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Taking the derivative of $sum_limitsn=1^inftyarctan(fracxn^2)$
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Problem: Study the possibility of taking the derivative of the following series:
$$sum_limitsn=1^inftyarctan(fracxn^2)::,xinmathbbR$$
I have studied the following theorem:
Theorem: Suppose that $sum_limitsn=k^inftyf_n$ converges uniformly to $F$ on $S=[a,b]$. Assume that $F$ and $f_n::,ngeqslant k$, are integrable on $[a,b]$. Then:
$int_limitsa^bF(x)dx=sum_limitsn=k^inftyint_limitsa^bf_n(x)dx$
Following the theorem I would need to check out if the derivative converges uniformly $sum_limitsn=1^infty(frac11+fracx^2n^4frac2xn^4)$ I tried to apply Dirichlet to latter. Once I know that $sum_limitsn=1^inftyfrac2xn^4$ by the integral test converges uniformly. However I was not able to apply it due to the fact that I could not prove $sum_limitsn=1^infty(frac11+fracx^2n^4leqslant M$
Question:
How should I solve the problem?
How should I prove the series $sum_limitsn=1^inftyarctan(fracxn^2)$ converge?
Thanks in advance!
calculus real-analysis sequences-and-series derivatives uniform-convergence
edited Jul 21 at 20:53
José Carlos Santos
114k1698177
asked Jul 21 at 18:21
Pedro Gomes
1,3192618
add a comment |Â
up vote
1
down vote
favorite
Problem: Study the possibility of taking the derivative of the following series:
$$sum_limitsn=1^inftyarctan(fracxn^2)::,xinmathbbR$$
I have studied the following theorem:
Theorem: Suppose that $sum_limitsn=k^inftyf_n$ converges uniformly to $F$ on $S=[a,b]$. Assume that $F$ and $f_n::,ngeqslant k$, are integrable on $[a,b]$. Then:
$int_limitsa^bF(x)dx=sum_limitsn=k^inftyint_limitsa^bf_n(x)dx$
Following the theorem I would need to check out if the derivative converges uniformly $sum_limitsn=1^infty(frac11+fracx^2n^4frac2xn^4)$ I tried to apply Dirichlet to latter. Once I know that $sum_limitsn=1^inftyfrac2xn^4$ by the integral test converges uniformly. However I was not able to apply it due to the fact that I could not prove $sum_limitsn=1^infty(frac11+fracx^2n^4leqslant M$
Question:
How should I solve the problem?
How should I prove the series $sum_limitsn=1^inftyarctan(fracxn^2)$ converge?
Thanks in advance!
calculus real-analysis sequences-and-series derivatives uniform-convergence
edited Jul 21 at 20:53
José Carlos Santos
114k1698177
asked Jul 21 at 18:21
Pedro Gomes
1,3192618
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Problem: Study the possibility of taking the derivative of the following series:
$$sum_limitsn=1^inftyarctan(fracxn^2)::,xinmathbbR$$
I have studied the following theorem:
Theorem: Suppose that $sum_limitsn=k^inftyf_n$ converges uniformly to $F$ on $S=[a,b]$. Assume that $F$ and $f_n::,ngeqslant k$, are integrable on $[a,b]$. Then:
$int_limitsa^bF(x)dx=sum_limitsn=k^inftyint_limitsa^bf_n(x)dx$
Following the theorem I would need to check out if the derivative converges uniformly $sum_limitsn=1^infty(frac11+fracx^2n^4frac2xn^4)$ I tried to apply Dirichlet to latter. Once I know that $sum_limitsn=1^inftyfrac2xn^4$ by the integral test converges uniformly. However I was not able to apply it due to the fact that I could not prove $sum_limitsn=1^infty(frac11+fracx^2n^4leqslant M$
Question:
How should I solve the problem?
How should I prove the series $sum_limitsn=1^inftyarctan(fracxn^2)$ converge?
Thanks in advance!
calculus real-analysis sequences-and-series derivatives uniform-convergence
edited Jul 21 at 20:53
José Carlos Santos
114k1698177
asked Jul 21 at 18:21
Pedro Gomes
1,3192618
Problem: Study the possibility of taking the derivative of the following series:
$$sum_limitsn=1^inftyarctan(fracxn^2)::,xinmathbbR$$
I have studied the following theorem:
Theorem: Suppose that $sum_limitsn=k^inftyf_n$ converges uniformly to $F$ on $S=[a,b]$. Assume that $F$ and $f_n::,ngeqslant k$, are integrable on $[a,b]$. Then:
$int_limitsa^bF(x)dx=sum_limitsn=k^inftyint_limitsa^bf_n(x)dx$
Following the theorem I would need to check out if the derivative converges uniformly $sum_limitsn=1^infty(frac11+fracx^2n^4frac2xn^4)$ I tried to apply Dirichlet to latter. Once I know that $sum_limitsn=1^inftyfrac2xn^4$ by the integral test converges uniformly. However I was not able to apply it due to the fact that I could not prove $sum_limitsn=1^infty(frac11+fracx^2n^4leqslant M$
Question:
How should I solve the problem?
How should I prove the series $sum_limitsn=1^inftyarctan(fracxn^2)$ converge?
Thanks in advance!
calculus real-analysis sequences-and-series derivatives uniform-convergence
edited Jul 21 at 20:53
José Carlos Santos
114k1698177
asked Jul 21 at 18:21
Pedro Gomes
1,3192618
edited Jul 21 at 20:53
José Carlos Santos
114k1698177
edited Jul 21 at 20:53
José Carlos Santos
114k1698177
edited Jul 21 at 20:53
José Carlos Santos
114k1698177
114k1698177
asked Jul 21 at 18:21
Pedro Gomes
1,3192618
asked Jul 21 at 18:21
Pedro Gomes
1,3192618
asked Jul 21 at 18:21
Pedro Gomes
1,3192618
1,3192618
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Note that, if $f_n(x)=arctanleft(frac xn^2right)$, then$$f_n'(x)=fracn^2n^4+x^2leqslantfracn^2n^4=frac1n^2.$$Besides, the series $sum_n=1^inftyfrac1n^2$ converges.
answered Jul 21 at 18:27
José Carlos Santos
114k1698177
Is there a way to verify if the series $sum_limitsn=1^inftyarctan(fracxn^2)$ converge? I have been trying to come up with one but it has been hard to deal with the $arctan$ function. Thanks for your answer!
– Pedro Gomes
Jul 21 at 20:35
Sure. For each $x$, $lim_ntoinftyfracarctanleft(frac xn^2right)frac1n^2=arctan'(0)=1$. Since $sum_n=1^inftyfrac1n^2$ converges, $sum_n=1^inftyarctanleft(frac xn^2right)$ converges too.
– José Carlos Santos
Jul 21 at 20:41
add a comment |Â
up vote
3
down vote
The standard theorem is this:
If $sum_n=1^inftyf_n(x)$ converges at least at one point, and the series of derivatives $sum_n=1^inftyf_n^'(x)$ converges uniformly in some interval $I$, then in that interval the series can be differentiated term by term, meaning that the sum of the derivatives converges to the derivative of the sum.
In our case, the series clearly converges for all $x$. Moreover, the series of derivatives is:
$$sum_n=1^inftyfracn^2n^4+x^2$$
which converges uniformly in $mathbbR$ because for every $xinmathbbR$, the general term is bounded by $1/n^2$. It follows that the original series can be differentiated term by term.
answered Jul 21 at 18:29
uniquesolution
7,690721
Could you please give me a link to the theorem proof? Thanks for your answer!
– Pedro Gomes
Jul 21 at 20:33
@PedroGomes I already used this theorem to answer one of your recent questions. The proof is linked in the answer.
– mechanodroid
Jul 22 at 0:02
I was asking about the proof of the theorem relating series not sequences. I do not know how to adapt the theorem to series. Thanks in advance!
– Pedro Gomes
Jul 22 at 19:32
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Note that, if $f_n(x)=arctanleft(frac xn^2right)$, then$$f_n'(x)=fracn^2n^4+x^2leqslantfracn^2n^4=frac1n^2.$$Besides, the series $sum_n=1^inftyfrac1n^2$ converges.
answered Jul 21 at 18:27
José Carlos Santos
114k1698177
Is there a way to verify if the series $sum_limitsn=1^inftyarctan(fracxn^2)$ converge? I have been trying to come up with one but it has been hard to deal with the $arctan$ function. Thanks for your answer!
– Pedro Gomes
Jul 21 at 20:35
Sure. For each $x$, $lim_ntoinftyfracarctanleft(frac xn^2right)frac1n^2=arctan'(0)=1$. Since $sum_n=1^inftyfrac1n^2$ converges, $sum_n=1^inftyarctanleft(frac xn^2right)$ converges too.
– José Carlos Santos
Jul 21 at 20:41
add a comment |Â
up vote
3
down vote
accepted
Note that, if $f_n(x)=arctanleft(frac xn^2right)$, then$$f_n'(x)=fracn^2n^4+x^2leqslantfracn^2n^4=frac1n^2.$$Besides, the series $sum_n=1^inftyfrac1n^2$ converges.
answered Jul 21 at 18:27
José Carlos Santos
114k1698177
Is there a way to verify if the series $sum_limitsn=1^inftyarctan(fracxn^2)$ converge? I have been trying to come up with one but it has been hard to deal with the $arctan$ function. Thanks for your answer!
– Pedro Gomes
Jul 21 at 20:35
Sure. For each $x$, $lim_ntoinftyfracarctanleft(frac xn^2right)frac1n^2=arctan'(0)=1$. Since $sum_n=1^inftyfrac1n^2$ converges, $sum_n=1^inftyarctanleft(frac xn^2right)$ converges too.
– José Carlos Santos
Jul 21 at 20:41
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Note that, if $f_n(x)=arctanleft(frac xn^2right)$, then$$f_n'(x)=fracn^2n^4+x^2leqslantfracn^2n^4=frac1n^2.$$Besides, the series $sum_n=1^inftyfrac1n^2$ converges.
answered Jul 21 at 18:27
José Carlos Santos
114k1698177
Note that, if $f_n(x)=arctanleft(frac xn^2right)$, then$$f_n'(x)=fracn^2n^4+x^2leqslantfracn^2n^4=frac1n^2.$$Besides, the series $sum_n=1^inftyfrac1n^2$ converges.
answered Jul 21 at 18:27
José Carlos Santos
114k1698177
answered Jul 21 at 18:27
José Carlos Santos
114k1698177
answered Jul 21 at 18:27
José Carlos Santos
114k1698177
answered Jul 21 at 18:27
José Carlos Santos
114k1698177
114k1698177
Is there a way to verify if the series $sum_limitsn=1^inftyarctan(fracxn^2)$ converge? I have been trying to come up with one but it has been hard to deal with the $arctan$ function. Thanks for your answer!
– Pedro Gomes
Jul 21 at 20:35
Sure. For each $x$, $lim_ntoinftyfracarctanleft(frac xn^2right)frac1n^2=arctan'(0)=1$. Since $sum_n=1^inftyfrac1n^2$ converges, $sum_n=1^inftyarctanleft(frac xn^2right)$ converges too.
– José Carlos Santos
Jul 21 at 20:41
add a comment |Â
Is there a way to verify if the series $sum_limitsn=1^inftyarctan(fracxn^2)$ converge? I have been trying to come up with one but it has been hard to deal with the $arctan$ function. Thanks for your answer!
– Pedro Gomes
Jul 21 at 20:35
Sure. For each $x$, $lim_ntoinftyfracarctanleft(frac xn^2right)frac1n^2=arctan'(0)=1$. Since $sum_n=1^inftyfrac1n^2$ converges, $sum_n=1^inftyarctanleft(frac xn^2right)$ converges too.
– José Carlos Santos
Jul 21 at 20:41
Is there a way to verify if the series $sum_limitsn=1^inftyarctan(fracxn^2)$ converge? I have been trying to come up with one but it has been hard to deal with the $arctan$ function. Thanks for your answer!
– Pedro Gomes
Jul 21 at 20:35
Is there a way to verify if the series $sum_limitsn=1^inftyarctan(fracxn^2)$ converge? I have been trying to come up with one but it has been hard to deal with the $arctan$ function. Thanks for your answer!
– Pedro Gomes
Jul 21 at 20:35
Sure. For each $x$, $lim_ntoinftyfracarctanleft(frac xn^2right)frac1n^2=arctan'(0)=1$. Since $sum_n=1^inftyfrac1n^2$ converges, $sum_n=1^inftyarctanleft(frac xn^2right)$ converges too.
– José Carlos Santos
Jul 21 at 20:41
Sure. For each $x$, $lim_ntoinftyfracarctanleft(frac xn^2right)frac1n^2=arctan'(0)=1$. Since $sum_n=1^inftyfrac1n^2$ converges, $sum_n=1^inftyarctanleft(frac xn^2right)$ converges too.
– José Carlos Santos
Jul 21 at 20:41
add a comment |Â
up vote
3
down vote
The standard theorem is this:
If $sum_n=1^inftyf_n(x)$ converges at least at one point, and the series of derivatives $sum_n=1^inftyf_n^'(x)$ converges uniformly in some interval $I$, then in that interval the series can be differentiated term by term, meaning that the sum of the derivatives converges to the derivative of the sum.
In our case, the series clearly converges for all $x$. Moreover, the series of derivatives is:
$$sum_n=1^inftyfracn^2n^4+x^2$$
which converges uniformly in $mathbbR$ because for every $xinmathbbR$, the general term is bounded by $1/n^2$. It follows that the original series can be differentiated term by term.
answered Jul 21 at 18:29
uniquesolution
7,690721
Could you please give me a link to the theorem proof? Thanks for your answer!
– Pedro Gomes
Jul 21 at 20:33
@PedroGomes I already used this theorem to answer one of your recent questions. The proof is linked in the answer.
– mechanodroid
Jul 22 at 0:02
I was asking about the proof of the theorem relating series not sequences. I do not know how to adapt the theorem to series. Thanks in advance!
– Pedro Gomes
Jul 22 at 19:32
add a comment |Â
up vote
3
down vote
The standard theorem is this:
If $sum_n=1^inftyf_n(x)$ converges at least at one point, and the series of derivatives $sum_n=1^inftyf_n^'(x)$ converges uniformly in some interval $I$, then in that interval the series can be differentiated term by term, meaning that the sum of the derivatives converges to the derivative of the sum.
In our case, the series clearly converges for all $x$. Moreover, the series of derivatives is:
$$sum_n=1^inftyfracn^2n^4+x^2$$
which converges uniformly in $mathbbR$ because for every $xinmathbbR$, the general term is bounded by $1/n^2$. It follows that the original series can be differentiated term by term.
answered Jul 21 at 18:29
uniquesolution
7,690721
Could you please give me a link to the theorem proof? Thanks for your answer!
– Pedro Gomes
Jul 21 at 20:33
@PedroGomes I already used this theorem to answer one of your recent questions. The proof is linked in the answer.
– mechanodroid
Jul 22 at 0:02
I was asking about the proof of the theorem relating series not sequences. I do not know how to adapt the theorem to series. Thanks in advance!
– Pedro Gomes
Jul 22 at 19:32
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The standard theorem is this:
If $sum_n=1^inftyf_n(x)$ converges at least at one point, and the series of derivatives $sum_n=1^inftyf_n^'(x)$ converges uniformly in some interval $I$, then in that interval the series can be differentiated term by term, meaning that the sum of the derivatives converges to the derivative of the sum.
In our case, the series clearly converges for all $x$. Moreover, the series of derivatives is:
$$sum_n=1^inftyfracn^2n^4+x^2$$
which converges uniformly in $mathbbR$ because for every $xinmathbbR$, the general term is bounded by $1/n^2$. It follows that the original series can be differentiated term by term.
answered Jul 21 at 18:29
uniquesolution
7,690721
The standard theorem is this:
If $sum_n=1^inftyf_n(x)$ converges at least at one point, and the series of derivatives $sum_n=1^inftyf_n^'(x)$ converges uniformly in some interval $I$, then in that interval the series can be differentiated term by term, meaning that the sum of the derivatives converges to the derivative of the sum.
In our case, the series clearly converges for all $x$. Moreover, the series of derivatives is:
$$sum_n=1^inftyfracn^2n^4+x^2$$
which converges uniformly in $mathbbR$ because for every $xinmathbbR$, the general term is bounded by $1/n^2$. It follows that the original series can be differentiated term by term.
answered Jul 21 at 18:29
uniquesolution
7,690721
answered Jul 21 at 18:29
uniquesolution
7,690721
answered Jul 21 at 18:29
uniquesolution
7,690721
answered Jul 21 at 18:29
uniquesolution
7,690721
7,690721
Could you please give me a link to the theorem proof? Thanks for your answer!
– Pedro Gomes
Jul 21 at 20:33
@PedroGomes I already used this theorem to answer one of your recent questions. The proof is linked in the answer.
– mechanodroid
Jul 22 at 0:02
I was asking about the proof of the theorem relating series not sequences. I do not know how to adapt the theorem to series. Thanks in advance!
– Pedro Gomes
Jul 22 at 19:32
add a comment |Â
Could you please give me a link to the theorem proof? Thanks for your answer!
– Pedro Gomes
Jul 21 at 20:33
@PedroGomes I already used this theorem to answer one of your recent questions. The proof is linked in the answer.
– mechanodroid
Jul 22 at 0:02
I was asking about the proof of the theorem relating series not sequences. I do not know how to adapt the theorem to series. Thanks in advance!
– Pedro Gomes
Jul 22 at 19:32
Could you please give me a link to the theorem proof? Thanks for your answer!
– Pedro Gomes
Jul 21 at 20:33
Could you please give me a link to the theorem proof? Thanks for your answer!
– Pedro Gomes
Jul 21 at 20:33
@PedroGomes I already used this theorem to answer one of your recent questions. The proof is linked in the answer.
– mechanodroid
Jul 22 at 0:02
@PedroGomes I already used this theorem to answer one of your recent questions. The proof is linked in the answer.
– mechanodroid
Jul 22 at 0:02
I was asking about the proof of the theorem relating series not sequences. I do not know how to adapt the theorem to series. Thanks in advance!
– Pedro Gomes
Jul 22 at 19:32
I was asking about the proof of the theorem relating series not sequences. I do not know how to adapt the theorem to series. Thanks in advance!
– Pedro Gomes
Jul 22 at 19:32
add a comment |Â
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