How to split a series into two other series.












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How do you take this $$sum_{n=0}^{100} frac{1}{a^n+b^n}:;text{(or)}:;sum_{n=0}^{100} frac{x}{a^n+b^n}$$ and split it into two sums, one having only $1$ (or $x$) and $a$ and the other having only $1$ (or $x$) and $b$.






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    $begingroup$


    How do you take this $$sum_{n=0}^{100} frac{1}{a^n+b^n}:;text{(or)}:;sum_{n=0}^{100} frac{x}{a^n+b^n}$$ and split it into two sums, one having only $1$ (or $x$) and $a$ and the other having only $1$ (or $x$) and $b$.






    sequences-and-series







    share|cite|improve this question











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      0












      0








      0





      $begingroup$


      How do you take this $$sum_{n=0}^{100} frac{1}{a^n+b^n}:;text{(or)}:;sum_{n=0}^{100} frac{x}{a^n+b^n}$$ and split it into two sums, one having only $1$ (or $x$) and $a$ and the other having only $1$ (or $x$) and $b$.






      sequences-and-series







      share|cite|improve this question











      $endgroup$




      How do you take this $$sum_{n=0}^{100} frac{1}{a^n+b^n}:;text{(or)}:;sum_{n=0}^{100} frac{x}{a^n+b^n}$$ and split it into two sums, one having only $1$ (or $x$) and $a$ and the other having only $1$ (or $x$) and $b$.






      sequences-and-series




      sequences-and-series






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      share|cite|improve this question













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      edited Dec 20 '18 at 1:53









      Yadati Kiran

      2,1121622




      2,1121622










      asked Dec 20 '18 at 1:46









      JamesJames

      218




      218






















          2 Answers
          2






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          6












          $begingroup$

          This is not possible. Let
          $$f(a,b)=sum_{n=0}^{100}frac{1}{a^n+b^n}.$$
          We have that
          $$frac{partial f}{partial b}=-sum_{n=0}^{100}frac{n b^{n-1}}{(a^n+b^n)^2}$$
          which clearly depends on $a$. If we were to have
          $$f(a,b)=g(a)+k(b)$$
          then we'd have that
          $$frac{partial f}{partial b}=k'(b)$$
          doesn't depend on $a$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            +1 you got there first. Welcome to stackexchange.
            $endgroup$
            – Ethan Bolker
            Dec 20 '18 at 1:55



















          1












          $begingroup$

          You can't. If that sum were of the form $f(a) + g(b)$ then the partial derivative with respect to $a$ would be independent of $b$. That's clearly not the case.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ok but what about $$sum_{n=0}^{100} {-1}^{n-1}*{3an+2bn}$$ How would you split this
            $endgroup$
            – James
            Dec 20 '18 at 2:05












          • $begingroup$
            @James That one is easy. It's just the distributive law. Write out the sum with ellipses or for an upper limit of $2$ instead of $100$ without the $Sigma$.
            $endgroup$
            – Ethan Bolker
            Dec 20 '18 at 13:03











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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          This is not possible. Let
          $$f(a,b)=sum_{n=0}^{100}frac{1}{a^n+b^n}.$$
          We have that
          $$frac{partial f}{partial b}=-sum_{n=0}^{100}frac{n b^{n-1}}{(a^n+b^n)^2}$$
          which clearly depends on $a$. If we were to have
          $$f(a,b)=g(a)+k(b)$$
          then we'd have that
          $$frac{partial f}{partial b}=k'(b)$$
          doesn't depend on $a$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            +1 you got there first. Welcome to stackexchange.
            $endgroup$
            – Ethan Bolker
            Dec 20 '18 at 1:55
















          6












          $begingroup$

          This is not possible. Let
          $$f(a,b)=sum_{n=0}^{100}frac{1}{a^n+b^n}.$$
          We have that
          $$frac{partial f}{partial b}=-sum_{n=0}^{100}frac{n b^{n-1}}{(a^n+b^n)^2}$$
          which clearly depends on $a$. If we were to have
          $$f(a,b)=g(a)+k(b)$$
          then we'd have that
          $$frac{partial f}{partial b}=k'(b)$$
          doesn't depend on $a$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            +1 you got there first. Welcome to stackexchange.
            $endgroup$
            – Ethan Bolker
            Dec 20 '18 at 1:55














          6












          6








          6





          $begingroup$

          This is not possible. Let
          $$f(a,b)=sum_{n=0}^{100}frac{1}{a^n+b^n}.$$
          We have that
          $$frac{partial f}{partial b}=-sum_{n=0}^{100}frac{n b^{n-1}}{(a^n+b^n)^2}$$
          which clearly depends on $a$. If we were to have
          $$f(a,b)=g(a)+k(b)$$
          then we'd have that
          $$frac{partial f}{partial b}=k'(b)$$
          doesn't depend on $a$.






          share|cite|improve this answer









          $endgroup$



          This is not possible. Let
          $$f(a,b)=sum_{n=0}^{100}frac{1}{a^n+b^n}.$$
          We have that
          $$frac{partial f}{partial b}=-sum_{n=0}^{100}frac{n b^{n-1}}{(a^n+b^n)^2}$$
          which clearly depends on $a$. If we were to have
          $$f(a,b)=g(a)+k(b)$$
          then we'd have that
          $$frac{partial f}{partial b}=k'(b)$$
          doesn't depend on $a$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 1:54









          user628024user628024

          762




          762












          • $begingroup$
            +1 you got there first. Welcome to stackexchange.
            $endgroup$
            – Ethan Bolker
            Dec 20 '18 at 1:55


















          • $begingroup$
            +1 you got there first. Welcome to stackexchange.
            $endgroup$
            – Ethan Bolker
            Dec 20 '18 at 1:55
















          $begingroup$
          +1 you got there first. Welcome to stackexchange.
          $endgroup$
          – Ethan Bolker
          Dec 20 '18 at 1:55




          $begingroup$
          +1 you got there first. Welcome to stackexchange.
          $endgroup$
          – Ethan Bolker
          Dec 20 '18 at 1:55











          1












          $begingroup$

          You can't. If that sum were of the form $f(a) + g(b)$ then the partial derivative with respect to $a$ would be independent of $b$. That's clearly not the case.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ok but what about $$sum_{n=0}^{100} {-1}^{n-1}*{3an+2bn}$$ How would you split this
            $endgroup$
            – James
            Dec 20 '18 at 2:05












          • $begingroup$
            @James That one is easy. It's just the distributive law. Write out the sum with ellipses or for an upper limit of $2$ instead of $100$ without the $Sigma$.
            $endgroup$
            – Ethan Bolker
            Dec 20 '18 at 13:03
















          1












          $begingroup$

          You can't. If that sum were of the form $f(a) + g(b)$ then the partial derivative with respect to $a$ would be independent of $b$. That's clearly not the case.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ok but what about $$sum_{n=0}^{100} {-1}^{n-1}*{3an+2bn}$$ How would you split this
            $endgroup$
            – James
            Dec 20 '18 at 2:05












          • $begingroup$
            @James That one is easy. It's just the distributive law. Write out the sum with ellipses or for an upper limit of $2$ instead of $100$ without the $Sigma$.
            $endgroup$
            – Ethan Bolker
            Dec 20 '18 at 13:03














          1












          1








          1





          $begingroup$

          You can't. If that sum were of the form $f(a) + g(b)$ then the partial derivative with respect to $a$ would be independent of $b$. That's clearly not the case.






          share|cite|improve this answer









          $endgroup$



          You can't. If that sum were of the form $f(a) + g(b)$ then the partial derivative with respect to $a$ would be independent of $b$. That's clearly not the case.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 1:54









          Ethan BolkerEthan Bolker

          45.1k553120




          45.1k553120












          • $begingroup$
            Ok but what about $$sum_{n=0}^{100} {-1}^{n-1}*{3an+2bn}$$ How would you split this
            $endgroup$
            – James
            Dec 20 '18 at 2:05












          • $begingroup$
            @James That one is easy. It's just the distributive law. Write out the sum with ellipses or for an upper limit of $2$ instead of $100$ without the $Sigma$.
            $endgroup$
            – Ethan Bolker
            Dec 20 '18 at 13:03


















          • $begingroup$
            Ok but what about $$sum_{n=0}^{100} {-1}^{n-1}*{3an+2bn}$$ How would you split this
            $endgroup$
            – James
            Dec 20 '18 at 2:05












          • $begingroup$
            @James That one is easy. It's just the distributive law. Write out the sum with ellipses or for an upper limit of $2$ instead of $100$ without the $Sigma$.
            $endgroup$
            – Ethan Bolker
            Dec 20 '18 at 13:03
















          $begingroup$
          Ok but what about $$sum_{n=0}^{100} {-1}^{n-1}*{3an+2bn}$$ How would you split this
          $endgroup$
          – James
          Dec 20 '18 at 2:05






          $begingroup$
          Ok but what about $$sum_{n=0}^{100} {-1}^{n-1}*{3an+2bn}$$ How would you split this
          $endgroup$
          – James
          Dec 20 '18 at 2:05














          $begingroup$
          @James That one is easy. It's just the distributive law. Write out the sum with ellipses or for an upper limit of $2$ instead of $100$ without the $Sigma$.
          $endgroup$
          – Ethan Bolker
          Dec 20 '18 at 13:03




          $begingroup$
          @James That one is easy. It's just the distributive law. Write out the sum with ellipses or for an upper limit of $2$ instead of $100$ without the $Sigma$.
          $endgroup$
          – Ethan Bolker
          Dec 20 '18 at 13:03


















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