Mixing
Prove that $sumlimits_{k=1}^{infty}frac{(-1)^k}{2k+1}x^{2k+1}$ uniformly converges on $[-1,1].$
$begingroup$
Prove that $sumlimits_{k=1}^{infty}dfrac{(-1)^k}{2k+1}x^{2k+1}$ uniformly converges on $[-1,1].$
My book says I have use alternating series test. I can see that the series converges for any $xin[-1,1]$ by the alternating series test but it doesn't tell us the series is uniformly convergent on $[-1,1]$. I tried to use Weierstrass M-Test instead but it fails to pass the M-test.
I let $f_k(x) = dfrac{(-1)^k}{2k+1}x^{2k+1}$ and found $M_k = sup{|dfrac{(-1)^k}{2k+1}x^{2k+1}|:xin[-1,1]} = dfrac{1}{2k+1}$. But $sumlimits_{k=1}^{infty}dfrac{1}{2k+1}$ is not convergent and does not pass the M-test. Hence, the series cannot be uniformly convergent.
Did I use the M-Test correctly?
real-analysis sequences-and-series power-series
edited Jan 2 at 21:44
Lorenzo B.
1,8402520
asked Apr 19 '17 at 20:01
user3000482user3000482
761417
$endgroup$
add a comment |
$begingroup$
Prove that $sumlimits_{k=1}^{infty}dfrac{(-1)^k}{2k+1}x^{2k+1}$ uniformly converges on $[-1,1].$
My book says I have use alternating series test. I can see that the series converges for any $xin[-1,1]$ by the alternating series test but it doesn't tell us the series is uniformly convergent on $[-1,1]$. I tried to use Weierstrass M-Test instead but it fails to pass the M-test.
I let $f_k(x) = dfrac{(-1)^k}{2k+1}x^{2k+1}$ and found $M_k = sup{|dfrac{(-1)^k}{2k+1}x^{2k+1}|:xin[-1,1]} = dfrac{1}{2k+1}$. But $sumlimits_{k=1}^{infty}dfrac{1}{2k+1}$ is not convergent and does not pass the M-test. Hence, the series cannot be uniformly convergent.
Did I use the M-Test correctly?
real-analysis sequences-and-series power-series
edited Jan 2 at 21:44
Lorenzo B.
1,8402520
asked Apr 19 '17 at 20:01
user3000482user3000482
761417
$endgroup$
$begingroup$
@πr8 Oops, I thought $x^k$, not $x^{2k+1}$ ...
$endgroup$
– Hagen von Eitzen
Apr 19 '17 at 20:19
1
$begingroup$
The Weierstrass $M$-test isn't a “test” in the way that the Ratio or Root Test is. You can't use it to conclude that a power series does not converge uniformly.
$endgroup$
– Matthew Leingang
Apr 19 '17 at 20:31
$begingroup$
@MatthewLeingang Then what would be the purpose of using the M-test then? I am really confused because I've been using it to test convergence for series problems. Are you saying if M-test fails, I can't conclude anything about the series?
$endgroup$
– user3000482
Apr 20 '17 at 0:27
$begingroup$
By "if the $M$-test fails", I guess you are referring to the fact that $sum_{k=1}^infty supleft{left|frac{(-1)^kx^{2k+1}}{2k+1}right| : x in [-1,1]right}$ does not converge. That's not a failure of the test; rather, the test does not apply. It just means you have to use another method.
$endgroup$
– Matthew Leingang
Apr 20 '17 at 0:54
add a comment |
$begingroup$
Prove that $sumlimits_{k=1}^{infty}dfrac{(-1)^k}{2k+1}x^{2k+1}$ uniformly converges on $[-1,1].$
My book says I have use alternating series test. I can see that the series converges for any $xin[-1,1]$ by the alternating series test but it doesn't tell us the series is uniformly convergent on $[-1,1]$. I tried to use Weierstrass M-Test instead but it fails to pass the M-test.
I let $f_k(x) = dfrac{(-1)^k}{2k+1}x^{2k+1}$ and found $M_k = sup{|dfrac{(-1)^k}{2k+1}x^{2k+1}|:xin[-1,1]} = dfrac{1}{2k+1}$. But $sumlimits_{k=1}^{infty}dfrac{1}{2k+1}$ is not convergent and does not pass the M-test. Hence, the series cannot be uniformly convergent.
Did I use the M-Test correctly?
real-analysis sequences-and-series power-series
edited Jan 2 at 21:44
Lorenzo B.
1,8402520
asked Apr 19 '17 at 20:01
user3000482user3000482
761417
$endgroup$
Prove that $sumlimits_{k=1}^{infty}dfrac{(-1)^k}{2k+1}x^{2k+1}$ uniformly converges on $[-1,1].$
My book says I have use alternating series test. I can see that the series converges for any $xin[-1,1]$ by the alternating series test but it doesn't tell us the series is uniformly convergent on $[-1,1]$. I tried to use Weierstrass M-Test instead but it fails to pass the M-test.
I let $f_k(x) = dfrac{(-1)^k}{2k+1}x^{2k+1}$ and found $M_k = sup{|dfrac{(-1)^k}{2k+1}x^{2k+1}|:xin[-1,1]} = dfrac{1}{2k+1}$. But $sumlimits_{k=1}^{infty}dfrac{1}{2k+1}$ is not convergent and does not pass the M-test. Hence, the series cannot be uniformly convergent.
Did I use the M-Test correctly?
real-analysis sequences-and-series power-series
real-analysis sequences-and-series power-series
edited Jan 2 at 21:44
Lorenzo B.
1,8402520
asked Apr 19 '17 at 20:01
user3000482user3000482
761417
edited Jan 2 at 21:44
Lorenzo B.
1,8402520
asked Apr 19 '17 at 20:01
user3000482user3000482
761417
edited Jan 2 at 21:44
Lorenzo B.
1,8402520
edited Jan 2 at 21:44
Lorenzo B.
1,8402520
edited Jan 2 at 21:44
Lorenzo B.
1,8402520
1,8402520
asked Apr 19 '17 at 20:01
user3000482user3000482
761417
asked Apr 19 '17 at 20:01
user3000482user3000482
761417
asked Apr 19 '17 at 20:01
user3000482user3000482
761417
761417
$begingroup$
@πr8 Oops, I thought $x^k$, not $x^{2k+1}$ ...
$endgroup$
– Hagen von Eitzen
Apr 19 '17 at 20:19
1
$begingroup$
The Weierstrass $M$-test isn't a “test” in the way that the Ratio or Root Test is. You can't use it to conclude that a power series does not converge uniformly.
$endgroup$
– Matthew Leingang
Apr 19 '17 at 20:31
$begingroup$
@MatthewLeingang Then what would be the purpose of using the M-test then? I am really confused because I've been using it to test convergence for series problems. Are you saying if M-test fails, I can't conclude anything about the series?
$endgroup$
– user3000482
Apr 20 '17 at 0:27
$begingroup$
By "if the $M$-test fails", I guess you are referring to the fact that $sum_{k=1}^infty supleft{left|frac{(-1)^kx^{2k+1}}{2k+1}right| : x in [-1,1]right}$ does not converge. That's not a failure of the test; rather, the test does not apply. It just means you have to use another method.
$endgroup$
– Matthew Leingang
Apr 20 '17 at 0:54
add a comment |
$begingroup$
@πr8 Oops, I thought $x^k$, not $x^{2k+1}$ ...
$endgroup$
– Hagen von Eitzen
Apr 19 '17 at 20:19
1
$begingroup$
The Weierstrass $M$-test isn't a “test” in the way that the Ratio or Root Test is. You can't use it to conclude that a power series does not converge uniformly.
$endgroup$
– Matthew Leingang
Apr 19 '17 at 20:31
$begingroup$
@MatthewLeingang Then what would be the purpose of using the M-test then? I am really confused because I've been using it to test convergence for series problems. Are you saying if M-test fails, I can't conclude anything about the series?
$endgroup$
– user3000482
Apr 20 '17 at 0:27
$begingroup$
By "if the $M$-test fails", I guess you are referring to the fact that $sum_{k=1}^infty supleft{left|frac{(-1)^kx^{2k+1}}{2k+1}right| : x in [-1,1]right}$ does not converge. That's not a failure of the test; rather, the test does not apply. It just means you have to use another method.
$endgroup$
– Matthew Leingang
Apr 20 '17 at 0:54
$begingroup$
@πr8 Oops, I thought $x^k$, not $x^{2k+1}$ ...
$endgroup$
– Hagen von Eitzen
Apr 19 '17 at 20:19
$begingroup$
@πr8 Oops, I thought $x^k$, not $x^{2k+1}$ ...
$endgroup$
– Hagen von Eitzen
Apr 19 '17 at 20:19
1
1
$begingroup$
The Weierstrass $M$-test isn't a “test” in the way that the Ratio or Root Test is. You can't use it to conclude that a power series does not converge uniformly.
$endgroup$
– Matthew Leingang
Apr 19 '17 at 20:31
$begingroup$
The Weierstrass $M$-test isn't a “test” in the way that the Ratio or Root Test is. You can't use it to conclude that a power series does not converge uniformly.
$endgroup$
– Matthew Leingang
Apr 19 '17 at 20:31
$begingroup$
@MatthewLeingang Then what would be the purpose of using the M-test then? I am really confused because I've been using it to test convergence for series problems. Are you saying if M-test fails, I can't conclude anything about the series?
$endgroup$
– user3000482
Apr 20 '17 at 0:27
$begingroup$
@MatthewLeingang Then what would be the purpose of using the M-test then? I am really confused because I've been using it to test convergence for series problems. Are you saying if M-test fails, I can't conclude anything about the series?
$endgroup$
– user3000482
Apr 20 '17 at 0:27
$begingroup$
By "if the $M$-test fails", I guess you are referring to the fact that $sum_{k=1}^infty supleft{left|frac{(-1)^kx^{2k+1}}{2k+1}right| : x in [-1,1]right}$ does not converge. That's not a failure of the test; rather, the test does not apply. It just means you have to use another method.
$endgroup$
– Matthew Leingang
Apr 20 '17 at 0:54
$begingroup$
By "if the $M$-test fails", I guess you are referring to the fact that $sum_{k=1}^infty supleft{left|frac{(-1)^kx^{2k+1}}{2k+1}right| : x in [-1,1]right}$ does not converge. That's not a failure of the test; rather, the test does not apply. It just means you have to use another method.
$endgroup$
– Matthew Leingang
Apr 20 '17 at 0:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that for $xin[0,1]$, the sequence $frac{x^{2k+1}}{2k+1}$ is monotone decreasing (and for $xin[-1,0]$, the negative of this sequence is monotone decreasing).
If $a_k$ is a monotone decreasing, nonnegative sequence, the alternating series test says that:
- $sum_{k ge 1} (-1)^k a_k$ converges to some $s$
- $|s - sum_{1 le k le n}(-1)^k a_k|<a_{n+1}$
Thus, if $f_n(x)=sum_{k=1}^{n} frac{(-1)^k}{2k+1} x^{2k+1}$ and $f(x)=sum_{k=1}^{infty} frac{(-1)^k}{2k+1} x^{2k+1}$, then $$|f(x)-f_n(x)|le frac{ |x|^{2n+3} }{2n+3}le frac{1}{2n+3} to 0$$
answered Apr 19 '17 at 20:43
πr8πr8
9,84831024
$endgroup$
$begingroup$
But if we use the alternating series test, don't we only know that the series is convergent not uniformly convergent?
$endgroup$
– user3000482
Apr 20 '17 at 21:30
$begingroup$
You have a uniform bound on $|f(x)-f_n(x)|$; this is what is needed for uniform convergence.
$endgroup$
– πr8
Apr 20 '17 at 22:48
$begingroup$
I'm sorry but I am not sure what you mean by uniform bound. Don't we need to make sure that $| f(x)-f_n(x) |_{sup} < epsilon$?
$endgroup$
– user3000482
Apr 21 '17 at 15:05
1
$begingroup$
A uniform bound is the same as a supremum bound, yes. $| f(x)-f_n(x) |_{sup} = sup_{xin [-1,1]} |f(x)-f_n(x)|$. Since $|f(x)-f_n(x)|<frac{1}{2n+3}$ for all relevant $x$, it is clear that $sup_{xin [-1,1]} |f(x)-f_n(x)|<frac{1}{2n+3}$ also. So yes, the series certainly is uniformly convergent.
$endgroup$
– πr8
Apr 21 '17 at 15:23
1
$begingroup$
Maybe - one could conceivably adapt the proof above to prove a result of this form for power series with decaying coefficients of alternating sign.
$endgroup$
– πr8
Apr 21 '17 at 18:35
|
show 3 more comments
$begingroup$
Let $$f_n(x)=sum_{k=1}^{n} frac{(-1)^k}{2k+1} x^{2k+1},$$ and call $$f(x)=sum_{k=1}^{infty} frac{(-1)^k}{2k+1} x^{2k+1}.$$
Note that $$f_n'(s)=-s^2frac{1-(-s^2)^n}{1+s^2} implies f(x)=- int_0^x s^2frac{1-(-s^2)^n}{1+s^2},ds$$ and $$f(x)=arctan(x)-x=-int_0^xfrac{s^2}{1+s^2},ds.$$
Thus:
$$|f_n(x)-f(x)|=left| int_0^x frac{s^{2(n+1)}}{1+s^2},dsright| le left| int_0^1 s^{2(n+1)},dsright|=frac{1}{2n+3}to0$$
answered Apr 19 '17 at 20:15
πr8πr8
9,84831024
$endgroup$
$begingroup$
I'm sorry but I dont' quite understand your answer. How can you take derivative of $f_n(x)$? And what is the variable $s$? Is it just an arbitrary element from $[-1,1]$? I feel frustrated because I don't know how am I supposed to come up with such crazy proof... if that's the only way to solve this. Can you comment on my attempt of using M-test? I think I used it properly but the series fails the M-test.
$endgroup$
– user3000482
Apr 19 '17 at 20:22
$begingroup$
1) You can take the derivative of $f_n$ by differentiating it term-by-term; $f_n$ is just a polynomial. 2) Yes, $s$ is just a dummy variable. 3) To show uniform convergence, you need to control the difference between the partial sums ($f_n$ in this question) and the true limit ($f$), and integrals often end up being a good way of doing so. 4) The M-test doesn't work here because, as you say, the terms in the resulting sum do not have a convergent sum.
$endgroup$
– πr8
Apr 19 '17 at 20:25
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Note that for $xin[0,1]$, the sequence $frac{x^{2k+1}}{2k+1}$ is monotone decreasing (and for $xin[-1,0]$, the negative of this sequence is monotone decreasing).
If $a_k$ is a monotone decreasing, nonnegative sequence, the alternating series test says that:
- $sum_{k ge 1} (-1)^k a_k$ converges to some $s$
- $|s - sum_{1 le k le n}(-1)^k a_k|<a_{n+1}$
Thus, if $f_n(x)=sum_{k=1}^{n} frac{(-1)^k}{2k+1} x^{2k+1}$ and $f(x)=sum_{k=1}^{infty} frac{(-1)^k}{2k+1} x^{2k+1}$, then $$|f(x)-f_n(x)|le frac{ |x|^{2n+3} }{2n+3}le frac{1}{2n+3} to 0$$
answered Apr 19 '17 at 20:43
πr8πr8
9,84831024
$endgroup$
$begingroup$
But if we use the alternating series test, don't we only know that the series is convergent not uniformly convergent?
$endgroup$
– user3000482
Apr 20 '17 at 21:30
$begingroup$
You have a uniform bound on $|f(x)-f_n(x)|$; this is what is needed for uniform convergence.
$endgroup$
– πr8
Apr 20 '17 at 22:48
$begingroup$
I'm sorry but I am not sure what you mean by uniform bound. Don't we need to make sure that $| f(x)-f_n(x) |_{sup} < epsilon$?
$endgroup$
– user3000482
Apr 21 '17 at 15:05
1
$begingroup$
A uniform bound is the same as a supremum bound, yes. $| f(x)-f_n(x) |_{sup} = sup_{xin [-1,1]} |f(x)-f_n(x)|$. Since $|f(x)-f_n(x)|<frac{1}{2n+3}$ for all relevant $x$, it is clear that $sup_{xin [-1,1]} |f(x)-f_n(x)|<frac{1}{2n+3}$ also. So yes, the series certainly is uniformly convergent.
$endgroup$
– πr8
Apr 21 '17 at 15:23
1
$begingroup$
Maybe - one could conceivably adapt the proof above to prove a result of this form for power series with decaying coefficients of alternating sign.
$endgroup$
– πr8
Apr 21 '17 at 18:35
|
show 3 more comments
$begingroup$
Note that for $xin[0,1]$, the sequence $frac{x^{2k+1}}{2k+1}$ is monotone decreasing (and for $xin[-1,0]$, the negative of this sequence is monotone decreasing).
If $a_k$ is a monotone decreasing, nonnegative sequence, the alternating series test says that:
- $sum_{k ge 1} (-1)^k a_k$ converges to some $s$
- $|s - sum_{1 le k le n}(-1)^k a_k|<a_{n+1}$
Thus, if $f_n(x)=sum_{k=1}^{n} frac{(-1)^k}{2k+1} x^{2k+1}$ and $f(x)=sum_{k=1}^{infty} frac{(-1)^k}{2k+1} x^{2k+1}$, then $$|f(x)-f_n(x)|le frac{ |x|^{2n+3} }{2n+3}le frac{1}{2n+3} to 0$$
answered Apr 19 '17 at 20:43
πr8πr8
9,84831024
$endgroup$
$begingroup$
But if we use the alternating series test, don't we only know that the series is convergent not uniformly convergent?
$endgroup$
– user3000482
Apr 20 '17 at 21:30
$begingroup$
You have a uniform bound on $|f(x)-f_n(x)|$; this is what is needed for uniform convergence.
$endgroup$
– πr8
Apr 20 '17 at 22:48
$begingroup$
I'm sorry but I am not sure what you mean by uniform bound. Don't we need to make sure that $| f(x)-f_n(x) |_{sup} < epsilon$?
$endgroup$
– user3000482
Apr 21 '17 at 15:05
1
$begingroup$
A uniform bound is the same as a supremum bound, yes. $| f(x)-f_n(x) |_{sup} = sup_{xin [-1,1]} |f(x)-f_n(x)|$. Since $|f(x)-f_n(x)|<frac{1}{2n+3}$ for all relevant $x$, it is clear that $sup_{xin [-1,1]} |f(x)-f_n(x)|<frac{1}{2n+3}$ also. So yes, the series certainly is uniformly convergent.
$endgroup$
– πr8
Apr 21 '17 at 15:23
1
$begingroup$
Maybe - one could conceivably adapt the proof above to prove a result of this form for power series with decaying coefficients of alternating sign.
$endgroup$
– πr8
Apr 21 '17 at 18:35
|
show 3 more comments
$begingroup$
Note that for $xin[0,1]$, the sequence $frac{x^{2k+1}}{2k+1}$ is monotone decreasing (and for $xin[-1,0]$, the negative of this sequence is monotone decreasing).
If $a_k$ is a monotone decreasing, nonnegative sequence, the alternating series test says that:
- $sum_{k ge 1} (-1)^k a_k$ converges to some $s$
- $|s - sum_{1 le k le n}(-1)^k a_k|<a_{n+1}$
Thus, if $f_n(x)=sum_{k=1}^{n} frac{(-1)^k}{2k+1} x^{2k+1}$ and $f(x)=sum_{k=1}^{infty} frac{(-1)^k}{2k+1} x^{2k+1}$, then $$|f(x)-f_n(x)|le frac{ |x|^{2n+3} }{2n+3}le frac{1}{2n+3} to 0$$
answered Apr 19 '17 at 20:43
πr8πr8
9,84831024
$endgroup$
Note that for $xin[0,1]$, the sequence $frac{x^{2k+1}}{2k+1}$ is monotone decreasing (and for $xin[-1,0]$, the negative of this sequence is monotone decreasing).
If $a_k$ is a monotone decreasing, nonnegative sequence, the alternating series test says that:
- $sum_{k ge 1} (-1)^k a_k$ converges to some $s$
- $|s - sum_{1 le k le n}(-1)^k a_k|<a_{n+1}$
Thus, if $f_n(x)=sum_{k=1}^{n} frac{(-1)^k}{2k+1} x^{2k+1}$ and $f(x)=sum_{k=1}^{infty} frac{(-1)^k}{2k+1} x^{2k+1}$, then $$|f(x)-f_n(x)|le frac{ |x|^{2n+3} }{2n+3}le frac{1}{2n+3} to 0$$
answered Apr 19 '17 at 20:43
πr8πr8
9,84831024
answered Apr 19 '17 at 20:43
πr8πr8
9,84831024
answered Apr 19 '17 at 20:43
πr8πr8
9,84831024
answered Apr 19 '17 at 20:43
πr8πr8
9,84831024
9,84831024
$begingroup$
But if we use the alternating series test, don't we only know that the series is convergent not uniformly convergent?
$endgroup$
– user3000482
Apr 20 '17 at 21:30
$begingroup$
You have a uniform bound on $|f(x)-f_n(x)|$; this is what is needed for uniform convergence.
$endgroup$
– πr8
Apr 20 '17 at 22:48
$begingroup$
I'm sorry but I am not sure what you mean by uniform bound. Don't we need to make sure that $| f(x)-f_n(x) |_{sup} < epsilon$?
$endgroup$
– user3000482
Apr 21 '17 at 15:05
1
$begingroup$
A uniform bound is the same as a supremum bound, yes. $| f(x)-f_n(x) |_{sup} = sup_{xin [-1,1]} |f(x)-f_n(x)|$. Since $|f(x)-f_n(x)|<frac{1}{2n+3}$ for all relevant $x$, it is clear that $sup_{xin [-1,1]} |f(x)-f_n(x)|<frac{1}{2n+3}$ also. So yes, the series certainly is uniformly convergent.
$endgroup$
– πr8
Apr 21 '17 at 15:23
1
$begingroup$
Maybe - one could conceivably adapt the proof above to prove a result of this form for power series with decaying coefficients of alternating sign.
$endgroup$
– πr8
Apr 21 '17 at 18:35
|
show 3 more comments
$begingroup$
But if we use the alternating series test, don't we only know that the series is convergent not uniformly convergent?
$endgroup$
– user3000482
Apr 20 '17 at 21:30
$begingroup$
You have a uniform bound on $|f(x)-f_n(x)|$; this is what is needed for uniform convergence.
$endgroup$
– πr8
Apr 20 '17 at 22:48
$begingroup$
I'm sorry but I am not sure what you mean by uniform bound. Don't we need to make sure that $| f(x)-f_n(x) |_{sup} < epsilon$?
$endgroup$
– user3000482
Apr 21 '17 at 15:05
1
$begingroup$
A uniform bound is the same as a supremum bound, yes. $| f(x)-f_n(x) |_{sup} = sup_{xin [-1,1]} |f(x)-f_n(x)|$. Since $|f(x)-f_n(x)|<frac{1}{2n+3}$ for all relevant $x$, it is clear that $sup_{xin [-1,1]} |f(x)-f_n(x)|<frac{1}{2n+3}$ also. So yes, the series certainly is uniformly convergent.
$endgroup$
– πr8
Apr 21 '17 at 15:23
1
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Maybe - one could conceivably adapt the proof above to prove a result of this form for power series with decaying coefficients of alternating sign.
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– πr8
Apr 21 '17 at 18:35
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But if we use the alternating series test, don't we only know that the series is convergent not uniformly convergent?
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– user3000482
Apr 20 '17 at 21:30
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But if we use the alternating series test, don't we only know that the series is convergent not uniformly convergent?
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– user3000482
Apr 20 '17 at 21:30
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You have a uniform bound on $|f(x)-f_n(x)|$; this is what is needed for uniform convergence.
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– πr8
Apr 20 '17 at 22:48
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You have a uniform bound on $|f(x)-f_n(x)|$; this is what is needed for uniform convergence.
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– πr8
Apr 20 '17 at 22:48
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I'm sorry but I am not sure what you mean by uniform bound. Don't we need to make sure that $| f(x)-f_n(x) |_{sup} < epsilon$?
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– user3000482
Apr 21 '17 at 15:05
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I'm sorry but I am not sure what you mean by uniform bound. Don't we need to make sure that $| f(x)-f_n(x) |_{sup} < epsilon$?
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– user3000482
Apr 21 '17 at 15:05
1
1
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A uniform bound is the same as a supremum bound, yes. $| f(x)-f_n(x) |_{sup} = sup_{xin [-1,1]} |f(x)-f_n(x)|$. Since $|f(x)-f_n(x)|<frac{1}{2n+3}$ for all relevant $x$, it is clear that $sup_{xin [-1,1]} |f(x)-f_n(x)|<frac{1}{2n+3}$ also. So yes, the series certainly is uniformly convergent.
$endgroup$
– πr8
Apr 21 '17 at 15:23
$begingroup$
A uniform bound is the same as a supremum bound, yes. $| f(x)-f_n(x) |_{sup} = sup_{xin [-1,1]} |f(x)-f_n(x)|$. Since $|f(x)-f_n(x)|<frac{1}{2n+3}$ for all relevant $x$, it is clear that $sup_{xin [-1,1]} |f(x)-f_n(x)|<frac{1}{2n+3}$ also. So yes, the series certainly is uniformly convergent.
$endgroup$
– πr8
Apr 21 '17 at 15:23
1
1
$begingroup$
Maybe - one could conceivably adapt the proof above to prove a result of this form for power series with decaying coefficients of alternating sign.
$endgroup$
– πr8
Apr 21 '17 at 18:35
$begingroup$
Maybe - one could conceivably adapt the proof above to prove a result of this form for power series with decaying coefficients of alternating sign.
$endgroup$
– πr8
Apr 21 '17 at 18:35
|
show 3 more comments
$begingroup$
Let $$f_n(x)=sum_{k=1}^{n} frac{(-1)^k}{2k+1} x^{2k+1},$$ and call $$f(x)=sum_{k=1}^{infty} frac{(-1)^k}{2k+1} x^{2k+1}.$$
Note that $$f_n'(s)=-s^2frac{1-(-s^2)^n}{1+s^2} implies f(x)=- int_0^x s^2frac{1-(-s^2)^n}{1+s^2},ds$$ and $$f(x)=arctan(x)-x=-int_0^xfrac{s^2}{1+s^2},ds.$$
Thus:
$$|f_n(x)-f(x)|=left| int_0^x frac{s^{2(n+1)}}{1+s^2},dsright| le left| int_0^1 s^{2(n+1)},dsright|=frac{1}{2n+3}to0$$
answered Apr 19 '17 at 20:15
πr8πr8
9,84831024
$endgroup$
$begingroup$
I'm sorry but I dont' quite understand your answer. How can you take derivative of $f_n(x)$? And what is the variable $s$? Is it just an arbitrary element from $[-1,1]$? I feel frustrated because I don't know how am I supposed to come up with such crazy proof... if that's the only way to solve this. Can you comment on my attempt of using M-test? I think I used it properly but the series fails the M-test.
$endgroup$
– user3000482
Apr 19 '17 at 20:22
$begingroup$
1) You can take the derivative of $f_n$ by differentiating it term-by-term; $f_n$ is just a polynomial. 2) Yes, $s$ is just a dummy variable. 3) To show uniform convergence, you need to control the difference between the partial sums ($f_n$ in this question) and the true limit ($f$), and integrals often end up being a good way of doing so. 4) The M-test doesn't work here because, as you say, the terms in the resulting sum do not have a convergent sum.
$endgroup$
– πr8
Apr 19 '17 at 20:25
add a comment |
$begingroup$
Let $$f_n(x)=sum_{k=1}^{n} frac{(-1)^k}{2k+1} x^{2k+1},$$ and call $$f(x)=sum_{k=1}^{infty} frac{(-1)^k}{2k+1} x^{2k+1}.$$
Note that $$f_n'(s)=-s^2frac{1-(-s^2)^n}{1+s^2} implies f(x)=- int_0^x s^2frac{1-(-s^2)^n}{1+s^2},ds$$ and $$f(x)=arctan(x)-x=-int_0^xfrac{s^2}{1+s^2},ds.$$
Thus:
$$|f_n(x)-f(x)|=left| int_0^x frac{s^{2(n+1)}}{1+s^2},dsright| le left| int_0^1 s^{2(n+1)},dsright|=frac{1}{2n+3}to0$$
answered Apr 19 '17 at 20:15
πr8πr8
9,84831024
$endgroup$
$begingroup$
I'm sorry but I dont' quite understand your answer. How can you take derivative of $f_n(x)$? And what is the variable $s$? Is it just an arbitrary element from $[-1,1]$? I feel frustrated because I don't know how am I supposed to come up with such crazy proof... if that's the only way to solve this. Can you comment on my attempt of using M-test? I think I used it properly but the series fails the M-test.
$endgroup$
– user3000482
Apr 19 '17 at 20:22
$begingroup$
1) You can take the derivative of $f_n$ by differentiating it term-by-term; $f_n$ is just a polynomial. 2) Yes, $s$ is just a dummy variable. 3) To show uniform convergence, you need to control the difference between the partial sums ($f_n$ in this question) and the true limit ($f$), and integrals often end up being a good way of doing so. 4) The M-test doesn't work here because, as you say, the terms in the resulting sum do not have a convergent sum.
$endgroup$
– πr8
Apr 19 '17 at 20:25
add a comment |
$begingroup$
Let $$f_n(x)=sum_{k=1}^{n} frac{(-1)^k}{2k+1} x^{2k+1},$$ and call $$f(x)=sum_{k=1}^{infty} frac{(-1)^k}{2k+1} x^{2k+1}.$$
Note that $$f_n'(s)=-s^2frac{1-(-s^2)^n}{1+s^2} implies f(x)=- int_0^x s^2frac{1-(-s^2)^n}{1+s^2},ds$$ and $$f(x)=arctan(x)-x=-int_0^xfrac{s^2}{1+s^2},ds.$$
Thus:
$$|f_n(x)-f(x)|=left| int_0^x frac{s^{2(n+1)}}{1+s^2},dsright| le left| int_0^1 s^{2(n+1)},dsright|=frac{1}{2n+3}to0$$
answered Apr 19 '17 at 20:15
πr8πr8
9,84831024
$endgroup$
Let $$f_n(x)=sum_{k=1}^{n} frac{(-1)^k}{2k+1} x^{2k+1},$$ and call $$f(x)=sum_{k=1}^{infty} frac{(-1)^k}{2k+1} x^{2k+1}.$$
Note that $$f_n'(s)=-s^2frac{1-(-s^2)^n}{1+s^2} implies f(x)=- int_0^x s^2frac{1-(-s^2)^n}{1+s^2},ds$$ and $$f(x)=arctan(x)-x=-int_0^xfrac{s^2}{1+s^2},ds.$$
Thus:
$$|f_n(x)-f(x)|=left| int_0^x frac{s^{2(n+1)}}{1+s^2},dsright| le left| int_0^1 s^{2(n+1)},dsright|=frac{1}{2n+3}to0$$
answered Apr 19 '17 at 20:15
πr8πr8
9,84831024
answered Apr 19 '17 at 20:15
πr8πr8
9,84831024
answered Apr 19 '17 at 20:15
πr8πr8
9,84831024
answered Apr 19 '17 at 20:15
πr8πr8
9,84831024
9,84831024
$begingroup$
I'm sorry but I dont' quite understand your answer. How can you take derivative of $f_n(x)$? And what is the variable $s$? Is it just an arbitrary element from $[-1,1]$? I feel frustrated because I don't know how am I supposed to come up with such crazy proof... if that's the only way to solve this. Can you comment on my attempt of using M-test? I think I used it properly but the series fails the M-test.
$endgroup$
– user3000482
Apr 19 '17 at 20:22
$begingroup$
1) You can take the derivative of $f_n$ by differentiating it term-by-term; $f_n$ is just a polynomial. 2) Yes, $s$ is just a dummy variable. 3) To show uniform convergence, you need to control the difference between the partial sums ($f_n$ in this question) and the true limit ($f$), and integrals often end up being a good way of doing so. 4) The M-test doesn't work here because, as you say, the terms in the resulting sum do not have a convergent sum.
$endgroup$
– πr8
Apr 19 '17 at 20:25
add a comment |
$begingroup$
I'm sorry but I dont' quite understand your answer. How can you take derivative of $f_n(x)$? And what is the variable $s$? Is it just an arbitrary element from $[-1,1]$? I feel frustrated because I don't know how am I supposed to come up with such crazy proof... if that's the only way to solve this. Can you comment on my attempt of using M-test? I think I used it properly but the series fails the M-test.
$endgroup$
– user3000482
Apr 19 '17 at 20:22
$begingroup$
1) You can take the derivative of $f_n$ by differentiating it term-by-term; $f_n$ is just a polynomial. 2) Yes, $s$ is just a dummy variable. 3) To show uniform convergence, you need to control the difference between the partial sums ($f_n$ in this question) and the true limit ($f$), and integrals often end up being a good way of doing so. 4) The M-test doesn't work here because, as you say, the terms in the resulting sum do not have a convergent sum.
$endgroup$
– πr8
Apr 19 '17 at 20:25
$begingroup$
I'm sorry but I dont' quite understand your answer. How can you take derivative of $f_n(x)$? And what is the variable $s$? Is it just an arbitrary element from $[-1,1]$? I feel frustrated because I don't know how am I supposed to come up with such crazy proof... if that's the only way to solve this. Can you comment on my attempt of using M-test? I think I used it properly but the series fails the M-test.
$endgroup$
– user3000482
Apr 19 '17 at 20:22
$begingroup$
I'm sorry but I dont' quite understand your answer. How can you take derivative of $f_n(x)$? And what is the variable $s$? Is it just an arbitrary element from $[-1,1]$? I feel frustrated because I don't know how am I supposed to come up with such crazy proof... if that's the only way to solve this. Can you comment on my attempt of using M-test? I think I used it properly but the series fails the M-test.
$endgroup$
– user3000482
Apr 19 '17 at 20:22
$begingroup$
1) You can take the derivative of $f_n$ by differentiating it term-by-term; $f_n$ is just a polynomial. 2) Yes, $s$ is just a dummy variable. 3) To show uniform convergence, you need to control the difference between the partial sums ($f_n$ in this question) and the true limit ($f$), and integrals often end up being a good way of doing so. 4) The M-test doesn't work here because, as you say, the terms in the resulting sum do not have a convergent sum.
$endgroup$
– πr8
Apr 19 '17 at 20:25
$begingroup$
1) You can take the derivative of $f_n$ by differentiating it term-by-term; $f_n$ is just a polynomial. 2) Yes, $s$ is just a dummy variable. 3) To show uniform convergence, you need to control the difference between the partial sums ($f_n$ in this question) and the true limit ($f$), and integrals often end up being a good way of doing so. 4) The M-test doesn't work here because, as you say, the terms in the resulting sum do not have a convergent sum.
$endgroup$
– πr8
Apr 19 '17 at 20:25
add a comment |
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$begingroup$
@πr8 Oops, I thought $x^k$, not $x^{2k+1}$ ...
$endgroup$
– Hagen von Eitzen
Apr 19 '17 at 20:19
1
$begingroup$
The Weierstrass $M$-test isn't a “test” in the way that the Ratio or Root Test is. You can't use it to conclude that a power series does not converge uniformly.
$endgroup$
– Matthew Leingang
Apr 19 '17 at 20:31
$begingroup$
@MatthewLeingang Then what would be the purpose of using the M-test then? I am really confused because I've been using it to test convergence for series problems. Are you saying if M-test fails, I can't conclude anything about the series?
$endgroup$
– user3000482
Apr 20 '17 at 0:27
$begingroup$
By "if the $M$-test fails", I guess you are referring to the fact that $sum_{k=1}^infty supleft{left|frac{(-1)^kx^{2k+1}}{2k+1}right| : x in [-1,1]right}$ does not converge. That's not a failure of the test; rather, the test does not apply. It just means you have to use another method.
$endgroup$
– Matthew Leingang
Apr 20 '17 at 0:54