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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 21 hours ago
Ryder Rude
393110
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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 21 hours ago
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
20 hours ago
But $sum_n 2^n x^n$ doesn't converge for any $x$
– Ryder Rude
19 hours ago
I think it does when $|x|<1/2$
– Richard Martin
19 hours ago
@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
19 hours ago
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
19 hours ago
|
show 1 more comment
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 21 hours ago
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 21 hours ago
Ryder Rude
393110
asked 21 hours ago
Ryder Rude
393110
edited 20 hours ago
asked 21 hours ago
Ryder Rude
393110
asked 21 hours ago
Ryder Rude
393110
asked 21 hours ago
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
20 hours ago
But $sum_n 2^n x^n$ doesn't converge for any $x$
– Ryder Rude
19 hours ago
I think it does when $|x|<1/2$
– Richard Martin
19 hours ago
@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
19 hours ago
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
19 hours ago
|
show 1 more comment
'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
20 hours ago
But $sum_n 2^n x^n$ doesn't converge for any $x$
– Ryder Rude
19 hours ago
I think it does when $|x|<1/2$
– Richard Martin
19 hours ago
@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
19 hours ago
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
19 hours ago
'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
20 hours ago
'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
20 hours ago
But $sum_n 2^n x^n$ doesn't converge for any $x$
– Ryder Rude
19 hours ago
But $sum_n 2^n x^n$ doesn't converge for any $x$
– Ryder Rude
19 hours ago
I think it does when $|x|<1/2$
– Richard Martin
19 hours ago
I think it does when $|x|<1/2$
– Richard Martin
19 hours ago
@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
19 hours ago
@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
19 hours ago
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
19 hours ago
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
19 hours ago
|
show 1 more comment
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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
20 hours ago
But $sum_n 2^n x^n$ doesn't converge for any $x$
– Ryder Rude
19 hours ago
I think it does when $|x|<1/2$
– Richard Martin
19 hours ago
@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
19 hours ago
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
19 hours ago