Order of growth of the entire function $sin(sqrt{z})/sqrt{z}$












6












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















6












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















6












6








6


4



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 9:37









Gaby Alfonso

736315




736315










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




















  • $begingroup$
    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












1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $f(-x)=frac{e^{-sqrt x} -e^{sqrt x}}{-2sqrt x}$, 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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $f(-x)=frac{e^{-sqrt x} -e^{sqrt x}}{-2sqrt x}$, 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
















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $f(-x)=frac{e^{-sqrt x} -e^{sqrt x}}{-2sqrt x}$, 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














1












1








1





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jun 27 '15 at 3:51







user147263

















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


















  • $begingroup$
    $f(-x)=frac{e^{-sqrt x} -e^{sqrt x}}{-2sqrt x}$, 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})$?
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– 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




$begingroup$
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




$begingroup$
@FedericaMaggioni exactly.
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– Antonio Vargas
Jan 9 '13 at 21:17




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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.
$endgroup$
– user147263
Jun 27 '15 at 3:52




$begingroup$
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


















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