What is the meaning of True in my result? [closed]What is the principal difference betwen two results of a seemingly similar sums?Question about sum and list in a functionProblem evaluating a complicated integralSumConvergence[((-1)^n)/(Sqrt[n] + (-1)^n), n] returns True in Version 10.2?Infinite Sum: $sum_k (ak^2+bk+c)^-1$Numerical result using summation limitLeave Kronecker deltas unevaluatedReturn analytic expressionComputing numerically infinite sum of some double seriesEfficiently define a function as the numerical result of infinite sumsWhy is the result of a Cesaro regularized sum dependant on a mathematically redundant parameter?

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What is the meaning of True in my result? [closed]


What is the principal difference betwen two results of a seemingly similar sums?Question about sum and list in a functionProblem evaluating a complicated integralSumConvergence[((-1)^n)/(Sqrt[n] + (-1)^n), n] returns True in Version 10.2?Infinite Sum: $sum_k (ak^2+bk+c)^-1$Numerical result using summation limitLeave Kronecker deltas unevaluatedReturn analytic expressionComputing numerically infinite sum of some double seriesEfficiently define a function as the numerical result of infinite sumsWhy is the result of a Cesaro regularized sum dependant on a mathematically redundant parameter?













3












$begingroup$


When I do the sum



Sum[(a + (b + π n)^2)^(-1), n, -∞, ∞]


the result reads



$$beginarraycc
{ &
beginarraycc
fraccoth left(sqrta+i bright)+coth left(sqrta-i bright)2 sqrta & arg left(sqrta-i bright)geq -fracpi 2land arg left(sqrta-i bright)leq fracpi 2 \
fraccoth left(sqrta+i bright)2 sqrta+fraccoth left(sqrta-i bright)2 sqrta-frac1sqrta & textTrue \
endarray
\
endarray
$$



What is the meaning of True in the second result?



If I get rid of $pi$, the result is simply:



$$ frac-pi cot left(pi sqrt-a+pi bright)-pi cot left(pi sqrt-a-pi bright)2 sqrt-a$$










share|improve this question











$endgroup$



closed as off-topic by Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB May 23 at 14:15


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB
If this question can be reworded to fit the rules in the help center, please edit the question.




















    3












    $begingroup$


    When I do the sum



    Sum[(a + (b + π n)^2)^(-1), n, -∞, ∞]


    the result reads



    $$beginarraycc
    { &
    beginarraycc
    fraccoth left(sqrta+i bright)+coth left(sqrta-i bright)2 sqrta & arg left(sqrta-i bright)geq -fracpi 2land arg left(sqrta-i bright)leq fracpi 2 \
    fraccoth left(sqrta+i bright)2 sqrta+fraccoth left(sqrta-i bright)2 sqrta-frac1sqrta & textTrue \
    endarray
    \
    endarray
    $$



    What is the meaning of True in the second result?



    If I get rid of $pi$, the result is simply:



    $$ frac-pi cot left(pi sqrt-a+pi bright)-pi cot left(pi sqrt-a-pi bright)2 sqrt-a$$










    share|improve this question











    $endgroup$



    closed as off-topic by Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB May 23 at 14:15


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB
    If this question can be reworded to fit the rules in the help center, please edit the question.


















      3












      3








      3





      $begingroup$


      When I do the sum



      Sum[(a + (b + π n)^2)^(-1), n, -∞, ∞]


      the result reads



      $$beginarraycc
      { &
      beginarraycc
      fraccoth left(sqrta+i bright)+coth left(sqrta-i bright)2 sqrta & arg left(sqrta-i bright)geq -fracpi 2land arg left(sqrta-i bright)leq fracpi 2 \
      fraccoth left(sqrta+i bright)2 sqrta+fraccoth left(sqrta-i bright)2 sqrta-frac1sqrta & textTrue \
      endarray
      \
      endarray
      $$



      What is the meaning of True in the second result?



      If I get rid of $pi$, the result is simply:



      $$ frac-pi cot left(pi sqrt-a+pi bright)-pi cot left(pi sqrt-a-pi bright)2 sqrt-a$$










      share|improve this question











      $endgroup$




      When I do the sum



      Sum[(a + (b + π n)^2)^(-1), n, -∞, ∞]


      the result reads



      $$beginarraycc
      { &
      beginarraycc
      fraccoth left(sqrta+i bright)+coth left(sqrta-i bright)2 sqrta & arg left(sqrta-i bright)geq -fracpi 2land arg left(sqrta-i bright)leq fracpi 2 \
      fraccoth left(sqrta+i bright)2 sqrta+fraccoth left(sqrta-i bright)2 sqrta-frac1sqrta & textTrue \
      endarray
      \
      endarray
      $$



      What is the meaning of True in the second result?



      If I get rid of $pi$, the result is simply:



      $$ frac-pi cot left(pi sqrt-a+pi bright)-pi cot left(pi sqrt-a-pi bright)2 sqrt-a$$







      summation






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited May 23 at 13:13









      m_goldberg

      90.5k873203




      90.5k873203










      asked May 22 at 18:40









      wonderingwondering

      295113




      295113




      closed as off-topic by Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB May 23 at 14:15


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB
      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB May 23 at 14:15


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Bob Hanlon, Carl Lange, m_goldberg, Daniel Lichtblau, MarcoB
      If this question can be reworded to fit the rules in the help center, please edit the question.




















          1 Answer
          1






          active

          oldest

          votes


















          6












          $begingroup$

          See Piecewise. It is like the cases environment in $LaTeX$. In native math, one would use "else" instead of True.



          By the way, version 12 returns



          $$begincases
          fraccoth left(sqrta-i bright)+coth left(sqrta+i
          bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
          fraccoth left(sqrta-i bright)2 sqrta+fraccoth
          left(sqrta+i bright)2 sqrta-frac1sqrta &
          textTrue
          endcases$$



          to be interpreted as



          $$begincases
          fraccoth left(sqrta-i bright)+coth left(sqrta+i
          bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
          fraccoth left(sqrta-i bright)2 sqrta+fraccoth
          left(sqrta+i bright)2 sqrta-frac1sqrta &
          textelse
          endcases$$






          share|improve this answer









          $endgroup$








          • 1




            $begingroup$
            Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
            $endgroup$
            – wondering
            May 22 at 18:44











          • $begingroup$
            Yes. That is correct.
            $endgroup$
            – Henrik Schumacher
            May 22 at 18:45

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          See Piecewise. It is like the cases environment in $LaTeX$. In native math, one would use "else" instead of True.



          By the way, version 12 returns



          $$begincases
          fraccoth left(sqrta-i bright)+coth left(sqrta+i
          bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
          fraccoth left(sqrta-i bright)2 sqrta+fraccoth
          left(sqrta+i bright)2 sqrta-frac1sqrta &
          textTrue
          endcases$$



          to be interpreted as



          $$begincases
          fraccoth left(sqrta-i bright)+coth left(sqrta+i
          bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
          fraccoth left(sqrta-i bright)2 sqrta+fraccoth
          left(sqrta+i bright)2 sqrta-frac1sqrta &
          textelse
          endcases$$






          share|improve this answer









          $endgroup$








          • 1




            $begingroup$
            Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
            $endgroup$
            – wondering
            May 22 at 18:44











          • $begingroup$
            Yes. That is correct.
            $endgroup$
            – Henrik Schumacher
            May 22 at 18:45















          6












          $begingroup$

          See Piecewise. It is like the cases environment in $LaTeX$. In native math, one would use "else" instead of True.



          By the way, version 12 returns



          $$begincases
          fraccoth left(sqrta-i bright)+coth left(sqrta+i
          bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
          fraccoth left(sqrta-i bright)2 sqrta+fraccoth
          left(sqrta+i bright)2 sqrta-frac1sqrta &
          textTrue
          endcases$$



          to be interpreted as



          $$begincases
          fraccoth left(sqrta-i bright)+coth left(sqrta+i
          bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
          fraccoth left(sqrta-i bright)2 sqrta+fraccoth
          left(sqrta+i bright)2 sqrta-frac1sqrta &
          textelse
          endcases$$






          share|improve this answer









          $endgroup$








          • 1




            $begingroup$
            Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
            $endgroup$
            – wondering
            May 22 at 18:44











          • $begingroup$
            Yes. That is correct.
            $endgroup$
            – Henrik Schumacher
            May 22 at 18:45













          6












          6








          6





          $begingroup$

          See Piecewise. It is like the cases environment in $LaTeX$. In native math, one would use "else" instead of True.



          By the way, version 12 returns



          $$begincases
          fraccoth left(sqrta-i bright)+coth left(sqrta+i
          bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
          fraccoth left(sqrta-i bright)2 sqrta+fraccoth
          left(sqrta+i bright)2 sqrta-frac1sqrta &
          textTrue
          endcases$$



          to be interpreted as



          $$begincases
          fraccoth left(sqrta-i bright)+coth left(sqrta+i
          bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
          fraccoth left(sqrta-i bright)2 sqrta+fraccoth
          left(sqrta+i bright)2 sqrta-frac1sqrta &
          textelse
          endcases$$






          share|improve this answer









          $endgroup$



          See Piecewise. It is like the cases environment in $LaTeX$. In native math, one would use "else" instead of True.



          By the way, version 12 returns



          $$begincases
          fraccoth left(sqrta-i bright)+coth left(sqrta+i
          bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
          fraccoth left(sqrta-i bright)2 sqrta+fraccoth
          left(sqrta+i bright)2 sqrta-frac1sqrta &
          textTrue
          endcases$$



          to be interpreted as



          $$begincases
          fraccoth left(sqrta-i bright)+coth left(sqrta+i
          bright)2 sqrta & arg left(b+i sqrtaright)geq 0 \
          fraccoth left(sqrta-i bright)2 sqrta+fraccoth
          left(sqrta+i bright)2 sqrta-frac1sqrta &
          textelse
          endcases$$







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered May 22 at 18:43









          Henrik SchumacherHenrik Schumacher

          63.8k590178




          63.8k590178







          • 1




            $begingroup$
            Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
            $endgroup$
            – wondering
            May 22 at 18:44











          • $begingroup$
            Yes. That is correct.
            $endgroup$
            – Henrik Schumacher
            May 22 at 18:45












          • 1




            $begingroup$
            Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
            $endgroup$
            – wondering
            May 22 at 18:44











          • $begingroup$
            Yes. That is correct.
            $endgroup$
            – Henrik Schumacher
            May 22 at 18:45







          1




          1




          $begingroup$
          Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
          $endgroup$
          – wondering
          May 22 at 18:44





          $begingroup$
          Ok, so it really simply means "otherwise" or "else" or "in all other cases"?
          $endgroup$
          – wondering
          May 22 at 18:44













          $begingroup$
          Yes. That is correct.
          $endgroup$
          – Henrik Schumacher
          May 22 at 18:45




          $begingroup$
          Yes. That is correct.
          $endgroup$
          – Henrik Schumacher
          May 22 at 18:45



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