How to find the sum with Mathematica?












4












$begingroup$


That hard problem was invented by V. P. Beshkarev (Russia) in 1971:



Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // FullSimplify


The result should be 45, but the command is running on my comp without any output for hours. I know its tricky calculation by hand which cannot be mimicked with Mathematica.






simplifying-expressions symbolic







share|improve this question











$endgroup$












  • $begingroup$
    N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]] gives 45.
    $endgroup$
    – Carl Lange
    5 hours ago










  • $begingroup$
    @Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
    $endgroup$
    – user64494
    5 hours ago










  • $begingroup$
    Sorry, it's not clear to me what you're expecting as a result except "The result should be 45". Why do you expect FullSimplify to do anything in this case?
    $endgroup$
    – Carl Lange
    5 hours ago












  • $begingroup$
    @Carl Lange: A simpler problem of such type is Sum[j,{j,1,100}], where the result should be 5050, not 5050.0 . Hope I am clear now.
    $endgroup$
    – user64494
    5 hours ago






  • 1




    $begingroup$
    @CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
    $endgroup$
    – Henrik Schumacher
    4 hours ago
















4












$begingroup$


That hard problem was invented by V. P. Beshkarev (Russia) in 1971:



Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // FullSimplify


The result should be 45, but the command is running on my comp without any output for hours. I know its tricky calculation by hand which cannot be mimicked with Mathematica.






simplifying-expressions symbolic







share|improve this question











$endgroup$












  • $begingroup$
    N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]] gives 45.
    $endgroup$
    – Carl Lange
    5 hours ago










  • $begingroup$
    @Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
    $endgroup$
    – user64494
    5 hours ago










  • $begingroup$
    Sorry, it's not clear to me what you're expecting as a result except "The result should be 45". Why do you expect FullSimplify to do anything in this case?
    $endgroup$
    – Carl Lange
    5 hours ago












  • $begingroup$
    @Carl Lange: A simpler problem of such type is Sum[j,{j,1,100}], where the result should be 5050, not 5050.0 . Hope I am clear now.
    $endgroup$
    – user64494
    5 hours ago






  • 1




    $begingroup$
    @CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
    $endgroup$
    – Henrik Schumacher
    4 hours ago














4












4








4





$begingroup$


That hard problem was invented by V. P. Beshkarev (Russia) in 1971:



Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // FullSimplify


The result should be 45, but the command is running on my comp without any output for hours. I know its tricky calculation by hand which cannot be mimicked with Mathematica.






simplifying-expressions symbolic







share|improve this question











$endgroup$




That hard problem was invented by V. P. Beshkarev (Russia) in 1971:



Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // FullSimplify


The result should be 45, but the command is running on my comp without any output for hours. I know its tricky calculation by hand which cannot be mimicked with Mathematica.






simplifying-expressions symbolic




simplifying-expressions symbolic






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 5 hours ago







user64494

















asked 5 hours ago









user64494user64494

3,37011021




3,37011021












  • $begingroup$
    N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]] gives 45.
    $endgroup$
    – Carl Lange
    5 hours ago










  • $begingroup$
    @Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
    $endgroup$
    – user64494
    5 hours ago










  • $begingroup$
    Sorry, it's not clear to me what you're expecting as a result except "The result should be 45". Why do you expect FullSimplify to do anything in this case?
    $endgroup$
    – Carl Lange
    5 hours ago












  • $begingroup$
    @Carl Lange: A simpler problem of such type is Sum[j,{j,1,100}], where the result should be 5050, not 5050.0 . Hope I am clear now.
    $endgroup$
    – user64494
    5 hours ago






  • 1




    $begingroup$
    @CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
    $endgroup$
    – Henrik Schumacher
    4 hours ago


















  • $begingroup$
    N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]] gives 45.
    $endgroup$
    – Carl Lange
    5 hours ago










  • $begingroup$
    @Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
    $endgroup$
    – user64494
    5 hours ago










  • $begingroup$
    Sorry, it's not clear to me what you're expecting as a result except "The result should be 45". Why do you expect FullSimplify to do anything in this case?
    $endgroup$
    – Carl Lange
    5 hours ago












  • $begingroup$
    @Carl Lange: A simpler problem of such type is Sum[j,{j,1,100}], where the result should be 5050, not 5050.0 . Hope I am clear now.
    $endgroup$
    – user64494
    5 hours ago






  • 1




    $begingroup$
    @CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
    $endgroup$
    – Henrik Schumacher
    4 hours ago
















$begingroup$
N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]] gives 45.
$endgroup$
– Carl Lange
5 hours ago




$begingroup$
N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]] gives 45.
$endgroup$
– Carl Lange
5 hours ago












$begingroup$
@Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
$endgroup$
– user64494
5 hours ago




$begingroup$
@Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
$endgroup$
– user64494
5 hours ago












$begingroup$
Sorry, it's not clear to me what you're expecting as a result except "The result should be 45". Why do you expect FullSimplify to do anything in this case?
$endgroup$
– Carl Lange
5 hours ago






$begingroup$
Sorry, it's not clear to me what you're expecting as a result except "The result should be 45". Why do you expect FullSimplify to do anything in this case?
$endgroup$
– Carl Lange
5 hours ago














$begingroup$
@Carl Lange: A simpler problem of such type is Sum[j,{j,1,100}], where the result should be 5050, not 5050.0 . Hope I am clear now.
$endgroup$
– user64494
5 hours ago




$begingroup$
@Carl Lange: A simpler problem of such type is Sum[j,{j,1,100}], where the result should be 5050, not 5050.0 . Hope I am clear now.
$endgroup$
– user64494
5 hours ago




1




1




$begingroup$
@CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
$endgroup$
– Henrik Schumacher
4 hours ago




$begingroup$
@CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
$endgroup$
– Henrik Schumacher
4 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

Sometimes the easiest approach is to just divide it up into steps and see which transformations can be done reasonably quickly. First, I define the expression:



expr = Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}];


Verify its result numerically:



N[expr]



45.




This is likely, but not necessarily, exact. Thus, the strategy will be trying to prove that some transformation of expr - 45 is 0 exactly. Since expr is primarily trigonometric, there's a few functions that come to mind immediately. TrigExpand does not evaluate quickly, but TrigToExp shows a fairly self-similar form of a group of fractions. I find fractions usually become easier to work with after Apart, and it turns out that transformation is also reasonably quick. However, after Apart the numbers do not precisely add up to anything specific, so the 45 would seem to be a residual effect of several independent parts of this expression.



At this point I tried to see if Simplify could sort it out:



Simplify[Apart[TrigToExp[expr]] - 45]



0




Which is an exact result, though derived through somewhat convoluted means, which shows that expr == 45 exactly, so long as no errors occurred during TrigToExp and Apart, which are both supposed to be complex safe.






share|improve this answer









$endgroup$













  • $begingroup$
    Sorry, but the code suggested by you is running without any output on my comp during several minutes. The same issue with Apart[TrigToExp[expr]] too. I will try to execute your code in cloud.
    $endgroup$
    – user64494
    4 hours ago










  • $begingroup$
    It takes about 5 seconds on an i7 4770K on Mathematica 10.1. It takes about 8.4 on a fresh start of 11.2 for me as well. Not a super easy computation, so I wouldn't be surprised if it takes a little bit, but I'd expect it to take less than 2 minutes on most machines.
    $endgroup$
    – eyorble
    4 hours ago










  • $begingroup$
    Reproduced in Mathematica online in 6 s.. Simply and strongly.
    $endgroup$
    – user64494
    4 hours ago






  • 2




    $begingroup$
    Hm. I am very curious. I am on version 11.3 on a 4980HQ (so the single-thread performance should be very similar to the 4770K) and this computation takes 47 seconds (returning the correct result). That's significant slow-down...
    $endgroup$
    – Henrik Schumacher
    4 hours ago






  • 1




    $begingroup$
    @HenrikSchumacher on the cloud it takes like 90s
    $endgroup$
    – b3m2a1
    1 hour ago



















4












$begingroup$

Use RootReduce



Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // RootReduce



(* 45 *)





share|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Sometimes the easiest approach is to just divide it up into steps and see which transformations can be done reasonably quickly. First, I define the expression:



    expr = Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}];


    Verify its result numerically:



    N[expr]



    45.




    This is likely, but not necessarily, exact. Thus, the strategy will be trying to prove that some transformation of expr - 45 is 0 exactly. Since expr is primarily trigonometric, there's a few functions that come to mind immediately. TrigExpand does not evaluate quickly, but TrigToExp shows a fairly self-similar form of a group of fractions. I find fractions usually become easier to work with after Apart, and it turns out that transformation is also reasonably quick. However, after Apart the numbers do not precisely add up to anything specific, so the 45 would seem to be a residual effect of several independent parts of this expression.



    At this point I tried to see if Simplify could sort it out:



    Simplify[Apart[TrigToExp[expr]] - 45]



    0




    Which is an exact result, though derived through somewhat convoluted means, which shows that expr == 45 exactly, so long as no errors occurred during TrigToExp and Apart, which are both supposed to be complex safe.






    share|improve this answer









    $endgroup$













    • $begingroup$
      Sorry, but the code suggested by you is running without any output on my comp during several minutes. The same issue with Apart[TrigToExp[expr]] too. I will try to execute your code in cloud.
      $endgroup$
      – user64494
      4 hours ago










    • $begingroup$
      It takes about 5 seconds on an i7 4770K on Mathematica 10.1. It takes about 8.4 on a fresh start of 11.2 for me as well. Not a super easy computation, so I wouldn't be surprised if it takes a little bit, but I'd expect it to take less than 2 minutes on most machines.
      $endgroup$
      – eyorble
      4 hours ago










    • $begingroup$
      Reproduced in Mathematica online in 6 s.. Simply and strongly.
      $endgroup$
      – user64494
      4 hours ago






    • 2




      $begingroup$
      Hm. I am very curious. I am on version 11.3 on a 4980HQ (so the single-thread performance should be very similar to the 4770K) and this computation takes 47 seconds (returning the correct result). That's significant slow-down...
      $endgroup$
      – Henrik Schumacher
      4 hours ago






    • 1




      $begingroup$
      @HenrikSchumacher on the cloud it takes like 90s
      $endgroup$
      – b3m2a1
      1 hour ago
















    4












    $begingroup$

    Sometimes the easiest approach is to just divide it up into steps and see which transformations can be done reasonably quickly. First, I define the expression:



    expr = Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}];


    Verify its result numerically:



    N[expr]



    45.




    This is likely, but not necessarily, exact. Thus, the strategy will be trying to prove that some transformation of expr - 45 is 0 exactly. Since expr is primarily trigonometric, there's a few functions that come to mind immediately. TrigExpand does not evaluate quickly, but TrigToExp shows a fairly self-similar form of a group of fractions. I find fractions usually become easier to work with after Apart, and it turns out that transformation is also reasonably quick. However, after Apart the numbers do not precisely add up to anything specific, so the 45 would seem to be a residual effect of several independent parts of this expression.



    At this point I tried to see if Simplify could sort it out:



    Simplify[Apart[TrigToExp[expr]] - 45]



    0




    Which is an exact result, though derived through somewhat convoluted means, which shows that expr == 45 exactly, so long as no errors occurred during TrigToExp and Apart, which are both supposed to be complex safe.






    share|improve this answer









    $endgroup$













    • $begingroup$
      Sorry, but the code suggested by you is running without any output on my comp during several minutes. The same issue with Apart[TrigToExp[expr]] too. I will try to execute your code in cloud.
      $endgroup$
      – user64494
      4 hours ago










    • $begingroup$
      It takes about 5 seconds on an i7 4770K on Mathematica 10.1. It takes about 8.4 on a fresh start of 11.2 for me as well. Not a super easy computation, so I wouldn't be surprised if it takes a little bit, but I'd expect it to take less than 2 minutes on most machines.
      $endgroup$
      – eyorble
      4 hours ago










    • $begingroup$
      Reproduced in Mathematica online in 6 s.. Simply and strongly.
      $endgroup$
      – user64494
      4 hours ago






    • 2




      $begingroup$
      Hm. I am very curious. I am on version 11.3 on a 4980HQ (so the single-thread performance should be very similar to the 4770K) and this computation takes 47 seconds (returning the correct result). That's significant slow-down...
      $endgroup$
      – Henrik Schumacher
      4 hours ago






    • 1




      $begingroup$
      @HenrikSchumacher on the cloud it takes like 90s
      $endgroup$
      – b3m2a1
      1 hour ago














    4












    4








    4





    $begingroup$

    Sometimes the easiest approach is to just divide it up into steps and see which transformations can be done reasonably quickly. First, I define the expression:



    expr = Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}];


    Verify its result numerically:



    N[expr]



    45.




    This is likely, but not necessarily, exact. Thus, the strategy will be trying to prove that some transformation of expr - 45 is 0 exactly. Since expr is primarily trigonometric, there's a few functions that come to mind immediately. TrigExpand does not evaluate quickly, but TrigToExp shows a fairly self-similar form of a group of fractions. I find fractions usually become easier to work with after Apart, and it turns out that transformation is also reasonably quick. However, after Apart the numbers do not precisely add up to anything specific, so the 45 would seem to be a residual effect of several independent parts of this expression.



    At this point I tried to see if Simplify could sort it out:



    Simplify[Apart[TrigToExp[expr]] - 45]



    0




    Which is an exact result, though derived through somewhat convoluted means, which shows that expr == 45 exactly, so long as no errors occurred during TrigToExp and Apart, which are both supposed to be complex safe.






    share|improve this answer









    $endgroup$



    Sometimes the easiest approach is to just divide it up into steps and see which transformations can be done reasonably quickly. First, I define the expression:



    expr = Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}];


    Verify its result numerically:



    N[expr]



    45.




    This is likely, but not necessarily, exact. Thus, the strategy will be trying to prove that some transformation of expr - 45 is 0 exactly. Since expr is primarily trigonometric, there's a few functions that come to mind immediately. TrigExpand does not evaluate quickly, but TrigToExp shows a fairly self-similar form of a group of fractions. I find fractions usually become easier to work with after Apart, and it turns out that transformation is also reasonably quick. However, after Apart the numbers do not precisely add up to anything specific, so the 45 would seem to be a residual effect of several independent parts of this expression.



    At this point I tried to see if Simplify could sort it out:



    Simplify[Apart[TrigToExp[expr]] - 45]



    0




    Which is an exact result, though derived through somewhat convoluted means, which shows that expr == 45 exactly, so long as no errors occurred during TrigToExp and Apart, which are both supposed to be complex safe.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 4 hours ago









    eyorbleeyorble

    5,2381826




    5,2381826












    • $begingroup$
      Sorry, but the code suggested by you is running without any output on my comp during several minutes. The same issue with Apart[TrigToExp[expr]] too. I will try to execute your code in cloud.
      $endgroup$
      – user64494
      4 hours ago










    • $begingroup$
      It takes about 5 seconds on an i7 4770K on Mathematica 10.1. It takes about 8.4 on a fresh start of 11.2 for me as well. Not a super easy computation, so I wouldn't be surprised if it takes a little bit, but I'd expect it to take less than 2 minutes on most machines.
      $endgroup$
      – eyorble
      4 hours ago










    • $begingroup$
      Reproduced in Mathematica online in 6 s.. Simply and strongly.
      $endgroup$
      – user64494
      4 hours ago






    • 2




      $begingroup$
      Hm. I am very curious. I am on version 11.3 on a 4980HQ (so the single-thread performance should be very similar to the 4770K) and this computation takes 47 seconds (returning the correct result). That's significant slow-down...
      $endgroup$
      – Henrik Schumacher
      4 hours ago






    • 1




      $begingroup$
      @HenrikSchumacher on the cloud it takes like 90s
      $endgroup$
      – b3m2a1
      1 hour ago


















    • $begingroup$
      Sorry, but the code suggested by you is running without any output on my comp during several minutes. The same issue with Apart[TrigToExp[expr]] too. I will try to execute your code in cloud.
      $endgroup$
      – user64494
      4 hours ago










    • $begingroup$
      It takes about 5 seconds on an i7 4770K on Mathematica 10.1. It takes about 8.4 on a fresh start of 11.2 for me as well. Not a super easy computation, so I wouldn't be surprised if it takes a little bit, but I'd expect it to take less than 2 minutes on most machines.
      $endgroup$
      – eyorble
      4 hours ago










    • $begingroup$
      Reproduced in Mathematica online in 6 s.. Simply and strongly.
      $endgroup$
      – user64494
      4 hours ago






    • 2




      $begingroup$
      Hm. I am very curious. I am on version 11.3 on a 4980HQ (so the single-thread performance should be very similar to the 4770K) and this computation takes 47 seconds (returning the correct result). That's significant slow-down...
      $endgroup$
      – Henrik Schumacher
      4 hours ago






    • 1




      $begingroup$
      @HenrikSchumacher on the cloud it takes like 90s
      $endgroup$
      – b3m2a1
      1 hour ago
















    $begingroup$
    Sorry, but the code suggested by you is running without any output on my comp during several minutes. The same issue with Apart[TrigToExp[expr]] too. I will try to execute your code in cloud.
    $endgroup$
    – user64494
    4 hours ago




    $begingroup$
    Sorry, but the code suggested by you is running without any output on my comp during several minutes. The same issue with Apart[TrigToExp[expr]] too. I will try to execute your code in cloud.
    $endgroup$
    – user64494
    4 hours ago












    $begingroup$
    It takes about 5 seconds on an i7 4770K on Mathematica 10.1. It takes about 8.4 on a fresh start of 11.2 for me as well. Not a super easy computation, so I wouldn't be surprised if it takes a little bit, but I'd expect it to take less than 2 minutes on most machines.
    $endgroup$
    – eyorble
    4 hours ago




    $begingroup$
    It takes about 5 seconds on an i7 4770K on Mathematica 10.1. It takes about 8.4 on a fresh start of 11.2 for me as well. Not a super easy computation, so I wouldn't be surprised if it takes a little bit, but I'd expect it to take less than 2 minutes on most machines.
    $endgroup$
    – eyorble
    4 hours ago












    $begingroup$
    Reproduced in Mathematica online in 6 s.. Simply and strongly.
    $endgroup$
    – user64494
    4 hours ago




    $begingroup$
    Reproduced in Mathematica online in 6 s.. Simply and strongly.
    $endgroup$
    – user64494
    4 hours ago




    2




    2




    $begingroup$
    Hm. I am very curious. I am on version 11.3 on a 4980HQ (so the single-thread performance should be very similar to the 4770K) and this computation takes 47 seconds (returning the correct result). That's significant slow-down...
    $endgroup$
    – Henrik Schumacher
    4 hours ago




    $begingroup$
    Hm. I am very curious. I am on version 11.3 on a 4980HQ (so the single-thread performance should be very similar to the 4770K) and this computation takes 47 seconds (returning the correct result). That's significant slow-down...
    $endgroup$
    – Henrik Schumacher
    4 hours ago




    1




    1




    $begingroup$
    @HenrikSchumacher on the cloud it takes like 90s
    $endgroup$
    – b3m2a1
    1 hour ago




    $begingroup$
    @HenrikSchumacher on the cloud it takes like 90s
    $endgroup$
    – b3m2a1
    1 hour ago











    4












    $begingroup$

    Use RootReduce



    Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // RootReduce
    

    
(* 45 *)





    share|improve this answer









    $endgroup$


















      4












      $begingroup$

      Use RootReduce



      Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // RootReduce
      

      
(* 45 *)





      share|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Use RootReduce



        Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // RootReduce
        

        
(* 45 *)





        share|improve this answer









        $endgroup$



        Use RootReduce



        Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // RootReduce
        

        
(* 45 *)






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 1 hour ago









        Bob HanlonBob Hanlon

        60k33596




        60k33596






























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