Is there is an entire function $f$ such that $f(1)=pi$ and $f'(z)=|z|f(z)$ for all $zinmathbb C$?
$begingroup$
Problem: Is there is an entire function $f$ such that $f(1)=pi$ and $f'(z)=|z|f(z)$ for all $zinmathbb C$?
Question: Is the following correct?
On the unit circle line $partial K_1(0)$ the above equation is equivalent to $f'(z)=f(z)$. According to Picard-Lindelof the IVP with respect to $f(1)=pi$ is uniquely solved by $f(z)=frac{pi}{e}e^z$. Since $partial K_1(0)$ has an accumulation point the identity theorem can be applied and $f(z)=frac{pi}{e}e^z$ on the entire $mathbb C$-plane. This however contradicts $f'(z)=|z|f(z)$ on each point outside of the unit circle line.
complex-analysis proof-verification entire-functions
edited Jan 5 at 11:43
José Carlos Santos
173k23133241
asked Jan 5 at 11:04
RedLanternRedLantern
517
$endgroup$
add a comment |
$begingroup$
Problem: Is there is an entire function $f$ such that $f(1)=pi$ and $f'(z)=|z|f(z)$ for all $zinmathbb C$?
Question: Is the following correct?
On the unit circle line $partial K_1(0)$ the above equation is equivalent to $f'(z)=f(z)$. According to Picard-Lindelof the IVP with respect to $f(1)=pi$ is uniquely solved by $f(z)=frac{pi}{e}e^z$. Since $partial K_1(0)$ has an accumulation point the identity theorem can be applied and $f(z)=frac{pi}{e}e^z$ on the entire $mathbb C$-plane. This however contradicts $f'(z)=|z|f(z)$ on each point outside of the unit circle line.
complex-analysis proof-verification entire-functions
edited Jan 5 at 11:43
José Carlos Santos
173k23133241
asked Jan 5 at 11:04
RedLanternRedLantern
517
$endgroup$
add a comment |
$begingroup$
Problem: Is there is an entire function $f$ such that $f(1)=pi$ and $f'(z)=|z|f(z)$ for all $zinmathbb C$?
Question: Is the following correct?
On the unit circle line $partial K_1(0)$ the above equation is equivalent to $f'(z)=f(z)$. According to Picard-Lindelof the IVP with respect to $f(1)=pi$ is uniquely solved by $f(z)=frac{pi}{e}e^z$. Since $partial K_1(0)$ has an accumulation point the identity theorem can be applied and $f(z)=frac{pi}{e}e^z$ on the entire $mathbb C$-plane. This however contradicts $f'(z)=|z|f(z)$ on each point outside of the unit circle line.
complex-analysis proof-verification entire-functions
edited Jan 5 at 11:43
José Carlos Santos
173k23133241
asked Jan 5 at 11:04
RedLanternRedLantern
517
$endgroup$
Problem: Is there is an entire function $f$ such that $f(1)=pi$ and $f'(z)=|z|f(z)$ for all $zinmathbb C$?
Question: Is the following correct?
On the unit circle line $partial K_1(0)$ the above equation is equivalent to $f'(z)=f(z)$. According to Picard-Lindelof the IVP with respect to $f(1)=pi$ is uniquely solved by $f(z)=frac{pi}{e}e^z$. Since $partial K_1(0)$ has an accumulation point the identity theorem can be applied and $f(z)=frac{pi}{e}e^z$ on the entire $mathbb C$-plane. This however contradicts $f'(z)=|z|f(z)$ on each point outside of the unit circle line.
complex-analysis proof-verification entire-functions
complex-analysis proof-verification entire-functions
edited Jan 5 at 11:43
José Carlos Santos
173k23133241
asked Jan 5 at 11:04
RedLanternRedLantern
517
edited Jan 5 at 11:43
José Carlos Santos
173k23133241
asked Jan 5 at 11:04
RedLanternRedLantern
517
edited Jan 5 at 11:43
José Carlos Santos
173k23133241
edited Jan 5 at 11:43
José Carlos Santos
173k23133241
edited Jan 5 at 11:43
José Carlos Santos
173k23133241
173k23133241
asked Jan 5 at 11:04
RedLanternRedLantern
517
asked Jan 5 at 11:04
RedLanternRedLantern
517
asked Jan 5 at 11:04
RedLanternRedLantern
517
517
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
It's much easier to do it as follows: the set $Z$ of zeros of $f$ is a closed and discrete subset of $mathbb C$. So, on $mathbb{C}setminus Z$, we would have $lvert zrvert=frac{f'(z)}{f(z)}$. But $zmapstolvert zrvert$ is nowhere differentiable.
answered Jan 5 at 11:11
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
1
$begingroup$
So... is my argumentation correct or not?
$endgroup$
– RedLantern
Jan 5 at 11:45
1
$begingroup$
It looks fine to me.
$endgroup$
– José Carlos Santos
Jan 5 at 11:49
$begingroup$
Okay, thanks a lot!
$endgroup$
– RedLantern
Jan 5 at 12:01
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's much easier to do it as follows: the set $Z$ of zeros of $f$ is a closed and discrete subset of $mathbb C$. So, on $mathbb{C}setminus Z$, we would have $lvert zrvert=frac{f'(z)}{f(z)}$. But $zmapstolvert zrvert$ is nowhere differentiable.
answered Jan 5 at 11:11
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
1
$begingroup$
So... is my argumentation correct or not?
$endgroup$
– RedLantern
Jan 5 at 11:45
1
$begingroup$
It looks fine to me.
$endgroup$
– José Carlos Santos
Jan 5 at 11:49
$begingroup$
Okay, thanks a lot!
$endgroup$
– RedLantern
Jan 5 at 12:01
add a comment |
$begingroup$
It's much easier to do it as follows: the set $Z$ of zeros of $f$ is a closed and discrete subset of $mathbb C$. So, on $mathbb{C}setminus Z$, we would have $lvert zrvert=frac{f'(z)}{f(z)}$. But $zmapstolvert zrvert$ is nowhere differentiable.
answered Jan 5 at 11:11
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
1
$begingroup$
So... is my argumentation correct or not?
$endgroup$
– RedLantern
Jan 5 at 11:45
1
$begingroup$
It looks fine to me.
$endgroup$
– José Carlos Santos
Jan 5 at 11:49
$begingroup$
Okay, thanks a lot!
$endgroup$
– RedLantern
Jan 5 at 12:01
add a comment |
$begingroup$
It's much easier to do it as follows: the set $Z$ of zeros of $f$ is a closed and discrete subset of $mathbb C$. So, on $mathbb{C}setminus Z$, we would have $lvert zrvert=frac{f'(z)}{f(z)}$. But $zmapstolvert zrvert$ is nowhere differentiable.
answered Jan 5 at 11:11
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
It's much easier to do it as follows: the set $Z$ of zeros of $f$ is a closed and discrete subset of $mathbb C$. So, on $mathbb{C}setminus Z$, we would have $lvert zrvert=frac{f'(z)}{f(z)}$. But $zmapstolvert zrvert$ is nowhere differentiable.
answered Jan 5 at 11:11
José Carlos SantosJosé Carlos Santos
173k23133241
edited Jan 5 at 11:47
answered Jan 5 at 11:11
José Carlos SantosJosé Carlos Santos
173k23133241
answered Jan 5 at 11:11
José Carlos SantosJosé Carlos Santos
173k23133241
answered Jan 5 at 11:11
José Carlos SantosJosé Carlos Santos
173k23133241
173k23133241
1
$begingroup$
So... is my argumentation correct or not?
$endgroup$
– RedLantern
Jan 5 at 11:45
1
$begingroup$
It looks fine to me.
$endgroup$
– José Carlos Santos
Jan 5 at 11:49
$begingroup$
Okay, thanks a lot!
$endgroup$
– RedLantern
Jan 5 at 12:01
add a comment |
1
$begingroup$
So... is my argumentation correct or not?
$endgroup$
– RedLantern
Jan 5 at 11:45
1
$begingroup$
It looks fine to me.
$endgroup$
– José Carlos Santos
Jan 5 at 11:49
$begingroup$
Okay, thanks a lot!
$endgroup$
– RedLantern
Jan 5 at 12:01
1
1
$begingroup$
So... is my argumentation correct or not?
$endgroup$
– RedLantern
Jan 5 at 11:45
$begingroup$
So... is my argumentation correct or not?
$endgroup$
– RedLantern
Jan 5 at 11:45
1
1
$begingroup$
It looks fine to me.
$endgroup$
– José Carlos Santos
Jan 5 at 11:49
$begingroup$
It looks fine to me.
$endgroup$
– José Carlos Santos
Jan 5 at 11:49
$begingroup$
Okay, thanks a lot!
$endgroup$
– RedLantern
Jan 5 at 12:01
$begingroup$
Okay, thanks a lot!
$endgroup$
– RedLantern
Jan 5 at 12:01
add a comment |
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