How to find the sum with Mathematica?
$begingroup$
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
asked yesterday
user64494user64494
3,37811021
$endgroup$
add a comment |
$begingroup$
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
asked yesterday
user64494user64494
3,37811021
$endgroup$
$begingroup$
N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]]gives 45.
$endgroup$
– Carl Lange
yesterday
$begingroup$
@Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
$endgroup$
– user64494
yesterday
$begingroup$
Sorry, it's not clear to me what you're expecting as a result except "The result should be 45". Why do you expectFullSimplifyto do anything in this case?
$endgroup$
– Carl Lange
yesterday
$begingroup$
@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
yesterday
1
$begingroup$
@CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
$endgroup$
– Henrik Schumacher
yesterday
add a comment |
$begingroup$
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
asked yesterday
user64494user64494
3,37811021
$endgroup$
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
asked yesterday
user64494user64494
3,37811021
asked yesterday
user64494user64494
3,37811021
edited yesterday
user64494
asked yesterday
user64494user64494
3,37811021
asked yesterday
user64494user64494
3,37811021
asked yesterday
user64494user64494
3,37811021
3,37811021
$begingroup$
N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]]gives 45.
$endgroup$
– Carl Lange
yesterday
$begingroup$
@Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
$endgroup$
– user64494
yesterday
$begingroup$
Sorry, it's not clear to me what you're expecting as a result except "The result should be 45". Why do you expectFullSimplifyto do anything in this case?
$endgroup$
– Carl Lange
yesterday
$begingroup$
@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
yesterday
1
$begingroup$
@CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
$endgroup$
– Henrik Schumacher
yesterday
add a comment |
$begingroup$
N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]]gives 45.
$endgroup$
– Carl Lange
yesterday
$begingroup$
@Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
$endgroup$
– user64494
yesterday
$begingroup$
Sorry, it's not clear to me what you're expecting as a result except "The result should be 45". Why do you expectFullSimplifyto do anything in this case?
$endgroup$
– Carl Lange
yesterday
$begingroup$
@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
yesterday
1
$begingroup$
@CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
$endgroup$
– Henrik Schumacher
yesterday
$begingroup$
N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]] gives 45.$endgroup$
– Carl Lange
yesterday
$begingroup$
N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]] gives 45.$endgroup$
– Carl Lange
yesterday
$begingroup$
@Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
$endgroup$
– user64494
yesterday
$begingroup$
@Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
$endgroup$
– user64494
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
@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
yesterday
$begingroup$
@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
yesterday
1
1
$begingroup$
@CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
$endgroup$
– Henrik Schumacher
yesterday
$begingroup$
@CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
$endgroup$
– Henrik Schumacher
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered yesterday
eyorbleeyorble
5,2731826
$endgroup$
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
Reproduced in Mathematica online in 6 s.. Simply and strongly.
$endgroup$
– user64494
yesterday
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
yesterday
1
$begingroup$
@HenrikSchumacher on the cloud it takes like 90s
$endgroup$
– b3m2a1
yesterday
|
show 2 more comments
$begingroup$
Use RootReduce
Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // RootReduce
(* 45 *)
answered yesterday
Bob HanlonBob Hanlon
60.2k33597
$endgroup$
1
$begingroup$
+1. It is unclear to me how it works.
$endgroup$
– user64494
21 hours ago
1
$begingroup$
@user64494 - Mostly only people at Wolfram know "how it works". It is one of the functions likeSimplify,FullSimplify,ComplexExpand,TrigReduce, etc. that are used to find alternate forms of expressions. Experimentation is often required to find which one or which combination is best for a given expression.
$endgroup$
– Bob Hanlon
20 hours ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered yesterday
eyorbleeyorble
5,2731826
$endgroup$
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
Reproduced in Mathematica online in 6 s.. Simply and strongly.
$endgroup$
– user64494
yesterday
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
yesterday
1
$begingroup$
@HenrikSchumacher on the cloud it takes like 90s
$endgroup$
– b3m2a1
yesterday
|
show 2 more comments
$begingroup$
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.
answered yesterday
eyorbleeyorble
5,2731826
$endgroup$
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
Reproduced in Mathematica online in 6 s.. Simply and strongly.
$endgroup$
– user64494
yesterday
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
yesterday
1
$begingroup$
@HenrikSchumacher on the cloud it takes like 90s
$endgroup$
– b3m2a1
yesterday
|
show 2 more comments
$begingroup$
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.
answered yesterday
eyorbleeyorble
5,2731826
$endgroup$
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.
answered yesterday
eyorbleeyorble
5,2731826
answered yesterday
eyorbleeyorble
5,2731826
answered yesterday
eyorbleeyorble
5,2731826
answered yesterday
eyorbleeyorble
5,2731826
5,2731826
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
Reproduced in Mathematica online in 6 s.. Simply and strongly.
$endgroup$
– user64494
yesterday
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
yesterday
1
$begingroup$
@HenrikSchumacher on the cloud it takes like 90s
$endgroup$
– b3m2a1
yesterday
|
show 2 more comments
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
Reproduced in Mathematica online in 6 s.. Simply and strongly.
$endgroup$
– user64494
yesterday
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
yesterday
1
$begingroup$
@HenrikSchumacher on the cloud it takes like 90s
$endgroup$
– b3m2a1
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
$begingroup$
Reproduced in Mathematica online in 6 s.. Simply and strongly.
$endgroup$
– user64494
yesterday
$begingroup$
Reproduced in Mathematica online in 6 s.. Simply and strongly.
$endgroup$
– user64494
yesterday
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
yesterday
$begingroup$
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
yesterday
1
1
$begingroup$
@HenrikSchumacher on the cloud it takes like 90s
$endgroup$
– b3m2a1
yesterday
$begingroup$
@HenrikSchumacher on the cloud it takes like 90s
$endgroup$
– b3m2a1
yesterday
|
show 2 more comments
$begingroup$
Use RootReduce
Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // RootReduce
(* 45 *)
answered yesterday
Bob HanlonBob Hanlon
60.2k33597
$endgroup$
1
$begingroup$
+1. It is unclear to me how it works.
$endgroup$
– user64494
21 hours ago
1
$begingroup$
@user64494 - Mostly only people at Wolfram know "how it works". It is one of the functions likeSimplify,FullSimplify,ComplexExpand,TrigReduce, etc. that are used to find alternate forms of expressions. Experimentation is often required to find which one or which combination is best for a given expression.
$endgroup$
– Bob Hanlon
20 hours ago
add a comment |
$begingroup$
Use RootReduce
Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // RootReduce
(* 45 *)
answered yesterday
Bob HanlonBob Hanlon
60.2k33597
$endgroup$
1
$begingroup$
+1. It is unclear to me how it works.
$endgroup$
– user64494
21 hours ago
1
$begingroup$
@user64494 - Mostly only people at Wolfram know "how it works". It is one of the functions likeSimplify,FullSimplify,ComplexExpand,TrigReduce, etc. that are used to find alternate forms of expressions. Experimentation is often required to find which one or which combination is best for a given expression.
$endgroup$
– Bob Hanlon
20 hours ago
add a comment |
$begingroup$
Use RootReduce
Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // RootReduce
(* 45 *)
answered yesterday
Bob HanlonBob Hanlon
60.2k33597
$endgroup$
Use RootReduce
Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // RootReduce
(* 45 *)
answered yesterday
Bob HanlonBob Hanlon
60.2k33597
answered yesterday
Bob HanlonBob Hanlon
60.2k33597
answered yesterday
Bob HanlonBob Hanlon
60.2k33597
answered yesterday
Bob HanlonBob Hanlon
60.2k33597
60.2k33597
1
$begingroup$
+1. It is unclear to me how it works.
$endgroup$
– user64494
21 hours ago
1
$begingroup$
@user64494 - Mostly only people at Wolfram know "how it works". It is one of the functions likeSimplify,FullSimplify,ComplexExpand,TrigReduce, etc. that are used to find alternate forms of expressions. Experimentation is often required to find which one or which combination is best for a given expression.
$endgroup$
– Bob Hanlon
20 hours ago
add a comment |
1
$begingroup$
+1. It is unclear to me how it works.
$endgroup$
– user64494
21 hours ago
1
$begingroup$
@user64494 - Mostly only people at Wolfram know "how it works". It is one of the functions likeSimplify,FullSimplify,ComplexExpand,TrigReduce, etc. that are used to find alternate forms of expressions. Experimentation is often required to find which one or which combination is best for a given expression.
$endgroup$
– Bob Hanlon
20 hours ago
1
1
$begingroup$
+1. It is unclear to me how it works.
$endgroup$
– user64494
21 hours ago
$begingroup$
+1. It is unclear to me how it works.
$endgroup$
– user64494
21 hours ago
1
1
$begingroup$
@user64494 - Mostly only people at Wolfram know "how it works". It is one of the functions like
Simplify, FullSimplify, ComplexExpand, TrigReduce, etc. that are used to find alternate forms of expressions. Experimentation is often required to find which one or which combination is best for a given expression.$endgroup$
– Bob Hanlon
20 hours ago
$begingroup$
@user64494 - Mostly only people at Wolfram know "how it works". It is one of the functions like
Simplify, FullSimplify, ComplexExpand, TrigReduce, etc. that are used to find alternate forms of expressions. Experimentation is often required to find which one or which combination is best for a given expression.$endgroup$
– Bob Hanlon
20 hours ago
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$begingroup$
N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]]gives 45.$endgroup$
– Carl Lange
yesterday
$begingroup$
@Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
$endgroup$
– user64494
yesterday
$begingroup$
Sorry, it's not clear to me what you're expecting as a result except "The result should be 45". Why do you expect
FullSimplifyto do anything in this case?$endgroup$
– Carl Lange
yesterday
$begingroup$
@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
yesterday
1
$begingroup$
@CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
$endgroup$
– Henrik Schumacher
yesterday