Differentiability of series












1












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Is series $$sum_{n=0}^infty e^{-nx} cos{nx}$$ differentiable on $(0,infty)$?




I had this question on my exam and didn't solve it.



Any help is welcome. Thanks in advance!






sequences-and-series analysis derivatives







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    1












    $begingroup$



    Is series $$sum_{n=0}^infty e^{-nx} cos{nx}$$ differentiable on $(0,infty)$?




    I had this question on my exam and didn't solve it.



    Any help is welcome. Thanks in advance!






    sequences-and-series analysis derivatives







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$



      Is series $$sum_{n=0}^infty e^{-nx} cos{nx}$$ differentiable on $(0,infty)$?




      I had this question on my exam and didn't solve it.



      Any help is welcome. Thanks in advance!






      sequences-and-series analysis derivatives







      share|cite|improve this question











      $endgroup$





      Is series $$sum_{n=0}^infty e^{-nx} cos{nx}$$ differentiable on $(0,infty)$?




      I had this question on my exam and didn't solve it.



      Any help is welcome. Thanks in advance!






      sequences-and-series analysis derivatives




      sequences-and-series analysis derivatives






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 10 '18 at 11:00









      Chinnapparaj R

      5,4331928




      5,4331928










      asked Dec 10 '18 at 9:49









      alansalans

      4,2441239




      4,2441239






















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          $begingroup$

          $sum cos (nx)e^{-nx}$ is uniformly convergent in $[a,b]$ if $0<a<b<infty$. So is the derived series $sum (-nsin (nx) e^{-nx} -ncos (nx) e^{-nx})$ because it it dominated by $sum 2ne^{-na}$ which converges by ratio test. These two fact are enough to conclude that $sum cos (nx)e^{-nx}$ is differentiable on $[a,b]$ and its derivative is $-sum nsin (nx) e^{-nx}$ if $0<a<b<infty$. Since $a$ and $b$ are arbitrary we are done.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            you didn't derive well.
            $endgroup$
            – alans
            Dec 10 '18 at 10:04












          • $begingroup$
            @alans Thank you very much. I have corrected the proof.
            $endgroup$
            – Kavi Rama Murthy
            Dec 10 '18 at 10:09











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          $begingroup$

          $sum cos (nx)e^{-nx}$ is uniformly convergent in $[a,b]$ if $0<a<b<infty$. So is the derived series $sum (-nsin (nx) e^{-nx} -ncos (nx) e^{-nx})$ because it it dominated by $sum 2ne^{-na}$ which converges by ratio test. These two fact are enough to conclude that $sum cos (nx)e^{-nx}$ is differentiable on $[a,b]$ and its derivative is $-sum nsin (nx) e^{-nx}$ if $0<a<b<infty$. Since $a$ and $b$ are arbitrary we are done.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            you didn't derive well.
            $endgroup$
            – alans
            Dec 10 '18 at 10:04












          • $begingroup$
            @alans Thank you very much. I have corrected the proof.
            $endgroup$
            – Kavi Rama Murthy
            Dec 10 '18 at 10:09
















          2












          $begingroup$

          $sum cos (nx)e^{-nx}$ is uniformly convergent in $[a,b]$ if $0<a<b<infty$. So is the derived series $sum (-nsin (nx) e^{-nx} -ncos (nx) e^{-nx})$ because it it dominated by $sum 2ne^{-na}$ which converges by ratio test. These two fact are enough to conclude that $sum cos (nx)e^{-nx}$ is differentiable on $[a,b]$ and its derivative is $-sum nsin (nx) e^{-nx}$ if $0<a<b<infty$. Since $a$ and $b$ are arbitrary we are done.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            you didn't derive well.
            $endgroup$
            – alans
            Dec 10 '18 at 10:04












          • $begingroup$
            @alans Thank you very much. I have corrected the proof.
            $endgroup$
            – Kavi Rama Murthy
            Dec 10 '18 at 10:09














          2












          2








          2





          $begingroup$

          $sum cos (nx)e^{-nx}$ is uniformly convergent in $[a,b]$ if $0<a<b<infty$. So is the derived series $sum (-nsin (nx) e^{-nx} -ncos (nx) e^{-nx})$ because it it dominated by $sum 2ne^{-na}$ which converges by ratio test. These two fact are enough to conclude that $sum cos (nx)e^{-nx}$ is differentiable on $[a,b]$ and its derivative is $-sum nsin (nx) e^{-nx}$ if $0<a<b<infty$. Since $a$ and $b$ are arbitrary we are done.






          share|cite|improve this answer











          $endgroup$



          $sum cos (nx)e^{-nx}$ is uniformly convergent in $[a,b]$ if $0<a<b<infty$. So is the derived series $sum (-nsin (nx) e^{-nx} -ncos (nx) e^{-nx})$ because it it dominated by $sum 2ne^{-na}$ which converges by ratio test. These two fact are enough to conclude that $sum cos (nx)e^{-nx}$ is differentiable on $[a,b]$ and its derivative is $-sum nsin (nx) e^{-nx}$ if $0<a<b<infty$. Since $a$ and $b$ are arbitrary we are done.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 10:08

























          answered Dec 10 '18 at 9:56









          Kavi Rama MurthyKavi Rama Murthy

          57.5k42160




          57.5k42160












          • $begingroup$
            you didn't derive well.
            $endgroup$
            – alans
            Dec 10 '18 at 10:04












          • $begingroup$
            @alans Thank you very much. I have corrected the proof.
            $endgroup$
            – Kavi Rama Murthy
            Dec 10 '18 at 10:09


















          • $begingroup$
            you didn't derive well.
            $endgroup$
            – alans
            Dec 10 '18 at 10:04












          • $begingroup$
            @alans Thank you very much. I have corrected the proof.
            $endgroup$
            – Kavi Rama Murthy
            Dec 10 '18 at 10:09
















          $begingroup$
          you didn't derive well.
          $endgroup$
          – alans
          Dec 10 '18 at 10:04






          $begingroup$
          you didn't derive well.
          $endgroup$
          – alans
          Dec 10 '18 at 10:04














          $begingroup$
          @alans Thank you very much. I have corrected the proof.
          $endgroup$
          – Kavi Rama Murthy
          Dec 10 '18 at 10:09




          $begingroup$
          @alans Thank you very much. I have corrected the proof.
          $endgroup$
          – Kavi Rama Murthy
          Dec 10 '18 at 10:09


















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