What is the radius of convergence for the taylor series for $fracsin(x)1-x$ expanded at 0Showing that some differential equation has an infinite dimensional solution space?For what values of $alpha,beta$ is $x^alphasinx^betain L^1((0,1])$?Calculate the radius of convergence of $sum^infty_k=1 frac(2k-1)^2k-12^2k(2k)!x^k$Pointwise and uniform convergence of series of functionsProving that a sequence has $a_n = a_n+2$ for $n$ sufficiently large.Limit question - L'Hopital's rule doesn't seem to workConvergence of $sum_k=1^infty log(k+1) frace^sin(k)e^k$Real analysis problems involving proofs with sequences and setsRadius of convergence for three seriesA slightly different sine product
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What is the radius of convergence for the taylor series for $fracsin(x)1-x$ expanded at 0
Showing that some differential equation has an infinite dimensional solution space?For what values of $alpha,beta$ is $x^alphasinx^betain L^1((0,1])$?Calculate the radius of convergence of $sum^infty_k=1 frac(2k-1)^2k-12^2k(2k)!x^k$Pointwise and uniform convergence of series of functionsProving that a sequence has $a_n = a_n+2$ for $n$ sufficiently large.Limit question - L'Hopital's rule doesn't seem to workConvergence of $sum_k=1^infty log(k+1) frace^sin(k)e^k$Real analysis problems involving proofs with sequences and setsRadius of convergence for three seriesA slightly different sine product
$begingroup$
Using the Cauchy product I found that $beta=lim|sum_k=1^lfloorfracn2rfloorfrac(-1)^k2k+1|^frac1n$ (if I remember correctly), and thought this was equivalent to the radius of convergence of $sin(x)$, but that doesn't seem logical since the original function isn't even defined at $x=1$. Does anyone know what I did wrong?
real-analysis
asked Mar 28 at 15:43
tomMtomM
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Using the Cauchy product I found that $beta=lim|sum_k=1^lfloorfracn2rfloorfrac(-1)^k2k+1|^frac1n$ (if I remember correctly), and thought this was equivalent to the radius of convergence of $sin(x)$, but that doesn't seem logical since the original function isn't even defined at $x=1$. Does anyone know what I did wrong?
real-analysis
asked Mar 28 at 15:43
tomMtomM
61
New contributor
tomM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
$begingroup$
The series' denominator should be $(2k+1)!$.
$endgroup$
– Chrystomath
Mar 28 at 15:53
$begingroup$
In this case it's better to use the ratio test, which gives 1 as the limit, and hence the radius of convergence.
$endgroup$
– Chrystomath
Mar 28 at 16:00
$begingroup$
@Chrystomath of course, that makes perfect sense. Thanks!
$endgroup$
– tomM
Mar 28 at 16:38
add a comment |
$begingroup$
Using the Cauchy product I found that $beta=lim|sum_k=1^lfloorfracn2rfloorfrac(-1)^k2k+1|^frac1n$ (if I remember correctly), and thought this was equivalent to the radius of convergence of $sin(x)$, but that doesn't seem logical since the original function isn't even defined at $x=1$. Does anyone know what I did wrong?
real-analysis
asked Mar 28 at 15:43
tomMtomM
61
New contributor
tomM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Using the Cauchy product I found that $beta=lim|sum_k=1^lfloorfracn2rfloorfrac(-1)^k2k+1|^frac1n$ (if I remember correctly), and thought this was equivalent to the radius of convergence of $sin(x)$, but that doesn't seem logical since the original function isn't even defined at $x=1$. Does anyone know what I did wrong?
real-analysis
real-analysis
asked Mar 28 at 15:43
tomMtomM
61
New contributor
tomM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 15:43
tomMtomM
61
New contributor
tomM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 15:43
tomMtomM
61
New contributor
tomM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 15:43
tomMtomM
61
asked Mar 28 at 15:43
tomMtomM
61
61
New contributor
tomM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
tomM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
The series' denominator should be $(2k+1)!$.
$endgroup$
– Chrystomath
Mar 28 at 15:53
$begingroup$
In this case it's better to use the ratio test, which gives 1 as the limit, and hence the radius of convergence.
$endgroup$
– Chrystomath
Mar 28 at 16:00
$begingroup$
@Chrystomath of course, that makes perfect sense. Thanks!
$endgroup$
– tomM
Mar 28 at 16:38
add a comment |
$begingroup$
The series' denominator should be $(2k+1)!$.
$endgroup$
– Chrystomath
Mar 28 at 15:53
$begingroup$
In this case it's better to use the ratio test, which gives 1 as the limit, and hence the radius of convergence.
$endgroup$
– Chrystomath
Mar 28 at 16:00
$begingroup$
@Chrystomath of course, that makes perfect sense. Thanks!
$endgroup$
– tomM
Mar 28 at 16:38
$begingroup$
The series' denominator should be $(2k+1)!$.
$endgroup$
– Chrystomath
Mar 28 at 15:53
$begingroup$
The series' denominator should be $(2k+1)!$.
$endgroup$
– Chrystomath
Mar 28 at 15:53
$begingroup$
In this case it's better to use the ratio test, which gives 1 as the limit, and hence the radius of convergence.
$endgroup$
– Chrystomath
Mar 28 at 16:00
$begingroup$
In this case it's better to use the ratio test, which gives 1 as the limit, and hence the radius of convergence.
$endgroup$
– Chrystomath
Mar 28 at 16:00
$begingroup$
@Chrystomath of course, that makes perfect sense. Thanks!
$endgroup$
– tomM
Mar 28 at 16:38
$begingroup$
@Chrystomath of course, that makes perfect sense. Thanks!
$endgroup$
– tomM
Mar 28 at 16:38
add a comment |
1 Answer
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The radius of convergence is the distance from the center of the expansion to the nearest singularity, which would be $1$. This distance is $|1-0|=1$, so the radius of convergence is $1$.
answered Mar 28 at 15:53
MPWMPW
31k12157
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$begingroup$
The radius of convergence is the distance from the center of the expansion to the nearest singularity, which would be $1$. This distance is $|1-0|=1$, so the radius of convergence is $1$.
answered Mar 28 at 15:53
MPWMPW
31k12157
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add a comment |
$begingroup$
The radius of convergence is the distance from the center of the expansion to the nearest singularity, which would be $1$. This distance is $|1-0|=1$, so the radius of convergence is $1$.
answered Mar 28 at 15:53
MPWMPW
31k12157
$endgroup$
add a comment |
$begingroup$
The radius of convergence is the distance from the center of the expansion to the nearest singularity, which would be $1$. This distance is $|1-0|=1$, so the radius of convergence is $1$.
answered Mar 28 at 15:53
MPWMPW
31k12157
$endgroup$
The radius of convergence is the distance from the center of the expansion to the nearest singularity, which would be $1$. This distance is $|1-0|=1$, so the radius of convergence is $1$.
answered Mar 28 at 15:53
MPWMPW
31k12157
answered Mar 28 at 15:53
MPWMPW
31k12157
answered Mar 28 at 15:53
MPWMPW
31k12157
answered Mar 28 at 15:53
MPWMPW
31k12157
31k12157
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$begingroup$
The series' denominator should be $(2k+1)!$.
$endgroup$
– Chrystomath
Mar 28 at 15:53
$begingroup$
In this case it's better to use the ratio test, which gives 1 as the limit, and hence the radius of convergence.
$endgroup$
– Chrystomath
Mar 28 at 16:00
$begingroup$
@Chrystomath of course, that makes perfect sense. Thanks!
$endgroup$
– tomM
Mar 28 at 16:38