converge of the series $sum_n=1^inftyfraccos(nx)n^3+x$ to continuously differentiable function

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$sum_n=1^inftyfraccos(nx)n^3+x$



show that the series converge in $(-1,infty)$ to a continuously differentiable function.



I know it's something with uniformly convergence but not sure how.




sequences-and-series uniform-convergence




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  • So you didn't try anything? Weierstrass M? The theorem on uniform convergence and differentiation? Hard to believe you have no idea here ...
    – zhw.
    Jul 17 at 19:46














up vote
1
down vote

favorite












$sum_n=1^inftyfraccos(nx)n^3+x$



show that the series converge in $(-1,infty)$ to a continuously differentiable function.



I know it's something with uniformly convergence but not sure how.




sequences-and-series uniform-convergence




share|cite|improve this question





















  • So you didn't try anything? Weierstrass M? The theorem on uniform convergence and differentiation? Hard to believe you have no idea here ...
    – zhw.
    Jul 17 at 19:46












up vote
1
down vote

favorite









up vote
1
down vote

favorite











$sum_n=1^inftyfraccos(nx)n^3+x$



show that the series converge in $(-1,infty)$ to a continuously differentiable function.



I know it's something with uniformly convergence but not sure how.




sequences-and-series uniform-convergence




share|cite|improve this question













$sum_n=1^inftyfraccos(nx)n^3+x$



show that the series converge in $(-1,infty)$ to a continuously differentiable function.



I know it's something with uniformly convergence but not sure how.





sequences-and-series uniform-convergence





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 17 at 16:18









Bernard

110k635103




110k635103









asked Jul 17 at 16:16









UltimateMath

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385











  • So you didn't try anything? Weierstrass M? The theorem on uniform convergence and differentiation? Hard to believe you have no idea here ...
    – zhw.
    Jul 17 at 19:46
















  • So you didn't try anything? Weierstrass M? The theorem on uniform convergence and differentiation? Hard to believe you have no idea here ...
    – zhw.
    Jul 17 at 19:46















So you didn't try anything? Weierstrass M? The theorem on uniform convergence and differentiation? Hard to believe you have no idea here ...
– zhw.
Jul 17 at 19:46




So you didn't try anything? Weierstrass M? The theorem on uniform convergence and differentiation? Hard to believe you have no idea here ...
– zhw.
Jul 17 at 19:46










1 Answer
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Hint: it is sufficient to show that the derivatives converge uniformly and that original series converges at (at least) one point. To show uniform convergence, consider using the Weierstrass M-test.



Here's a further hint for uniform convergence: the derivative of the $n$th term is $$-fracnsin(nx)n^3 + x + fraccos(nx)(n^3 + x)^2,.$$



In order to show uniform convergence, it is sufficient to bound this term above for $x in (-1,infty)$ by a constant $M_n$ so that $sum M_n < infty$.






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  • can you show me the part with the uniformly converge of the derivatives? I tried it but got stuck.
    – UltimateMath
    Jul 17 at 16:30










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1 Answer
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active

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oldest

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up vote
0
down vote













Hint: it is sufficient to show that the derivatives converge uniformly and that original series converges at (at least) one point. To show uniform convergence, consider using the Weierstrass M-test.



Here's a further hint for uniform convergence: the derivative of the $n$th term is $$-fracnsin(nx)n^3 + x + fraccos(nx)(n^3 + x)^2,.$$



In order to show uniform convergence, it is sufficient to bound this term above for $x in (-1,infty)$ by a constant $M_n$ so that $sum M_n < infty$.






share|cite|improve this answer























  • can you show me the part with the uniformly converge of the derivatives? I tried it but got stuck.
    – UltimateMath
    Jul 17 at 16:30














up vote
0
down vote













Hint: it is sufficient to show that the derivatives converge uniformly and that original series converges at (at least) one point. To show uniform convergence, consider using the Weierstrass M-test.



Here's a further hint for uniform convergence: the derivative of the $n$th term is $$-fracnsin(nx)n^3 + x + fraccos(nx)(n^3 + x)^2,.$$



In order to show uniform convergence, it is sufficient to bound this term above for $x in (-1,infty)$ by a constant $M_n$ so that $sum M_n < infty$.






share|cite|improve this answer























  • can you show me the part with the uniformly converge of the derivatives? I tried it but got stuck.
    – UltimateMath
    Jul 17 at 16:30












up vote
0
down vote










up vote
0
down vote









Hint: it is sufficient to show that the derivatives converge uniformly and that original series converges at (at least) one point. To show uniform convergence, consider using the Weierstrass M-test.



Here's a further hint for uniform convergence: the derivative of the $n$th term is $$-fracnsin(nx)n^3 + x + fraccos(nx)(n^3 + x)^2,.$$



In order to show uniform convergence, it is sufficient to bound this term above for $x in (-1,infty)$ by a constant $M_n$ so that $sum M_n < infty$.






share|cite|improve this answer















Hint: it is sufficient to show that the derivatives converge uniformly and that original series converges at (at least) one point. To show uniform convergence, consider using the Weierstrass M-test.



Here's a further hint for uniform convergence: the derivative of the $n$th term is $$-fracnsin(nx)n^3 + x + fraccos(nx)(n^3 + x)^2,.$$



In order to show uniform convergence, it is sufficient to bound this term above for $x in (-1,infty)$ by a constant $M_n$ so that $sum M_n < infty$.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 17 at 16:34


























answered Jul 17 at 16:23









Marcus M

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  • can you show me the part with the uniformly converge of the derivatives? I tried it but got stuck.
    – UltimateMath
    Jul 17 at 16:30
















  • can you show me the part with the uniformly converge of the derivatives? I tried it but got stuck.
    – UltimateMath
    Jul 17 at 16:30















can you show me the part with the uniformly converge of the derivatives? I tried it but got stuck.
– UltimateMath
Jul 17 at 16:30




can you show me the part with the uniformly converge of the derivatives? I tried it but got stuck.
– UltimateMath
Jul 17 at 16:30












 

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