Is a function still analytic if its infinite sequence of subsequent derivatives, at every point of its...
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Taylor series have the nth derivative ($f^{(n)}(a)$) in their numerator and n! in their denominator. I was wondering what if the derivatives (for any point $a$) grows faster than the factorial. Does the function still equal it's Taylor series? Also, give examples of such functions.
derivatives taylor-expansion analyticity
asked Nov 19 at 11:14
Ryder Rude
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Taylor series have the nth derivative ($f^{(n)}(a)$) in their numerator and n! in their denominator. I was wondering what if the derivatives (for any point $a$) grows faster than the factorial. Does the function still equal it's Taylor series? Also, give examples of such functions.
derivatives taylor-expansion analyticity
asked Nov 19 at 11:14
Ryder Rude
393110
'Faster than the factorial' is not a precise enough statement. E.g. the $n$th derivative is $2^n n!$: in that case, the Taylor series still exists (and the radius of convergence is $1/2$).
– Richard Martin
Nov 19 at 12:04
But $sum_n 2^n x^n$ doesn't converge for any $x$
– Ryder Rude
Nov 19 at 12:41
I think it does when $|x|<1/2$
– Richard Martin
Nov 19 at 12:42
@RichardMartin I got it. I thought the condition 'derivatives grows faster than factorial' should have been enough for the radius of convergence to be zero everywhere.
– Ryder Rude
Nov 19 at 12:43
Well colloquially you are right. If they are like $(2n)!$ or $n^n$ for example, then yes. But you need to be clearer about how much faster, that's all.
– Richard Martin
Nov 19 at 12:45
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Taylor series have the nth derivative ($f^{(n)}(a)$) in their numerator and n! in their denominator. I was wondering what if the derivatives (for any point $a$) grows faster than the factorial. Does the function still equal it's Taylor series? Also, give examples of such functions.
derivatives taylor-expansion analyticity
asked Nov 19 at 11:14
Ryder Rude
393110
Taylor series have the nth derivative ($f^{(n)}(a)$) in their numerator and n! in their denominator. I was wondering what if the derivatives (for any point $a$) grows faster than the factorial. Does the function still equal it's Taylor series? Also, give examples of such functions.
derivatives taylor-expansion analyticity
derivatives taylor-expansion analyticity
asked Nov 19 at 11:14
Ryder Rude
393110
asked Nov 19 at 11:14
Ryder Rude
393110
edited Nov 19 at 11:54
asked Nov 19 at 11:14
Ryder Rude
393110
asked Nov 19 at 11:14
Ryder Rude
393110
asked Nov 19 at 11:14
Ryder Rude
393110
393110
'Faster than the factorial' is not a precise enough statement. E.g. the $n$th derivative is $2^n n!$: in that case, the Taylor series still exists (and the radius of convergence is $1/2$).
– Richard Martin
Nov 19 at 12:04
But $sum_n 2^n x^n$ doesn't converge for any $x$
– Ryder Rude
Nov 19 at 12:41
I think it does when $|x|<1/2$
– Richard Martin
Nov 19 at 12:42
@RichardMartin I got it. I thought the condition 'derivatives grows faster than factorial' should have been enough for the radius of convergence to be zero everywhere.
– Ryder Rude
Nov 19 at 12:43
Well colloquially you are right. If they are like $(2n)!$ or $n^n$ for example, then yes. But you need to be clearer about how much faster, that's all.
– Richard Martin
Nov 19 at 12:45
|
show 2 more comments
'Faster than the factorial' is not a precise enough statement. E.g. the $n$th derivative is $2^n n!$: in that case, the Taylor series still exists (and the radius of convergence is $1/2$).
– Richard Martin
Nov 19 at 12:04
But $sum_n 2^n x^n$ doesn't converge for any $x$
– Ryder Rude
Nov 19 at 12:41
I think it does when $|x|<1/2$
– Richard Martin
Nov 19 at 12:42
@RichardMartin I got it. I thought the condition 'derivatives grows faster than factorial' should have been enough for the radius of convergence to be zero everywhere.
– Ryder Rude
Nov 19 at 12:43
Well colloquially you are right. If they are like $(2n)!$ or $n^n$ for example, then yes. But you need to be clearer about how much faster, that's all.
– Richard Martin
Nov 19 at 12:45
'Faster than the factorial' is not a precise enough statement. E.g. the $n$th derivative is $2^n n!$: in that case, the Taylor series still exists (and the radius of convergence is $1/2$).
– Richard Martin
Nov 19 at 12:04
'Faster than the factorial' is not a precise enough statement. E.g. the $n$th derivative is $2^n n!$: in that case, the Taylor series still exists (and the radius of convergence is $1/2$).
– Richard Martin
Nov 19 at 12:04
But $sum_n 2^n x^n$ doesn't converge for any $x$
– Ryder Rude
Nov 19 at 12:41
But $sum_n 2^n x^n$ doesn't converge for any $x$
– Ryder Rude
Nov 19 at 12:41
I think it does when $|x|<1/2$
– Richard Martin
Nov 19 at 12:42
I think it does when $|x|<1/2$
– Richard Martin
Nov 19 at 12:42
@RichardMartin I got it. I thought the condition 'derivatives grows faster than factorial' should have been enough for the radius of convergence to be zero everywhere.
– Ryder Rude
Nov 19 at 12:43
@RichardMartin I got it. I thought the condition 'derivatives grows faster than factorial' should have been enough for the radius of convergence to be zero everywhere.
– Ryder Rude
Nov 19 at 12:43
Well colloquially you are right. If they are like $(2n)!$ or $n^n$ for example, then yes. But you need to be clearer about how much faster, that's all.
– Richard Martin
Nov 19 at 12:45
Well colloquially you are right. If they are like $(2n)!$ or $n^n$ for example, then yes. But you need to be clearer about how much faster, that's all.
– Richard Martin
Nov 19 at 12:45
|
show 2 more comments
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'Faster than the factorial' is not a precise enough statement. E.g. the $n$th derivative is $2^n n!$: in that case, the Taylor series still exists (and the radius of convergence is $1/2$).
– Richard Martin
Nov 19 at 12:04
But $sum_n 2^n x^n$ doesn't converge for any $x$
– Ryder Rude
Nov 19 at 12:41
I think it does when $|x|<1/2$
– Richard Martin
Nov 19 at 12:42
@RichardMartin I got it. I thought the condition 'derivatives grows faster than factorial' should have been enough for the radius of convergence to be zero everywhere.
– Ryder Rude
Nov 19 at 12:43
Well colloquially you are right. If they are like $(2n)!$ or $n^n$ for example, then yes. But you need to be clearer about how much faster, that's all.
– Richard Martin
Nov 19 at 12:45