Order of growth of the entire function $sin(sqrt{z})/sqrt{z}$
$begingroup$
Show that
$$f(z)=frac{sinsqrt z}{sqrt z}$$
is an entire function of finite order $rho$ and determine $rho$.
I observed that the two determinations of the square root differ only for the signum. Since $sin(-z)=-sin z$, we have that $f(z)$ is well defined, and entire because it's the ratio of two entire functions with denominator never vanishing.
For the order i use the Taylor expansion
$$sin z=sum_{n=0}^{infty}(-1)^nfrac{z^{2n+1}}{(2n+1)!}$$
which for $z=sqrt z$ gives
$$sinsqrt z=sum_{n=0}^{infty}(-1)^nfrac{z^{n}sqrt z}{(2n+1)!}$$
Thus
$$frac{sinsqrt z}{sqrt z}=sum_{n=0}^{infty}(-1)^nfrac{z^{n}}{(2n+1)!}$$
Then we have
$$(2n+1)!geq2^n n!$$
hence
$$bigg|frac{sinsqrt z}{sqrt z}bigg|leqlargesum_{n=0}^{infty}left(frac{|z|}{2}right)^{n}cdotfrac{1}{n!}=e^{|z|/2}$$
This (if correct) shows that $rholeqfrac{1}{2}$. How can be shown the identity?
complex-analysis asymptotics entire-functions
edited Dec 10 '18 at 9:37
Gaby Alfonso
736315
asked Jan 9 '13 at 19:51
Federica MaggioniFederica Maggioni
5,96111336
$endgroup$
add a comment |
$begingroup$
Show that
$$f(z)=frac{sinsqrt z}{sqrt z}$$
is an entire function of finite order $rho$ and determine $rho$.
I observed that the two determinations of the square root differ only for the signum. Since $sin(-z)=-sin z$, we have that $f(z)$ is well defined, and entire because it's the ratio of two entire functions with denominator never vanishing.
For the order i use the Taylor expansion
$$sin z=sum_{n=0}^{infty}(-1)^nfrac{z^{2n+1}}{(2n+1)!}$$
which for $z=sqrt z$ gives
$$sinsqrt z=sum_{n=0}^{infty}(-1)^nfrac{z^{n}sqrt z}{(2n+1)!}$$
Thus
$$frac{sinsqrt z}{sqrt z}=sum_{n=0}^{infty}(-1)^nfrac{z^{n}}{(2n+1)!}$$
Then we have
$$(2n+1)!geq2^n n!$$
hence
$$bigg|frac{sinsqrt z}{sqrt z}bigg|leqlargesum_{n=0}^{infty}left(frac{|z|}{2}right)^{n}cdotfrac{1}{n!}=e^{|z|/2}$$
This (if correct) shows that $rholeqfrac{1}{2}$. How can be shown the identity?
complex-analysis asymptotics entire-functions
edited Dec 10 '18 at 9:37
Gaby Alfonso
736315
asked Jan 9 '13 at 19:51
Federica MaggioniFederica Maggioni
5,96111336
$endgroup$
$begingroup$
There are a few mistakes here. The series for $sin z$ is not quite right. You mean: $$sin z = sum_{n=0}^{infty} frac{(-1)^nz^{n+1}}{(2n+1)!} , . $$ When you change from the series for $sin z$ to the series for $sin sqrt{z}$, you seem to change $z^{2n}$ to $z^nsqrt{z}$. This isn't correct. $(sqrt{z})^{2n} = (z^{1/2})^{2n} = z^{2n/2} = z^n$. In fact: $$sum_{n=0}^{infty} frac{(-1)^nz^nsqrt{z}}{(2n+1)!} = frac{sqrt{z}sinh sqrt{-z}}{sqrt{-z}} , . $$
$endgroup$
– Fly by Night
Jan 9 '13 at 20:08
$begingroup$
sorry, there was an error in the $sin z$ series, i've edited. thank you
$endgroup$
– Federica Maggioni
Jan 9 '13 at 20:09
$begingroup$
Remember: You can use sin cos tan sec cot and csc for the trig' functions in LaTeX, i.e. $sin z$ instead of $sin z$.
$endgroup$
– Fly by Night
Jan 9 '13 at 20:14
add a comment |
$begingroup$
Show that
$$f(z)=frac{sinsqrt z}{sqrt z}$$
is an entire function of finite order $rho$ and determine $rho$.
I observed that the two determinations of the square root differ only for the signum. Since $sin(-z)=-sin z$, we have that $f(z)$ is well defined, and entire because it's the ratio of two entire functions with denominator never vanishing.
For the order i use the Taylor expansion
$$sin z=sum_{n=0}^{infty}(-1)^nfrac{z^{2n+1}}{(2n+1)!}$$
which for $z=sqrt z$ gives
$$sinsqrt z=sum_{n=0}^{infty}(-1)^nfrac{z^{n}sqrt z}{(2n+1)!}$$
Thus
$$frac{sinsqrt z}{sqrt z}=sum_{n=0}^{infty}(-1)^nfrac{z^{n}}{(2n+1)!}$$
Then we have
$$(2n+1)!geq2^n n!$$
hence
$$bigg|frac{sinsqrt z}{sqrt z}bigg|leqlargesum_{n=0}^{infty}left(frac{|z|}{2}right)^{n}cdotfrac{1}{n!}=e^{|z|/2}$$
This (if correct) shows that $rholeqfrac{1}{2}$. How can be shown the identity?
complex-analysis asymptotics entire-functions
edited Dec 10 '18 at 9:37
Gaby Alfonso
736315
asked Jan 9 '13 at 19:51
Federica MaggioniFederica Maggioni
5,96111336
$endgroup$
Show that
$$f(z)=frac{sinsqrt z}{sqrt z}$$
is an entire function of finite order $rho$ and determine $rho$.
I observed that the two determinations of the square root differ only for the signum. Since $sin(-z)=-sin z$, we have that $f(z)$ is well defined, and entire because it's the ratio of two entire functions with denominator never vanishing.
For the order i use the Taylor expansion
$$sin z=sum_{n=0}^{infty}(-1)^nfrac{z^{2n+1}}{(2n+1)!}$$
which for $z=sqrt z$ gives
$$sinsqrt z=sum_{n=0}^{infty}(-1)^nfrac{z^{n}sqrt z}{(2n+1)!}$$
Thus
$$frac{sinsqrt z}{sqrt z}=sum_{n=0}^{infty}(-1)^nfrac{z^{n}}{(2n+1)!}$$
Then we have
$$(2n+1)!geq2^n n!$$
hence
$$bigg|frac{sinsqrt z}{sqrt z}bigg|leqlargesum_{n=0}^{infty}left(frac{|z|}{2}right)^{n}cdotfrac{1}{n!}=e^{|z|/2}$$
This (if correct) shows that $rholeqfrac{1}{2}$. How can be shown the identity?
complex-analysis asymptotics entire-functions
complex-analysis asymptotics entire-functions
edited Dec 10 '18 at 9:37
Gaby Alfonso
736315
asked Jan 9 '13 at 19:51
Federica MaggioniFederica Maggioni
5,96111336
edited Dec 10 '18 at 9:37
Gaby Alfonso
736315
asked Jan 9 '13 at 19:51
Federica MaggioniFederica Maggioni
5,96111336
edited Dec 10 '18 at 9:37
Gaby Alfonso
736315
edited Dec 10 '18 at 9:37
Gaby Alfonso
736315
edited Dec 10 '18 at 9:37
Gaby Alfonso
736315
736315
asked Jan 9 '13 at 19:51
Federica MaggioniFederica Maggioni
5,96111336
asked Jan 9 '13 at 19:51
Federica MaggioniFederica Maggioni
5,96111336
asked Jan 9 '13 at 19:51
Federica MaggioniFederica Maggioni
5,96111336
5,96111336
$begingroup$
There are a few mistakes here. The series for $sin z$ is not quite right. You mean: $$sin z = sum_{n=0}^{infty} frac{(-1)^nz^{n+1}}{(2n+1)!} , . $$ When you change from the series for $sin z$ to the series for $sin sqrt{z}$, you seem to change $z^{2n}$ to $z^nsqrt{z}$. This isn't correct. $(sqrt{z})^{2n} = (z^{1/2})^{2n} = z^{2n/2} = z^n$. In fact: $$sum_{n=0}^{infty} frac{(-1)^nz^nsqrt{z}}{(2n+1)!} = frac{sqrt{z}sinh sqrt{-z}}{sqrt{-z}} , . $$
$endgroup$
– Fly by Night
Jan 9 '13 at 20:08
$begingroup$
sorry, there was an error in the $sin z$ series, i've edited. thank you
$endgroup$
– Federica Maggioni
Jan 9 '13 at 20:09
$begingroup$
Remember: You can use sin cos tan sec cot and csc for the trig' functions in LaTeX, i.e. $sin z$ instead of $sin z$.
$endgroup$
– Fly by Night
Jan 9 '13 at 20:14
add a comment |
$begingroup$
There are a few mistakes here. The series for $sin z$ is not quite right. You mean: $$sin z = sum_{n=0}^{infty} frac{(-1)^nz^{n+1}}{(2n+1)!} , . $$ When you change from the series for $sin z$ to the series for $sin sqrt{z}$, you seem to change $z^{2n}$ to $z^nsqrt{z}$. This isn't correct. $(sqrt{z})^{2n} = (z^{1/2})^{2n} = z^{2n/2} = z^n$. In fact: $$sum_{n=0}^{infty} frac{(-1)^nz^nsqrt{z}}{(2n+1)!} = frac{sqrt{z}sinh sqrt{-z}}{sqrt{-z}} , . $$
$endgroup$
– Fly by Night
Jan 9 '13 at 20:08
$begingroup$
sorry, there was an error in the $sin z$ series, i've edited. thank you
$endgroup$
– Federica Maggioni
Jan 9 '13 at 20:09
$begingroup$
Remember: You can use sin cos tan sec cot and csc for the trig' functions in LaTeX, i.e. $sin z$ instead of $sin z$.
$endgroup$
– Fly by Night
Jan 9 '13 at 20:14
$begingroup$
There are a few mistakes here. The series for $sin z$ is not quite right. You mean: $$sin z = sum_{n=0}^{infty} frac{(-1)^nz^{n+1}}{(2n+1)!} , . $$ When you change from the series for $sin z$ to the series for $sin sqrt{z}$, you seem to change $z^{2n}$ to $z^nsqrt{z}$. This isn't correct. $(sqrt{z})^{2n} = (z^{1/2})^{2n} = z^{2n/2} = z^n$. In fact: $$sum_{n=0}^{infty} frac{(-1)^nz^nsqrt{z}}{(2n+1)!} = frac{sqrt{z}sinh sqrt{-z}}{sqrt{-z}} , . $$
$endgroup$
– Fly by Night
Jan 9 '13 at 20:08
$begingroup$
There are a few mistakes here. The series for $sin z$ is not quite right. You mean: $$sin z = sum_{n=0}^{infty} frac{(-1)^nz^{n+1}}{(2n+1)!} , . $$ When you change from the series for $sin z$ to the series for $sin sqrt{z}$, you seem to change $z^{2n}$ to $z^nsqrt{z}$. This isn't correct. $(sqrt{z})^{2n} = (z^{1/2})^{2n} = z^{2n/2} = z^n$. In fact: $$sum_{n=0}^{infty} frac{(-1)^nz^nsqrt{z}}{(2n+1)!} = frac{sqrt{z}sinh sqrt{-z}}{sqrt{-z}} , . $$
$endgroup$
– Fly by Night
Jan 9 '13 at 20:08
$begingroup$
sorry, there was an error in the $sin z$ series, i've edited. thank you
$endgroup$
– Federica Maggioni
Jan 9 '13 at 20:09
$begingroup$
sorry, there was an error in the $sin z$ series, i've edited. thank you
$endgroup$
– Federica Maggioni
Jan 9 '13 at 20:09
$begingroup$
Remember: You can use sin cos tan sec cot and csc for the trig' functions in LaTeX, i.e. $sin z$ instead of $sin z$.
$endgroup$
– Fly by Night
Jan 9 '13 at 20:14
$begingroup$
Remember: You can use sin cos tan sec cot and csc for the trig' functions in LaTeX, i.e. $sin z$ instead of $sin z$.
$endgroup$
– Fly by Night
Jan 9 '13 at 20:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To show $rho geq 1/2$, write
$$
f(z) = frac{sin sqrt{z}}{sqrt{z}} = frac{e^{isqrt{z}}-e^{-isqrt{z}}}{2isqrt{z}}
$$
and show that
$$
f(-x) sim frac{e^{sqrt{x}}}{2sqrt{x}}
$$
as $x to infty$ with $x > 0$.
Clarification: If $f$ is of order $rho'$ then for any $rho>rho'$ there is $C$ such that $$|f(z)| leq C exp(|z|^rho)$$ so that $|f(z)| exp(-|z|^rho)$ is bounded. In particular, $$|f(-x)| exp(-x^rho)$$ is bounded as $x to infty$. Now suppose $rho'< 1/2$, pick $rhoin (rho',1/2)$ and use the asymptotic estimate above.
answered Jan 9 '13 at 20:43
Antonio VargasAntonio Vargas
20.7k245111
$endgroup$
$begingroup$
$f(-x)=frac{e^{-sqrt x} -e^{sqrt x}}{-2sqrt x}$, dividing this by $-frac{e^{sqrt x}}{2sqrt x}$ i get something which tends to 1 as x goes to $infty$. Hence $f(-x)$ is asymptotic to what you said. Now, why this implies the order of growth is greater then one-half? does asymptotic implies $ O(x^{1/2})$?
$endgroup$
– Federica Maggioni
Jan 9 '13 at 20:57
$begingroup$
@FedericaMaggioni, If $f$ is of order $rho$ then $$|f(z)| leq C exp(|z|^rho)$$ for $|z|$ large enough, so that $$|f(z)| exp(-|z|^rho)$$ is bounded. In particular, $$|f(-x)| exp(-x^rho)$$ is bounded as $x to infty$. Now suppose $rho < 1/2$ and use the asymptotic estimate.
$endgroup$
– Antonio Vargas
Jan 9 '13 at 21:10
$begingroup$
substituting the asymptotic estimate in the last equation i get something which goes to infinite at numerator faster than denominator, contraddicting the boundness.
$endgroup$
– Federica Maggioni
Jan 9 '13 at 21:17
$begingroup$
@FedericaMaggioni exactly.
$endgroup$
– Antonio Vargas
Jan 9 '13 at 21:17
1
$begingroup$
But a correct form of a comment can be moved to the post, where it's better placed anyway. I made such an edit just now; please review.
$endgroup$
– user147263
Jun 27 '15 at 3:52
|
show 2 more comments
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1 Answer
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1 Answer
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$begingroup$
To show $rho geq 1/2$, write
$$
f(z) = frac{sin sqrt{z}}{sqrt{z}} = frac{e^{isqrt{z}}-e^{-isqrt{z}}}{2isqrt{z}}
$$
and show that
$$
f(-x) sim frac{e^{sqrt{x}}}{2sqrt{x}}
$$
as $x to infty$ with $x > 0$.
Clarification: If $f$ is of order $rho'$ then for any $rho>rho'$ there is $C$ such that $$|f(z)| leq C exp(|z|^rho)$$ so that $|f(z)| exp(-|z|^rho)$ is bounded. In particular, $$|f(-x)| exp(-x^rho)$$ is bounded as $x to infty$. Now suppose $rho'< 1/2$, pick $rhoin (rho',1/2)$ and use the asymptotic estimate above.
answered Jan 9 '13 at 20:43
Antonio VargasAntonio Vargas
20.7k245111
$endgroup$
$begingroup$
$f(-x)=frac{e^{-sqrt x} -e^{sqrt x}}{-2sqrt x}$, dividing this by $-frac{e^{sqrt x}}{2sqrt x}$ i get something which tends to 1 as x goes to $infty$. Hence $f(-x)$ is asymptotic to what you said. Now, why this implies the order of growth is greater then one-half? does asymptotic implies $ O(x^{1/2})$?
$endgroup$
– Federica Maggioni
Jan 9 '13 at 20:57
$begingroup$
@FedericaMaggioni, If $f$ is of order $rho$ then $$|f(z)| leq C exp(|z|^rho)$$ for $|z|$ large enough, so that $$|f(z)| exp(-|z|^rho)$$ is bounded. In particular, $$|f(-x)| exp(-x^rho)$$ is bounded as $x to infty$. Now suppose $rho < 1/2$ and use the asymptotic estimate.
$endgroup$
– Antonio Vargas
Jan 9 '13 at 21:10
$begingroup$
substituting the asymptotic estimate in the last equation i get something which goes to infinite at numerator faster than denominator, contraddicting the boundness.
$endgroup$
– Federica Maggioni
Jan 9 '13 at 21:17
$begingroup$
@FedericaMaggioni exactly.
$endgroup$
– Antonio Vargas
Jan 9 '13 at 21:17
1
$begingroup$
But a correct form of a comment can be moved to the post, where it's better placed anyway. I made such an edit just now; please review.
$endgroup$
– user147263
Jun 27 '15 at 3:52
|
show 2 more comments
$begingroup$
To show $rho geq 1/2$, write
$$
f(z) = frac{sin sqrt{z}}{sqrt{z}} = frac{e^{isqrt{z}}-e^{-isqrt{z}}}{2isqrt{z}}
$$
and show that
$$
f(-x) sim frac{e^{sqrt{x}}}{2sqrt{x}}
$$
as $x to infty$ with $x > 0$.
Clarification: If $f$ is of order $rho'$ then for any $rho>rho'$ there is $C$ such that $$|f(z)| leq C exp(|z|^rho)$$ so that $|f(z)| exp(-|z|^rho)$ is bounded. In particular, $$|f(-x)| exp(-x^rho)$$ is bounded as $x to infty$. Now suppose $rho'< 1/2$, pick $rhoin (rho',1/2)$ and use the asymptotic estimate above.
answered Jan 9 '13 at 20:43
Antonio VargasAntonio Vargas
20.7k245111
$endgroup$
$begingroup$
$f(-x)=frac{e^{-sqrt x} -e^{sqrt x}}{-2sqrt x}$, dividing this by $-frac{e^{sqrt x}}{2sqrt x}$ i get something which tends to 1 as x goes to $infty$. Hence $f(-x)$ is asymptotic to what you said. Now, why this implies the order of growth is greater then one-half? does asymptotic implies $ O(x^{1/2})$?
$endgroup$
– Federica Maggioni
Jan 9 '13 at 20:57
$begingroup$
@FedericaMaggioni, If $f$ is of order $rho$ then $$|f(z)| leq C exp(|z|^rho)$$ for $|z|$ large enough, so that $$|f(z)| exp(-|z|^rho)$$ is bounded. In particular, $$|f(-x)| exp(-x^rho)$$ is bounded as $x to infty$. Now suppose $rho < 1/2$ and use the asymptotic estimate.
$endgroup$
– Antonio Vargas
Jan 9 '13 at 21:10
$begingroup$
substituting the asymptotic estimate in the last equation i get something which goes to infinite at numerator faster than denominator, contraddicting the boundness.
$endgroup$
– Federica Maggioni
Jan 9 '13 at 21:17
$begingroup$
@FedericaMaggioni exactly.
$endgroup$
– Antonio Vargas
Jan 9 '13 at 21:17
1
$begingroup$
But a correct form of a comment can be moved to the post, where it's better placed anyway. I made such an edit just now; please review.
$endgroup$
– user147263
Jun 27 '15 at 3:52
|
show 2 more comments
$begingroup$
To show $rho geq 1/2$, write
$$
f(z) = frac{sin sqrt{z}}{sqrt{z}} = frac{e^{isqrt{z}}-e^{-isqrt{z}}}{2isqrt{z}}
$$
and show that
$$
f(-x) sim frac{e^{sqrt{x}}}{2sqrt{x}}
$$
as $x to infty$ with $x > 0$.
Clarification: If $f$ is of order $rho'$ then for any $rho>rho'$ there is $C$ such that $$|f(z)| leq C exp(|z|^rho)$$ so that $|f(z)| exp(-|z|^rho)$ is bounded. In particular, $$|f(-x)| exp(-x^rho)$$ is bounded as $x to infty$. Now suppose $rho'< 1/2$, pick $rhoin (rho',1/2)$ and use the asymptotic estimate above.
answered Jan 9 '13 at 20:43
Antonio VargasAntonio Vargas
20.7k245111
$endgroup$
To show $rho geq 1/2$, write
$$
f(z) = frac{sin sqrt{z}}{sqrt{z}} = frac{e^{isqrt{z}}-e^{-isqrt{z}}}{2isqrt{z}}
$$
and show that
$$
f(-x) sim frac{e^{sqrt{x}}}{2sqrt{x}}
$$
as $x to infty$ with $x > 0$.
Clarification: If $f$ is of order $rho'$ then for any $rho>rho'$ there is $C$ such that $$|f(z)| leq C exp(|z|^rho)$$ so that $|f(z)| exp(-|z|^rho)$ is bounded. In particular, $$|f(-x)| exp(-x^rho)$$ is bounded as $x to infty$. Now suppose $rho'< 1/2$, pick $rhoin (rho',1/2)$ and use the asymptotic estimate above.
answered Jan 9 '13 at 20:43
Antonio VargasAntonio Vargas
20.7k245111
edited Jun 27 '15 at 3:51
user147263
answered Jan 9 '13 at 20:43
Antonio VargasAntonio Vargas
20.7k245111
answered Jan 9 '13 at 20:43
Antonio VargasAntonio Vargas
20.7k245111
answered Jan 9 '13 at 20:43
Antonio VargasAntonio Vargas
20.7k245111
20.7k245111
$begingroup$
$f(-x)=frac{e^{-sqrt x} -e^{sqrt x}}{-2sqrt x}$, dividing this by $-frac{e^{sqrt x}}{2sqrt x}$ i get something which tends to 1 as x goes to $infty$. Hence $f(-x)$ is asymptotic to what you said. Now, why this implies the order of growth is greater then one-half? does asymptotic implies $ O(x^{1/2})$?
$endgroup$
– Federica Maggioni
Jan 9 '13 at 20:57
$begingroup$
@FedericaMaggioni, If $f$ is of order $rho$ then $$|f(z)| leq C exp(|z|^rho)$$ for $|z|$ large enough, so that $$|f(z)| exp(-|z|^rho)$$ is bounded. In particular, $$|f(-x)| exp(-x^rho)$$ is bounded as $x to infty$. Now suppose $rho < 1/2$ and use the asymptotic estimate.
$endgroup$
– Antonio Vargas
Jan 9 '13 at 21:10
$begingroup$
substituting the asymptotic estimate in the last equation i get something which goes to infinite at numerator faster than denominator, contraddicting the boundness.
$endgroup$
– Federica Maggioni
Jan 9 '13 at 21:17
$begingroup$
@FedericaMaggioni exactly.
$endgroup$
– Antonio Vargas
Jan 9 '13 at 21:17
1
$begingroup$
But a correct form of a comment can be moved to the post, where it's better placed anyway. I made such an edit just now; please review.
$endgroup$
– user147263
Jun 27 '15 at 3:52
|
show 2 more comments
$begingroup$
$f(-x)=frac{e^{-sqrt x} -e^{sqrt x}}{-2sqrt x}$, dividing this by $-frac{e^{sqrt x}}{2sqrt x}$ i get something which tends to 1 as x goes to $infty$. Hence $f(-x)$ is asymptotic to what you said. Now, why this implies the order of growth is greater then one-half? does asymptotic implies $ O(x^{1/2})$?
$endgroup$
– Federica Maggioni
Jan 9 '13 at 20:57
$begingroup$
@FedericaMaggioni, If $f$ is of order $rho$ then $$|f(z)| leq C exp(|z|^rho)$$ for $|z|$ large enough, so that $$|f(z)| exp(-|z|^rho)$$ is bounded. In particular, $$|f(-x)| exp(-x^rho)$$ is bounded as $x to infty$. Now suppose $rho < 1/2$ and use the asymptotic estimate.
$endgroup$
– Antonio Vargas
Jan 9 '13 at 21:10
$begingroup$
substituting the asymptotic estimate in the last equation i get something which goes to infinite at numerator faster than denominator, contraddicting the boundness.
$endgroup$
– Federica Maggioni
Jan 9 '13 at 21:17
$begingroup$
@FedericaMaggioni exactly.
$endgroup$
– Antonio Vargas
Jan 9 '13 at 21:17
1
$begingroup$
But a correct form of a comment can be moved to the post, where it's better placed anyway. I made such an edit just now; please review.
$endgroup$
– user147263
Jun 27 '15 at 3:52
$begingroup$
$f(-x)=frac{e^{-sqrt x} -e^{sqrt x}}{-2sqrt x}$, dividing this by $-frac{e^{sqrt x}}{2sqrt x}$ i get something which tends to 1 as x goes to $infty$. Hence $f(-x)$ is asymptotic to what you said. Now, why this implies the order of growth is greater then one-half? does asymptotic implies $ O(x^{1/2})$?
$endgroup$
– Federica Maggioni
Jan 9 '13 at 20:57
$begingroup$
$f(-x)=frac{e^{-sqrt x} -e^{sqrt x}}{-2sqrt x}$, dividing this by $-frac{e^{sqrt x}}{2sqrt x}$ i get something which tends to 1 as x goes to $infty$. Hence $f(-x)$ is asymptotic to what you said. Now, why this implies the order of growth is greater then one-half? does asymptotic implies $ O(x^{1/2})$?
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– Federica Maggioni
Jan 9 '13 at 20:57
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@FedericaMaggioni, If $f$ is of order $rho$ then $$|f(z)| leq C exp(|z|^rho)$$ for $|z|$ large enough, so that $$|f(z)| exp(-|z|^rho)$$ is bounded. In particular, $$|f(-x)| exp(-x^rho)$$ is bounded as $x to infty$. Now suppose $rho < 1/2$ and use the asymptotic estimate.
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– Antonio Vargas
Jan 9 '13 at 21:10
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@FedericaMaggioni, If $f$ is of order $rho$ then $$|f(z)| leq C exp(|z|^rho)$$ for $|z|$ large enough, so that $$|f(z)| exp(-|z|^rho)$$ is bounded. In particular, $$|f(-x)| exp(-x^rho)$$ is bounded as $x to infty$. Now suppose $rho < 1/2$ and use the asymptotic estimate.
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– Antonio Vargas
Jan 9 '13 at 21:10
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substituting the asymptotic estimate in the last equation i get something which goes to infinite at numerator faster than denominator, contraddicting the boundness.
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– Federica Maggioni
Jan 9 '13 at 21:17
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substituting the asymptotic estimate in the last equation i get something which goes to infinite at numerator faster than denominator, contraddicting the boundness.
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– Federica Maggioni
Jan 9 '13 at 21:17
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@FedericaMaggioni exactly.
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– Antonio Vargas
Jan 9 '13 at 21:17
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@FedericaMaggioni exactly.
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– Antonio Vargas
Jan 9 '13 at 21:17
1
1
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But a correct form of a comment can be moved to the post, where it's better placed anyway. I made such an edit just now; please review.
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– user147263
Jun 27 '15 at 3:52
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But a correct form of a comment can be moved to the post, where it's better placed anyway. I made such an edit just now; please review.
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– user147263
Jun 27 '15 at 3:52
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There are a few mistakes here. The series for $sin z$ is not quite right. You mean: $$sin z = sum_{n=0}^{infty} frac{(-1)^nz^{n+1}}{(2n+1)!} , . $$ When you change from the series for $sin z$ to the series for $sin sqrt{z}$, you seem to change $z^{2n}$ to $z^nsqrt{z}$. This isn't correct. $(sqrt{z})^{2n} = (z^{1/2})^{2n} = z^{2n/2} = z^n$. In fact: $$sum_{n=0}^{infty} frac{(-1)^nz^nsqrt{z}}{(2n+1)!} = frac{sqrt{z}sinh sqrt{-z}}{sqrt{-z}} , . $$
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– Fly by Night
Jan 9 '13 at 20:08
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sorry, there was an error in the $sin z$ series, i've edited. thank you
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– Federica Maggioni
Jan 9 '13 at 20:09
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Remember: You can use sin cos tan sec cot and csc for the trig' functions in LaTeX, i.e. $sin z$ instead of $sin z$.
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– Fly by Night
Jan 9 '13 at 20:14