, for all $n ge 0$, we have
$$f^{(n)}(0) = frac{n!}{2pi i}int_{|z|=1} frac{f(z)}{z^{n+1}}dz$$
Whenever $n > max{ 1, M }$, this leads to
$$|f^{(n)}(0)| le frac{n!}{2pi} int_{|z|=1} |f(z)| |dz| le
frac{n!}{2pi} int_{|z|=1} M |dz| = n!M
< n! n < n!n^n
$$
answered Dec 3 '18 at 7:20
achille hui
95.4k5130256
Let $M = sup { |f(z)| : |z| = 1}$. Since $f$ is entire, for all $n ge 0$, we have
$$f^{(n)}(0) = frac{n!}{2pi i}int_{|z|=1} frac{f(z)}{z^{n+1}}dz$$
Whenever $n > max{ 1, M }$, this leads to
$$|f^{(n)}(0)| le frac{n!}{2pi} int_{|z|=1} |f(z)| |dz| le
frac{n!}{2pi} int_{|z|=1} M |dz| = n!M
< n! n < n!n^n
$$
answered Dec 3 '18 at 7:20
achille hui
95.4k5130256
answered Dec 3 '18 at 7:20
achille hui
95.4k5130256
answered Dec 3 '18 at 7:20
achille hui
95.4k5130256
answered Dec 3 '18 at 7:20
achille hui
95.4k5130256
95.4k5130256
add a comment |
add a comment |
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