Mixing
Why is $cos sqrt z$ entire but $sin sqrt{z}$ isn't?
$begingroup$
I've been trying to formulate a way of comparing these two functions, in order to find out why the function $sin sqrt z$ is not entire, but I couldn't find a good way of doing that. What I tried so far:
I wrote the series of both functions:
begin{align}cos(sqrt{z})&=sum_{n=0}^infty frac{(-1)^nz^n}{(2n)!}\
sin(sqrt{z})&=sum_{n=0}^infty frac{(-1)^nz^ncdotsqrt{z}}{(2n+1)!}end{align}The $e$ version:
begin{align}cos(sqrt{z})&=frac{e^{isqrt{z}}+e^{-isqrt{z}}}{2}\
sin(sqrt{z})&=frac{e^{isqrt{z}}-e^{-isqrt{z}}}{2i} ,end{align}
but they didn't help me on solving the problem. How could I progress from here? I could not find a way of writing Cauchy-Riemann equations for $cos$ or $sin$.
complex-analysis power-series entire-functions
edited Jan 3 at 10:10
Paul Frost
9,9153932
asked Jan 2 at 22:19
Lincon RibeiroLincon Ribeiro
506
$endgroup$
|
show 7 more comments
$begingroup$
I've been trying to formulate a way of comparing these two functions, in order to find out why the function $sin sqrt z$ is not entire, but I couldn't find a good way of doing that. What I tried so far:
I wrote the series of both functions:
begin{align}cos(sqrt{z})&=sum_{n=0}^infty frac{(-1)^nz^n}{(2n)!}\
sin(sqrt{z})&=sum_{n=0}^infty frac{(-1)^nz^ncdotsqrt{z}}{(2n+1)!}end{align}The $e$ version:
begin{align}cos(sqrt{z})&=frac{e^{isqrt{z}}+e^{-isqrt{z}}}{2}\
sin(sqrt{z})&=frac{e^{isqrt{z}}-e^{-isqrt{z}}}{2i} ,end{align}
but they didn't help me on solving the problem. How could I progress from here? I could not find a way of writing Cauchy-Riemann equations for $cos$ or $sin$.
complex-analysis power-series entire-functions
edited Jan 3 at 10:10
Paul Frost
9,9153932
asked Jan 2 at 22:19
Lincon RibeiroLincon Ribeiro
506
$endgroup$
4
$begingroup$
Because the series for $cossqrt{z}$ has no pure $sqrt{z}$ (which is not analytical), only $z$.
$endgroup$
– A.Γ.
Jan 2 at 22:22
5
$begingroup$
Welcome to MSE! You almost formatted your question perfectly, which we appreciate (especially from users inexperienced with latex or mathjax). Your exponents hadn't been displaying correctly. This is becausee^ixrenders as $e^ix$ --- mathjax only interprets the first symbol after^as the exponent. To fix this, I added braces:e^{ix}yields $e^{ix}$. Good luck!
$endgroup$
– davidlowryduda♦
Jan 2 at 22:30
2
$begingroup$
Notice that the series representation of $sin sqrt{z}$ contains nonintegral powers of $z$, so it is not a power series in $z$---this is the key difference between the two functions here.
$endgroup$
– Travis
Jan 2 at 22:37
1
$begingroup$
@PaulFrost: $sqrt z$ denote the square root with argument in $(-pi/2,pi/2]$, so the question makes sense.
$endgroup$
– idm
Jan 2 at 23:09
2
$begingroup$
Here is a way of thinking about the question without reference to a square root function. Let $g(z)$ be a holomorphic function. You want to know when there is another holomorphic function $h(z)$ with $g(z) = h(z^2)$. The answer is that this exists iff $g(z)$ is even.
$endgroup$
– hunter
Jan 3 at 0:50
|
show 7 more comments
$begingroup$
I've been trying to formulate a way of comparing these two functions, in order to find out why the function $sin sqrt z$ is not entire, but I couldn't find a good way of doing that. What I tried so far:
I wrote the series of both functions:
begin{align}cos(sqrt{z})&=sum_{n=0}^infty frac{(-1)^nz^n}{(2n)!}\
sin(sqrt{z})&=sum_{n=0}^infty frac{(-1)^nz^ncdotsqrt{z}}{(2n+1)!}end{align}The $e$ version:
begin{align}cos(sqrt{z})&=frac{e^{isqrt{z}}+e^{-isqrt{z}}}{2}\
sin(sqrt{z})&=frac{e^{isqrt{z}}-e^{-isqrt{z}}}{2i} ,end{align}
but they didn't help me on solving the problem. How could I progress from here? I could not find a way of writing Cauchy-Riemann equations for $cos$ or $sin$.
complex-analysis power-series entire-functions
edited Jan 3 at 10:10
Paul Frost
9,9153932
asked Jan 2 at 22:19
Lincon RibeiroLincon Ribeiro
506
$endgroup$
I've been trying to formulate a way of comparing these two functions, in order to find out why the function $sin sqrt z$ is not entire, but I couldn't find a good way of doing that. What I tried so far:
I wrote the series of both functions:
begin{align}cos(sqrt{z})&=sum_{n=0}^infty frac{(-1)^nz^n}{(2n)!}\
sin(sqrt{z})&=sum_{n=0}^infty frac{(-1)^nz^ncdotsqrt{z}}{(2n+1)!}end{align}The $e$ version:
begin{align}cos(sqrt{z})&=frac{e^{isqrt{z}}+e^{-isqrt{z}}}{2}\
sin(sqrt{z})&=frac{e^{isqrt{z}}-e^{-isqrt{z}}}{2i} ,end{align}
but they didn't help me on solving the problem. How could I progress from here? I could not find a way of writing Cauchy-Riemann equations for $cos$ or $sin$.
complex-analysis power-series entire-functions
complex-analysis power-series entire-functions
edited Jan 3 at 10:10
Paul Frost
9,9153932
asked Jan 2 at 22:19
Lincon RibeiroLincon Ribeiro
506
edited Jan 3 at 10:10
Paul Frost
9,9153932
asked Jan 2 at 22:19
Lincon RibeiroLincon Ribeiro
506
edited Jan 3 at 10:10
Paul Frost
9,9153932
edited Jan 3 at 10:10
Paul Frost
9,9153932
edited Jan 3 at 10:10
Paul Frost
9,9153932
9,9153932
asked Jan 2 at 22:19
Lincon RibeiroLincon Ribeiro
506
asked Jan 2 at 22:19
Lincon RibeiroLincon Ribeiro
506
asked Jan 2 at 22:19
Lincon RibeiroLincon Ribeiro
506
506
4
$begingroup$
Because the series for $cossqrt{z}$ has no pure $sqrt{z}$ (which is not analytical), only $z$.
$endgroup$
– A.Γ.
Jan 2 at 22:22
5
$begingroup$
Welcome to MSE! You almost formatted your question perfectly, which we appreciate (especially from users inexperienced with latex or mathjax). Your exponents hadn't been displaying correctly. This is becausee^ixrenders as $e^ix$ --- mathjax only interprets the first symbol after^as the exponent. To fix this, I added braces:e^{ix}yields $e^{ix}$. Good luck!
$endgroup$
– davidlowryduda♦
Jan 2 at 22:30
2
$begingroup$
Notice that the series representation of $sin sqrt{z}$ contains nonintegral powers of $z$, so it is not a power series in $z$---this is the key difference between the two functions here.
$endgroup$
– Travis
Jan 2 at 22:37
1
$begingroup$
@PaulFrost: $sqrt z$ denote the square root with argument in $(-pi/2,pi/2]$, so the question makes sense.
$endgroup$
– idm
Jan 2 at 23:09
2
$begingroup$
Here is a way of thinking about the question without reference to a square root function. Let $g(z)$ be a holomorphic function. You want to know when there is another holomorphic function $h(z)$ with $g(z) = h(z^2)$. The answer is that this exists iff $g(z)$ is even.
$endgroup$
– hunter
Jan 3 at 0:50
|
show 7 more comments
4
$begingroup$
Because the series for $cossqrt{z}$ has no pure $sqrt{z}$ (which is not analytical), only $z$.
$endgroup$
– A.Γ.
Jan 2 at 22:22
5
$begingroup$
Welcome to MSE! You almost formatted your question perfectly, which we appreciate (especially from users inexperienced with latex or mathjax). Your exponents hadn't been displaying correctly. This is becausee^ixrenders as $e^ix$ --- mathjax only interprets the first symbol after^as the exponent. To fix this, I added braces:e^{ix}yields $e^{ix}$. Good luck!
$endgroup$
– davidlowryduda♦
Jan 2 at 22:30
2
$begingroup$
Notice that the series representation of $sin sqrt{z}$ contains nonintegral powers of $z$, so it is not a power series in $z$---this is the key difference between the two functions here.
$endgroup$
– Travis
Jan 2 at 22:37
1
$begingroup$
@PaulFrost: $sqrt z$ denote the square root with argument in $(-pi/2,pi/2]$, so the question makes sense.
$endgroup$
– idm
Jan 2 at 23:09
2
$begingroup$
Here is a way of thinking about the question without reference to a square root function. Let $g(z)$ be a holomorphic function. You want to know when there is another holomorphic function $h(z)$ with $g(z) = h(z^2)$. The answer is that this exists iff $g(z)$ is even.
$endgroup$
– hunter
Jan 3 at 0:50
4
4
$begingroup$
Because the series for $cossqrt{z}$ has no pure $sqrt{z}$ (which is not analytical), only $z$.
$endgroup$
– A.Γ.
Jan 2 at 22:22
$begingroup$
Because the series for $cossqrt{z}$ has no pure $sqrt{z}$ (which is not analytical), only $z$.
$endgroup$
– A.Γ.
Jan 2 at 22:22
5
5
$begingroup$
Welcome to MSE! You almost formatted your question perfectly, which we appreciate (especially from users inexperienced with latex or mathjax). Your exponents hadn't been displaying correctly. This is because
e^ix renders as $e^ix$ --- mathjax only interprets the first symbol after ^ as the exponent. To fix this, I added braces: e^{ix} yields $e^{ix}$. Good luck!$endgroup$
– davidlowryduda♦
Jan 2 at 22:30
$begingroup$
Welcome to MSE! You almost formatted your question perfectly, which we appreciate (especially from users inexperienced with latex or mathjax). Your exponents hadn't been displaying correctly. This is because
e^ix renders as $e^ix$ --- mathjax only interprets the first symbol after ^ as the exponent. To fix this, I added braces: e^{ix} yields $e^{ix}$. Good luck!$endgroup$
– davidlowryduda♦
Jan 2 at 22:30
2
2
$begingroup$
Notice that the series representation of $sin sqrt{z}$ contains nonintegral powers of $z$, so it is not a power series in $z$---this is the key difference between the two functions here.
$endgroup$
– Travis
Jan 2 at 22:37
$begingroup$
Notice that the series representation of $sin sqrt{z}$ contains nonintegral powers of $z$, so it is not a power series in $z$---this is the key difference between the two functions here.
$endgroup$
– Travis
Jan 2 at 22:37
1
1
$begingroup$
@PaulFrost: $sqrt z$ denote the square root with argument in $(-pi/2,pi/2]$, so the question makes sense.
$endgroup$
– idm
Jan 2 at 23:09
$begingroup$
@PaulFrost: $sqrt z$ denote the square root with argument in $(-pi/2,pi/2]$, so the question makes sense.
$endgroup$
– idm
Jan 2 at 23:09
2
2
$begingroup$
Here is a way of thinking about the question without reference to a square root function. Let $g(z)$ be a holomorphic function. You want to know when there is another holomorphic function $h(z)$ with $g(z) = h(z^2)$. The answer is that this exists iff $g(z)$ is even.
$endgroup$
– hunter
Jan 3 at 0:50
$begingroup$
Here is a way of thinking about the question without reference to a square root function. Let $g(z)$ be a holomorphic function. You want to know when there is another holomorphic function $h(z)$ with $g(z) = h(z^2)$. The answer is that this exists iff $g(z)$ is even.
$endgroup$
– hunter
Jan 3 at 0:50
|
show 7 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Hint Suppose $F(z) := f(sqrt{z})$ is analytic at $z = 0$ for some choice of branch cut of $sqrt{cdot}$, say, $F(w)$ has power series $$F(w) sim a_0 + a_1 w + a_2 w^2 + cdots$$ at $w = 0$. What is the power series of $f(w) = F(w^2)$ at $w = 0$?
Additional hint We have $$f(w) sim a_0 + a_1 w^2 + a_2 w^4 + cdots .$$
This shows that not only is $sin sqrt{z}$ not entire, it's not analytic in any neighborhood of $z = 0$ (for any choice of branch cut).
answered Jan 2 at 22:31
TravisTravis
59.9k767146
$endgroup$
$begingroup$
What do you mean by : "for some branch cut of $sqrtcdot $ ? $sqrt z$ denote the square root of $z$ with argument in $(-pi/2,pi/2]$.
$endgroup$
– idm
Jan 2 at 23:06
$begingroup$
If you're not familiar with branch cuts yet, feel free to disregard that phrase---the answer applies when using the particular convention you mention.
$endgroup$
– Travis
Jan 3 at 3:32
add a comment |
$begingroup$
Given any argument $z$, there are exactly two values for the square root, $w$ and $-w$. If the sine of $w$ equals the sine of $-w$ for all $z$, then the two different possible values for the square root of $z$ lead to only a single value of the sine and thus no branch cut is required. But, if $w$ and $-w$ give different sines for any $z$, then there must be a branch cut to separate the two sine values therefore you can't have an entire function.
It turns out this doesn't work because $sin (-w)=-sin w$ and thus $ne sin w$ except where the sine value happens to be zero. We are forced to accept branch cuts between the zeroes of $sin w$, meaning $w=sqrt{z}$ is a multiple of $pi$ and $z$ itself us a multiple of $pi^2$. Can you diagram a relatively simple choice for the needed branch cuts In the complex plane?
Now try it with the cosine function using the same reasoning above, but there is one little difference. Cosine is an even function, not odd, so $cos (-w)=+cos w$ instead of $-cos w$. How does that change the rest of your conclusions above for the cosine function, as opposed to the sine?
Always watch your signs.
answered Jan 3 at 0:12
Oscar LanziOscar Lanzi
12.2k12036
$endgroup$
1
$begingroup$
In "...where the wine value..." perhaps you meant "sine"?
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 0:25
$begingroup$
Please correct this. My device often auticorrects words to the wrong word and I cannot chase it down 100% of the time. Thanks.
$endgroup$
– Oscar Lanzi
Jan 3 at 0:36
1
$begingroup$
Um... don't worry. You can always edit your own posts.
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 1:07
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint Suppose $F(z) := f(sqrt{z})$ is analytic at $z = 0$ for some choice of branch cut of $sqrt{cdot}$, say, $F(w)$ has power series $$F(w) sim a_0 + a_1 w + a_2 w^2 + cdots$$ at $w = 0$. What is the power series of $f(w) = F(w^2)$ at $w = 0$?
Additional hint We have $$f(w) sim a_0 + a_1 w^2 + a_2 w^4 + cdots .$$
This shows that not only is $sin sqrt{z}$ not entire, it's not analytic in any neighborhood of $z = 0$ (for any choice of branch cut).
answered Jan 2 at 22:31
TravisTravis
59.9k767146
$endgroup$
$begingroup$
What do you mean by : "for some branch cut of $sqrtcdot $ ? $sqrt z$ denote the square root of $z$ with argument in $(-pi/2,pi/2]$.
$endgroup$
– idm
Jan 2 at 23:06
$begingroup$
If you're not familiar with branch cuts yet, feel free to disregard that phrase---the answer applies when using the particular convention you mention.
$endgroup$
– Travis
Jan 3 at 3:32
add a comment |
$begingroup$
Hint Suppose $F(z) := f(sqrt{z})$ is analytic at $z = 0$ for some choice of branch cut of $sqrt{cdot}$, say, $F(w)$ has power series $$F(w) sim a_0 + a_1 w + a_2 w^2 + cdots$$ at $w = 0$. What is the power series of $f(w) = F(w^2)$ at $w = 0$?
Additional hint We have $$f(w) sim a_0 + a_1 w^2 + a_2 w^4 + cdots .$$
This shows that not only is $sin sqrt{z}$ not entire, it's not analytic in any neighborhood of $z = 0$ (for any choice of branch cut).
answered Jan 2 at 22:31
TravisTravis
59.9k767146
$endgroup$
$begingroup$
What do you mean by : "for some branch cut of $sqrtcdot $ ? $sqrt z$ denote the square root of $z$ with argument in $(-pi/2,pi/2]$.
$endgroup$
– idm
Jan 2 at 23:06
$begingroup$
If you're not familiar with branch cuts yet, feel free to disregard that phrase---the answer applies when using the particular convention you mention.
$endgroup$
– Travis
Jan 3 at 3:32
add a comment |
$begingroup$
Hint Suppose $F(z) := f(sqrt{z})$ is analytic at $z = 0$ for some choice of branch cut of $sqrt{cdot}$, say, $F(w)$ has power series $$F(w) sim a_0 + a_1 w + a_2 w^2 + cdots$$ at $w = 0$. What is the power series of $f(w) = F(w^2)$ at $w = 0$?
Additional hint We have $$f(w) sim a_0 + a_1 w^2 + a_2 w^4 + cdots .$$
This shows that not only is $sin sqrt{z}$ not entire, it's not analytic in any neighborhood of $z = 0$ (for any choice of branch cut).
answered Jan 2 at 22:31
TravisTravis
59.9k767146
$endgroup$
Hint Suppose $F(z) := f(sqrt{z})$ is analytic at $z = 0$ for some choice of branch cut of $sqrt{cdot}$, say, $F(w)$ has power series $$F(w) sim a_0 + a_1 w + a_2 w^2 + cdots$$ at $w = 0$. What is the power series of $f(w) = F(w^2)$ at $w = 0$?
Additional hint We have $$f(w) sim a_0 + a_1 w^2 + a_2 w^4 + cdots .$$
This shows that not only is $sin sqrt{z}$ not entire, it's not analytic in any neighborhood of $z = 0$ (for any choice of branch cut).
answered Jan 2 at 22:31
TravisTravis
59.9k767146
edited Jan 3 at 3:35
answered Jan 2 at 22:31
TravisTravis
59.9k767146
answered Jan 2 at 22:31
TravisTravis
59.9k767146
answered Jan 2 at 22:31
TravisTravis
59.9k767146
59.9k767146
$begingroup$
What do you mean by : "for some branch cut of $sqrtcdot $ ? $sqrt z$ denote the square root of $z$ with argument in $(-pi/2,pi/2]$.
$endgroup$
– idm
Jan 2 at 23:06
$begingroup$
If you're not familiar with branch cuts yet, feel free to disregard that phrase---the answer applies when using the particular convention you mention.
$endgroup$
– Travis
Jan 3 at 3:32
add a comment |
$begingroup$
What do you mean by : "for some branch cut of $sqrtcdot $ ? $sqrt z$ denote the square root of $z$ with argument in $(-pi/2,pi/2]$.
$endgroup$
– idm
Jan 2 at 23:06
$begingroup$
If you're not familiar with branch cuts yet, feel free to disregard that phrase---the answer applies when using the particular convention you mention.
$endgroup$
– Travis
Jan 3 at 3:32
$begingroup$
What do you mean by : "for some branch cut of $sqrtcdot $ ? $sqrt z$ denote the square root of $z$ with argument in $(-pi/2,pi/2]$.
$endgroup$
– idm
Jan 2 at 23:06
$begingroup$
What do you mean by : "for some branch cut of $sqrtcdot $ ? $sqrt z$ denote the square root of $z$ with argument in $(-pi/2,pi/2]$.
$endgroup$
– idm
Jan 2 at 23:06
$begingroup$
If you're not familiar with branch cuts yet, feel free to disregard that phrase---the answer applies when using the particular convention you mention.
$endgroup$
– Travis
Jan 3 at 3:32
$begingroup$
If you're not familiar with branch cuts yet, feel free to disregard that phrase---the answer applies when using the particular convention you mention.
$endgroup$
– Travis
Jan 3 at 3:32
add a comment |
$begingroup$
Given any argument $z$, there are exactly two values for the square root, $w$ and $-w$. If the sine of $w$ equals the sine of $-w$ for all $z$, then the two different possible values for the square root of $z$ lead to only a single value of the sine and thus no branch cut is required. But, if $w$ and $-w$ give different sines for any $z$, then there must be a branch cut to separate the two sine values therefore you can't have an entire function.
It turns out this doesn't work because $sin (-w)=-sin w$ and thus $ne sin w$ except where the sine value happens to be zero. We are forced to accept branch cuts between the zeroes of $sin w$, meaning $w=sqrt{z}$ is a multiple of $pi$ and $z$ itself us a multiple of $pi^2$. Can you diagram a relatively simple choice for the needed branch cuts In the complex plane?
Now try it with the cosine function using the same reasoning above, but there is one little difference. Cosine is an even function, not odd, so $cos (-w)=+cos w$ instead of $-cos w$. How does that change the rest of your conclusions above for the cosine function, as opposed to the sine?
Always watch your signs.
answered Jan 3 at 0:12
Oscar LanziOscar Lanzi
12.2k12036
$endgroup$
1
$begingroup$
In "...where the wine value..." perhaps you meant "sine"?
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 0:25
$begingroup$
Please correct this. My device often auticorrects words to the wrong word and I cannot chase it down 100% of the time. Thanks.
$endgroup$
– Oscar Lanzi
Jan 3 at 0:36
1
$begingroup$
Um... don't worry. You can always edit your own posts.
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 1:07
add a comment |
$begingroup$
Given any argument $z$, there are exactly two values for the square root, $w$ and $-w$. If the sine of $w$ equals the sine of $-w$ for all $z$, then the two different possible values for the square root of $z$ lead to only a single value of the sine and thus no branch cut is required. But, if $w$ and $-w$ give different sines for any $z$, then there must be a branch cut to separate the two sine values therefore you can't have an entire function.
It turns out this doesn't work because $sin (-w)=-sin w$ and thus $ne sin w$ except where the sine value happens to be zero. We are forced to accept branch cuts between the zeroes of $sin w$, meaning $w=sqrt{z}$ is a multiple of $pi$ and $z$ itself us a multiple of $pi^2$. Can you diagram a relatively simple choice for the needed branch cuts In the complex plane?
Now try it with the cosine function using the same reasoning above, but there is one little difference. Cosine is an even function, not odd, so $cos (-w)=+cos w$ instead of $-cos w$. How does that change the rest of your conclusions above for the cosine function, as opposed to the sine?
Always watch your signs.
answered Jan 3 at 0:12
Oscar LanziOscar Lanzi
12.2k12036
$endgroup$
1
$begingroup$
In "...where the wine value..." perhaps you meant "sine"?
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 0:25
$begingroup$
Please correct this. My device often auticorrects words to the wrong word and I cannot chase it down 100% of the time. Thanks.
$endgroup$
– Oscar Lanzi
Jan 3 at 0:36
1
$begingroup$
Um... don't worry. You can always edit your own posts.
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 1:07
add a comment |
$begingroup$
Given any argument $z$, there are exactly two values for the square root, $w$ and $-w$. If the sine of $w$ equals the sine of $-w$ for all $z$, then the two different possible values for the square root of $z$ lead to only a single value of the sine and thus no branch cut is required. But, if $w$ and $-w$ give different sines for any $z$, then there must be a branch cut to separate the two sine values therefore you can't have an entire function.
It turns out this doesn't work because $sin (-w)=-sin w$ and thus $ne sin w$ except where the sine value happens to be zero. We are forced to accept branch cuts between the zeroes of $sin w$, meaning $w=sqrt{z}$ is a multiple of $pi$ and $z$ itself us a multiple of $pi^2$. Can you diagram a relatively simple choice for the needed branch cuts In the complex plane?
Now try it with the cosine function using the same reasoning above, but there is one little difference. Cosine is an even function, not odd, so $cos (-w)=+cos w$ instead of $-cos w$. How does that change the rest of your conclusions above for the cosine function, as opposed to the sine?
Always watch your signs.
answered Jan 3 at 0:12
Oscar LanziOscar Lanzi
12.2k12036
$endgroup$
Given any argument $z$, there are exactly two values for the square root, $w$ and $-w$. If the sine of $w$ equals the sine of $-w$ for all $z$, then the two different possible values for the square root of $z$ lead to only a single value of the sine and thus no branch cut is required. But, if $w$ and $-w$ give different sines for any $z$, then there must be a branch cut to separate the two sine values therefore you can't have an entire function.
It turns out this doesn't work because $sin (-w)=-sin w$ and thus $ne sin w$ except where the sine value happens to be zero. We are forced to accept branch cuts between the zeroes of $sin w$, meaning $w=sqrt{z}$ is a multiple of $pi$ and $z$ itself us a multiple of $pi^2$. Can you diagram a relatively simple choice for the needed branch cuts In the complex plane?
Now try it with the cosine function using the same reasoning above, but there is one little difference. Cosine is an even function, not odd, so $cos (-w)=+cos w$ instead of $-cos w$. How does that change the rest of your conclusions above for the cosine function, as opposed to the sine?
Always watch your signs.
answered Jan 3 at 0:12
Oscar LanziOscar Lanzi
12.2k12036
edited Jan 3 at 13:49
answered Jan 3 at 0:12
Oscar LanziOscar Lanzi
12.2k12036
answered Jan 3 at 0:12
Oscar LanziOscar Lanzi
12.2k12036
answered Jan 3 at 0:12
Oscar LanziOscar Lanzi
12.2k12036
12.2k12036
1
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In "...where the wine value..." perhaps you meant "sine"?
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– Pedro Tamaroff♦
Jan 3 at 0:25
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Please correct this. My device often auticorrects words to the wrong word and I cannot chase it down 100% of the time. Thanks.
$endgroup$
– Oscar Lanzi
Jan 3 at 0:36
1
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Um... don't worry. You can always edit your own posts.
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 1:07
add a comment |
1
$begingroup$
In "...where the wine value..." perhaps you meant "sine"?
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 0:25
$begingroup$
Please correct this. My device often auticorrects words to the wrong word and I cannot chase it down 100% of the time. Thanks.
$endgroup$
– Oscar Lanzi
Jan 3 at 0:36
1
$begingroup$
Um... don't worry. You can always edit your own posts.
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 1:07
1
1
$begingroup$
In "...where the wine value..." perhaps you meant "sine"?
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 0:25
$begingroup$
In "...where the wine value..." perhaps you meant "sine"?
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 0:25
$begingroup$
Please correct this. My device often auticorrects words to the wrong word and I cannot chase it down 100% of the time. Thanks.
$endgroup$
– Oscar Lanzi
Jan 3 at 0:36
$begingroup$
Please correct this. My device often auticorrects words to the wrong word and I cannot chase it down 100% of the time. Thanks.
$endgroup$
– Oscar Lanzi
Jan 3 at 0:36
1
1
$begingroup$
Um... don't worry. You can always edit your own posts.
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 1:07
$begingroup$
Um... don't worry. You can always edit your own posts.
$endgroup$
– Pedro Tamaroff♦
Jan 3 at 1:07
add a comment |
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Because the series for $cossqrt{z}$ has no pure $sqrt{z}$ (which is not analytical), only $z$.
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– A.Γ.
Jan 2 at 22:22
5
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Welcome to MSE! You almost formatted your question perfectly, which we appreciate (especially from users inexperienced with latex or mathjax). Your exponents hadn't been displaying correctly. This is because
e^ixrenders as $e^ix$ --- mathjax only interprets the first symbol after^as the exponent. To fix this, I added braces:e^{ix}yields $e^{ix}$. Good luck!$endgroup$
– davidlowryduda♦
Jan 2 at 22:30
2
$begingroup$
Notice that the series representation of $sin sqrt{z}$ contains nonintegral powers of $z$, so it is not a power series in $z$---this is the key difference between the two functions here.
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– Travis
Jan 2 at 22:37
1
$begingroup$
@PaulFrost: $sqrt z$ denote the square root with argument in $(-pi/2,pi/2]$, so the question makes sense.
$endgroup$
– idm
Jan 2 at 23:09
2
$begingroup$
Here is a way of thinking about the question without reference to a square root function. Let $g(z)$ be a holomorphic function. You want to know when there is another holomorphic function $h(z)$ with $g(z) = h(z^2)$. The answer is that this exists iff $g(z)$ is even.
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– hunter
Jan 3 at 0:50