prove that $;f(z) = g(z);$ for all $zin mathbb{C}$
0
The problem is:
Let $f(z)$ and $g(z)$ be entire such that $r>0 ,; f(z) = g(z)$ for all $|z| < r. $
- Prove that $f(z) = g(z)$ for all $z in mathbb{C}$
Does that mean I should prove that the function is polynomial ?
I am thinking that's because it's bounded by a polynomial ??
Since $f$ is entire, it is equal to a power series centered at zero with radius of convergence $infty$, which must match its Taylor series there.
calculus complex-analysis complex-numbers entire-functions
edited Dec 27 '18 at 21:38
user376343
2,8582823
asked Dec 27 '18 at 13:15
Smb Youz
285
add a comment |
0
The problem is:
Let $f(z)$ and $g(z)$ be entire such that $r>0 ,; f(z) = g(z)$ for all $|z| < r. $
- Prove that $f(z) = g(z)$ for all $z in mathbb{C}$
Does that mean I should prove that the function is polynomial ?
I am thinking that's because it's bounded by a polynomial ??
Since $f$ is entire, it is equal to a power series centered at zero with radius of convergence $infty$, which must match its Taylor series there.
calculus complex-analysis complex-numbers entire-functions
edited Dec 27 '18 at 21:38
user376343
2,8582823
asked Dec 27 '18 at 13:15
Smb Youz
285
4
Look up “identity theorem for holomorphic functions.”
– Martin R
Dec 27 '18 at 13:19
en.wikipedia.org/wiki/Principle_of_permanence
– metamorphy
Dec 27 '18 at 13:57
thanks @MartinR that's really useful
– Smb Youz
yesterday
add a comment |
0
0
0
The problem is:
Let $f(z)$ and $g(z)$ be entire such that $r>0 ,; f(z) = g(z)$ for all $|z| < r. $
- Prove that $f(z) = g(z)$ for all $z in mathbb{C}$
Does that mean I should prove that the function is polynomial ?
I am thinking that's because it's bounded by a polynomial ??
Since $f$ is entire, it is equal to a power series centered at zero with radius of convergence $infty$, which must match its Taylor series there.
calculus complex-analysis complex-numbers entire-functions
edited Dec 27 '18 at 21:38
user376343
2,8582823
asked Dec 27 '18 at 13:15
Smb Youz
285
The problem is:
Let $f(z)$ and $g(z)$ be entire such that $r>0 ,; f(z) = g(z)$ for all $|z| < r. $
- Prove that $f(z) = g(z)$ for all $z in mathbb{C}$
Does that mean I should prove that the function is polynomial ?
I am thinking that's because it's bounded by a polynomial ??
Since $f$ is entire, it is equal to a power series centered at zero with radius of convergence $infty$, which must match its Taylor series there.
calculus complex-analysis complex-numbers entire-functions
calculus complex-analysis complex-numbers entire-functions
edited Dec 27 '18 at 21:38
user376343
2,8582823
asked Dec 27 '18 at 13:15
Smb Youz
285
edited Dec 27 '18 at 21:38
user376343
2,8582823
asked Dec 27 '18 at 13:15
Smb Youz
285
edited Dec 27 '18 at 21:38
user376343
2,8582823
edited Dec 27 '18 at 21:38
user376343
2,8582823
edited Dec 27 '18 at 21:38
user376343
2,8582823
2,8582823
asked Dec 27 '18 at 13:15
Smb Youz
285
asked Dec 27 '18 at 13:15
Smb Youz
285
asked Dec 27 '18 at 13:15
Smb Youz
285
285
4
Look up “identity theorem for holomorphic functions.”
– Martin R
Dec 27 '18 at 13:19
en.wikipedia.org/wiki/Principle_of_permanence
– metamorphy
Dec 27 '18 at 13:57
thanks @MartinR that's really useful
– Smb Youz
yesterday
add a comment |
4
Look up “identity theorem for holomorphic functions.”
– Martin R
Dec 27 '18 at 13:19
en.wikipedia.org/wiki/Principle_of_permanence
– metamorphy
Dec 27 '18 at 13:57
thanks @MartinR that's really useful
– Smb Youz
yesterday
4
4
Look up “identity theorem for holomorphic functions.”
– Martin R
Dec 27 '18 at 13:19
Look up “identity theorem for holomorphic functions.”
– Martin R
Dec 27 '18 at 13:19
en.wikipedia.org/wiki/Principle_of_permanence
– metamorphy
Dec 27 '18 at 13:57
en.wikipedia.org/wiki/Principle_of_permanence
– metamorphy
Dec 27 '18 at 13:57
thanks @MartinR that's really useful
– Smb Youz
yesterday
thanks @MartinR that's really useful
– Smb Youz
yesterday
add a comment |
1 Answer
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4
If you consider the function $h(z)=f(z)-g(z)$, then $h(z)$ will also be analytic on the entire $mathbb{C}$. By the hypothesis, $h(z)=0$ in a disc $|z|<r$. Since the zeroes of an analytic function which does not vanish identically are $textit{isolated}$, and the disc $|z|<r$ has accumulation points in $mathbb{C}$ hence the function $h(z)$ is identically $0$ in $mathbb{C}$.
edited Dec 28 '18 at 21:02
Duchamp Gérard H. E.
2,527918
answered Dec 27 '18 at 14:28
Dèö
16113
1
Welcome new contributor. What do you mean by [in fact in the complement of the disc] ? the set of accumulation points of ${|z|<r}$ is in fact the closed disk ${|z|leq r}$
– Duchamp Gérard H. E.
Dec 27 '18 at 21:38
Actually, it doesn't have much to do. I made that point to stress the fact that even if the set, $S$ on which $f(z)$ and $g(z)$ agree, is not as nice as the disc, still we can have the above result. If the set just has one accumulation point in $mathbb{C}-S$, i.e. there is a sequence of zeroes which converges to a point which is outside $S$, then the function must agree entirely. It would be a nice exercise to prove that the number of zeroes of an analytic function can't be infinite. :-)
– Dèö
Dec 28 '18 at 17:07
[It would be a nice exercise to prove that the number of zeroes of an analytic function can't be infinite]--->This can occur (i.e. the exercise would be false) take $f(z)=e^z-1$ which has $2ipimathbb{Z}$ as set of zeroes.
– Duchamp Gérard H. E.
Dec 28 '18 at 20:59
I have edited your answer, now it is correct (+1) In mathematics, not only the intention counts, the text must be locally exact (well, as much as possible, we are humans :)
– Duchamp Gérard H. E.
Dec 28 '18 at 21:04
Thankyou @Duchamp Gérard H. E. Yeah the infinite number of zeros can occur. Well I am new to this forum, so I need to be precise in my upcoming answers. Thank you 🙂
– Dèö
Dec 29 '18 at 10:18
add a comment |
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1 Answer
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1 Answer
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4
If you consider the function $h(z)=f(z)-g(z)$, then $h(z)$ will also be analytic on the entire $mathbb{C}$. By the hypothesis, $h(z)=0$ in a disc $|z|<r$. Since the zeroes of an analytic function which does not vanish identically are $textit{isolated}$, and the disc $|z|<r$ has accumulation points in $mathbb{C}$ hence the function $h(z)$ is identically $0$ in $mathbb{C}$.
edited Dec 28 '18 at 21:02
Duchamp Gérard H. E.
2,527918
answered Dec 27 '18 at 14:28
Dèö
16113
1
Welcome new contributor. What do you mean by [in fact in the complement of the disc] ? the set of accumulation points of ${|z|<r}$ is in fact the closed disk ${|z|leq r}$
– Duchamp Gérard H. E.
Dec 27 '18 at 21:38
Actually, it doesn't have much to do. I made that point to stress the fact that even if the set, $S$ on which $f(z)$ and $g(z)$ agree, is not as nice as the disc, still we can have the above result. If the set just has one accumulation point in $mathbb{C}-S$, i.e. there is a sequence of zeroes which converges to a point which is outside $S$, then the function must agree entirely. It would be a nice exercise to prove that the number of zeroes of an analytic function can't be infinite. :-)
– Dèö
Dec 28 '18 at 17:07
[It would be a nice exercise to prove that the number of zeroes of an analytic function can't be infinite]--->This can occur (i.e. the exercise would be false) take $f(z)=e^z-1$ which has $2ipimathbb{Z}$ as set of zeroes.
– Duchamp Gérard H. E.
Dec 28 '18 at 20:59
I have edited your answer, now it is correct (+1) In mathematics, not only the intention counts, the text must be locally exact (well, as much as possible, we are humans :)
– Duchamp Gérard H. E.
Dec 28 '18 at 21:04
Thankyou @Duchamp Gérard H. E. Yeah the infinite number of zeros can occur. Well I am new to this forum, so I need to be precise in my upcoming answers. Thank you 🙂
– Dèö
Dec 29 '18 at 10:18
add a comment |
4
If you consider the function $h(z)=f(z)-g(z)$, then $h(z)$ will also be analytic on the entire $mathbb{C}$. By the hypothesis, $h(z)=0$ in a disc $|z|<r$. Since the zeroes of an analytic function which does not vanish identically are $textit{isolated}$, and the disc $|z|<r$ has accumulation points in $mathbb{C}$ hence the function $h(z)$ is identically $0$ in $mathbb{C}$.
edited Dec 28 '18 at 21:02
Duchamp Gérard H. E.
2,527918
answered Dec 27 '18 at 14:28
Dèö
16113
1
Welcome new contributor. What do you mean by [in fact in the complement of the disc] ? the set of accumulation points of ${|z|<r}$ is in fact the closed disk ${|z|leq r}$
– Duchamp Gérard H. E.
Dec 27 '18 at 21:38
Actually, it doesn't have much to do. I made that point to stress the fact that even if the set, $S$ on which $f(z)$ and $g(z)$ agree, is not as nice as the disc, still we can have the above result. If the set just has one accumulation point in $mathbb{C}-S$, i.e. there is a sequence of zeroes which converges to a point which is outside $S$, then the function must agree entirely. It would be a nice exercise to prove that the number of zeroes of an analytic function can't be infinite. :-)
– Dèö
Dec 28 '18 at 17:07
[It would be a nice exercise to prove that the number of zeroes of an analytic function can't be infinite]--->This can occur (i.e. the exercise would be false) take $f(z)=e^z-1$ which has $2ipimathbb{Z}$ as set of zeroes.
– Duchamp Gérard H. E.
Dec 28 '18 at 20:59
I have edited your answer, now it is correct (+1) In mathematics, not only the intention counts, the text must be locally exact (well, as much as possible, we are humans :)
– Duchamp Gérard H. E.
Dec 28 '18 at 21:04
Thankyou @Duchamp Gérard H. E. Yeah the infinite number of zeros can occur. Well I am new to this forum, so I need to be precise in my upcoming answers. Thank you 🙂
– Dèö
Dec 29 '18 at 10:18
add a comment |
4
4
4
If you consider the function $h(z)=f(z)-g(z)$, then $h(z)$ will also be analytic on the entire $mathbb{C}$. By the hypothesis, $h(z)=0$ in a disc $|z|<r$. Since the zeroes of an analytic function which does not vanish identically are $textit{isolated}$, and the disc $|z|<r$ has accumulation points in $mathbb{C}$ hence the function $h(z)$ is identically $0$ in $mathbb{C}$.
edited Dec 28 '18 at 21:02
Duchamp Gérard H. E.
2,527918
answered Dec 27 '18 at 14:28
Dèö
16113
If you consider the function $h(z)=f(z)-g(z)$, then $h(z)$ will also be analytic on the entire $mathbb{C}$. By the hypothesis, $h(z)=0$ in a disc $|z|<r$. Since the zeroes of an analytic function which does not vanish identically are $textit{isolated}$, and the disc $|z|<r$ has accumulation points in $mathbb{C}$ hence the function $h(z)$ is identically $0$ in $mathbb{C}$.
edited Dec 28 '18 at 21:02
Duchamp Gérard H. E.
2,527918
answered Dec 27 '18 at 14:28
Dèö
16113
edited Dec 28 '18 at 21:02
Duchamp Gérard H. E.
2,527918
edited Dec 28 '18 at 21:02
Duchamp Gérard H. E.
2,527918
edited Dec 28 '18 at 21:02
Duchamp Gérard H. E.
2,527918
2,527918
answered Dec 27 '18 at 14:28
Dèö
16113
answered Dec 27 '18 at 14:28
Dèö
16113
answered Dec 27 '18 at 14:28
Dèö
16113
16113
1
Welcome new contributor. What do you mean by [in fact in the complement of the disc] ? the set of accumulation points of ${|z|<r}$ is in fact the closed disk ${|z|leq r}$
– Duchamp Gérard H. E.
Dec 27 '18 at 21:38
Actually, it doesn't have much to do. I made that point to stress the fact that even if the set, $S$ on which $f(z)$ and $g(z)$ agree, is not as nice as the disc, still we can have the above result. If the set just has one accumulation point in $mathbb{C}-S$, i.e. there is a sequence of zeroes which converges to a point which is outside $S$, then the function must agree entirely. It would be a nice exercise to prove that the number of zeroes of an analytic function can't be infinite. :-)
– Dèö
Dec 28 '18 at 17:07
[It would be a nice exercise to prove that the number of zeroes of an analytic function can't be infinite]--->This can occur (i.e. the exercise would be false) take $f(z)=e^z-1$ which has $2ipimathbb{Z}$ as set of zeroes.
– Duchamp Gérard H. E.
Dec 28 '18 at 20:59
I have edited your answer, now it is correct (+1) In mathematics, not only the intention counts, the text must be locally exact (well, as much as possible, we are humans :)
– Duchamp Gérard H. E.
Dec 28 '18 at 21:04
Thankyou @Duchamp Gérard H. E. Yeah the infinite number of zeros can occur. Well I am new to this forum, so I need to be precise in my upcoming answers. Thank you 🙂
– Dèö
Dec 29 '18 at 10:18
add a comment |
1
Welcome new contributor. What do you mean by [in fact in the complement of the disc] ? the set of accumulation points of ${|z|<r}$ is in fact the closed disk ${|z|leq r}$
– Duchamp Gérard H. E.
Dec 27 '18 at 21:38
Actually, it doesn't have much to do. I made that point to stress the fact that even if the set, $S$ on which $f(z)$ and $g(z)$ agree, is not as nice as the disc, still we can have the above result. If the set just has one accumulation point in $mathbb{C}-S$, i.e. there is a sequence of zeroes which converges to a point which is outside $S$, then the function must agree entirely. It would be a nice exercise to prove that the number of zeroes of an analytic function can't be infinite. :-)
– Dèö
Dec 28 '18 at 17:07
[It would be a nice exercise to prove that the number of zeroes of an analytic function can't be infinite]--->This can occur (i.e. the exercise would be false) take $f(z)=e^z-1$ which has $2ipimathbb{Z}$ as set of zeroes.
– Duchamp Gérard H. E.
Dec 28 '18 at 20:59
I have edited your answer, now it is correct (+1) In mathematics, not only the intention counts, the text must be locally exact (well, as much as possible, we are humans :)
– Duchamp Gérard H. E.
Dec 28 '18 at 21:04
Thankyou @Duchamp Gérard H. E. Yeah the infinite number of zeros can occur. Well I am new to this forum, so I need to be precise in my upcoming answers. Thank you 🙂
– Dèö
Dec 29 '18 at 10:18
1
1
Welcome new contributor. What do you mean by [in fact in the complement of the disc] ? the set of accumulation points of ${|z|<r}$ is in fact the closed disk ${|z|leq r}$
– Duchamp Gérard H. E.
Dec 27 '18 at 21:38
Welcome new contributor. What do you mean by [in fact in the complement of the disc] ? the set of accumulation points of ${|z|<r}$ is in fact the closed disk ${|z|leq r}$
– Duchamp Gérard H. E.
Dec 27 '18 at 21:38
Actually, it doesn't have much to do. I made that point to stress the fact that even if the set, $S$ on which $f(z)$ and $g(z)$ agree, is not as nice as the disc, still we can have the above result. If the set just has one accumulation point in $mathbb{C}-S$, i.e. there is a sequence of zeroes which converges to a point which is outside $S$, then the function must agree entirely. It would be a nice exercise to prove that the number of zeroes of an analytic function can't be infinite. :-)
– Dèö
Dec 28 '18 at 17:07
Actually, it doesn't have much to do. I made that point to stress the fact that even if the set, $S$ on which $f(z)$ and $g(z)$ agree, is not as nice as the disc, still we can have the above result. If the set just has one accumulation point in $mathbb{C}-S$, i.e. there is a sequence of zeroes which converges to a point which is outside $S$, then the function must agree entirely. It would be a nice exercise to prove that the number of zeroes of an analytic function can't be infinite. :-)
– Dèö
Dec 28 '18 at 17:07
[It would be a nice exercise to prove that the number of zeroes of an analytic function can't be infinite]--->This can occur (i.e. the exercise would be false) take $f(z)=e^z-1$ which has $2ipimathbb{Z}$ as set of zeroes.
– Duchamp Gérard H. E.
Dec 28 '18 at 20:59
[It would be a nice exercise to prove that the number of zeroes of an analytic function can't be infinite]--->This can occur (i.e. the exercise would be false) take $f(z)=e^z-1$ which has $2ipimathbb{Z}$ as set of zeroes.
– Duchamp Gérard H. E.
Dec 28 '18 at 20:59
I have edited your answer, now it is correct (+1) In mathematics, not only the intention counts, the text must be locally exact (well, as much as possible, we are humans :)
– Duchamp Gérard H. E.
Dec 28 '18 at 21:04
I have edited your answer, now it is correct (+1) In mathematics, not only the intention counts, the text must be locally exact (well, as much as possible, we are humans :)
– Duchamp Gérard H. E.
Dec 28 '18 at 21:04
Thankyou @Duchamp Gérard H. E. Yeah the infinite number of zeros can occur. Well I am new to this forum, so I need to be precise in my upcoming answers. Thank you 🙂
– Dèö
Dec 29 '18 at 10:18
Thankyou @Duchamp Gérard H. E. Yeah the infinite number of zeros can occur. Well I am new to this forum, so I need to be precise in my upcoming answers. Thank you 🙂
– Dèö
Dec 29 '18 at 10:18
add a comment |
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4
Look up “identity theorem for holomorphic functions.”
– Martin R
Dec 27 '18 at 13:19
en.wikipedia.org/wiki/Principle_of_permanence
– metamorphy
Dec 27 '18 at 13:57
thanks @MartinR that's really useful
– Smb Youz
yesterday