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 Feb 7 at 20:01
user64494user64494
3,44311021
$endgroup$
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 Feb 7 at 20:01
user64494user64494
3,44311021
$endgroup$
$begingroup$
N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]]gives 45.
$endgroup$
– Carl Lange
Feb 7 at 20:02
$begingroup$
@Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
$endgroup$
– user64494
Feb 7 at 20:04
$begingroup$
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
Feb 7 at 20:11
$begingroup$
@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
Feb 7 at 20:14
1
$begingroup$
@CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
$endgroup$
– Henrik Schumacher
Feb 7 at 20:27
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 Feb 7 at 20:01
user64494user64494
3,44311021
$endgroup$
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 Feb 7 at 20:01
user64494user64494
3,44311021
asked Feb 7 at 20:01
user64494user64494
3,44311021
edited Feb 7 at 20:05
user64494
asked Feb 7 at 20:01
user64494user64494
3,44311021
asked Feb 7 at 20:01
user64494user64494
3,44311021
asked Feb 7 at 20:01
user64494user64494
3,44311021
3,44311021
$begingroup$
N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]]gives 45.
$endgroup$
– Carl Lange
Feb 7 at 20:02
$begingroup$
@Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
$endgroup$
– user64494
Feb 7 at 20:04
$begingroup$
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
Feb 7 at 20:11
$begingroup$
@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
Feb 7 at 20:14
1
$begingroup$
@CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
$endgroup$
– Henrik Schumacher
Feb 7 at 20:27
add a comment |
$begingroup$
N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]]gives 45.
$endgroup$
– Carl Lange
Feb 7 at 20:02
$begingroup$
@Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
$endgroup$
– user64494
Feb 7 at 20:04
$begingroup$
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
Feb 7 at 20:11
$begingroup$
@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
Feb 7 at 20:14
1
$begingroup$
@CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
$endgroup$
– Henrik Schumacher
Feb 7 at 20:27
$begingroup$
N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]] gives 45.$endgroup$
– Carl Lange
Feb 7 at 20:02
$begingroup$
N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]] gives 45.$endgroup$
– Carl Lange
Feb 7 at 20:02
$begingroup$
@Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
$endgroup$
– user64494
Feb 7 at 20:04
$begingroup$
@Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
$endgroup$
– user64494
Feb 7 at 20:04
$begingroup$
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
Feb 7 at 20:11
$begingroup$
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
Feb 7 at 20:11
$begingroup$
@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
Feb 7 at 20:14
$begingroup$
@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
Feb 7 at 20:14
1
1
$begingroup$
@CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
$endgroup$
– Henrik Schumacher
Feb 7 at 20:27
$begingroup$
@CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
$endgroup$
– Henrik Schumacher
Feb 7 at 20:27
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 Feb 7 at 20:27
eyorbleeyorble
5,64311028
$endgroup$
$begingroup$
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
Feb 7 at 20:37
$begingroup$
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
Feb 7 at 20:40
$begingroup$
Reproduced in Mathematica online in 6 s.. Simply and strongly.
$endgroup$
– user64494
Feb 7 at 20:47
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
Feb 7 at 21:21
1
$begingroup$
@HenrikSchumacher on the cloud it takes like 90s
$endgroup$
– b3m2a1
Feb 8 at 0:01
|
show 2 more comments
$begingroup$
Use RootReduce
Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // RootReduce
(* 45 *)
answered Feb 8 at 0:16
Bob HanlonBob Hanlon
60.7k33597
$endgroup$
1
$begingroup$
+1. It is unclear to me how it works.
$endgroup$
– user64494
Feb 8 at 3:50
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
Feb 8 at 4:16
add a comment |
Your Answer
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2 Answers
2
active
oldest
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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 Feb 7 at 20:27
eyorbleeyorble
5,64311028
$endgroup$
$begingroup$
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
Feb 7 at 20:37
$begingroup$
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
Feb 7 at 20:40
$begingroup$
Reproduced in Mathematica online in 6 s.. Simply and strongly.
$endgroup$
– user64494
Feb 7 at 20:47
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
Feb 7 at 21:21
1
$begingroup$
@HenrikSchumacher on the cloud it takes like 90s
$endgroup$
– b3m2a1
Feb 8 at 0:01
|
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 Feb 7 at 20:27
eyorbleeyorble
5,64311028
$endgroup$
$begingroup$
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
Feb 7 at 20:37
$begingroup$
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
Feb 7 at 20:40
$begingroup$
Reproduced in Mathematica online in 6 s.. Simply and strongly.
$endgroup$
– user64494
Feb 7 at 20:47
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
Feb 7 at 21:21
1
$begingroup$
@HenrikSchumacher on the cloud it takes like 90s
$endgroup$
– b3m2a1
Feb 8 at 0:01
|
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 Feb 7 at 20:27
eyorbleeyorble
5,64311028
$endgroup$
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 Feb 7 at 20:27
eyorbleeyorble
5,64311028
answered Feb 7 at 20:27
eyorbleeyorble
5,64311028
answered Feb 7 at 20:27
eyorbleeyorble
5,64311028
answered Feb 7 at 20:27
eyorbleeyorble
5,64311028
5,64311028
$begingroup$
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
Feb 7 at 20:37
$begingroup$
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
Feb 7 at 20:40
$begingroup$
Reproduced in Mathematica online in 6 s.. Simply and strongly.
$endgroup$
– user64494
Feb 7 at 20:47
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
Feb 7 at 21:21
1
$begingroup$
@HenrikSchumacher on the cloud it takes like 90s
$endgroup$
– b3m2a1
Feb 8 at 0:01
|
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
Feb 7 at 20:37
$begingroup$
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
Feb 7 at 20:40
$begingroup$
Reproduced in Mathematica online in 6 s.. Simply and strongly.
$endgroup$
– user64494
Feb 7 at 20:47
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
Feb 7 at 21:21
1
$begingroup$
@HenrikSchumacher on the cloud it takes like 90s
$endgroup$
– b3m2a1
Feb 8 at 0:01
$begingroup$
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
Feb 7 at 20:37
$begingroup$
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
Feb 7 at 20:37
$begingroup$
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
Feb 7 at 20:40
$begingroup$
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
Feb 7 at 20:40
$begingroup$
Reproduced in Mathematica online in 6 s.. Simply and strongly.
$endgroup$
– user64494
Feb 7 at 20:47
$begingroup$
Reproduced in Mathematica online in 6 s.. Simply and strongly.
$endgroup$
– user64494
Feb 7 at 20:47
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
Feb 7 at 21:21
$begingroup$
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
Feb 7 at 21:21
1
1
$begingroup$
@HenrikSchumacher on the cloud it takes like 90s
$endgroup$
– b3m2a1
Feb 8 at 0:01
$begingroup$
@HenrikSchumacher on the cloud it takes like 90s
$endgroup$
– b3m2a1
Feb 8 at 0:01
|
show 2 more comments
$begingroup$
Use RootReduce
Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // RootReduce
(* 45 *)
answered Feb 8 at 0:16
Bob HanlonBob Hanlon
60.7k33597
$endgroup$
1
$begingroup$
+1. It is unclear to me how it works.
$endgroup$
– user64494
Feb 8 at 3:50
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
Feb 8 at 4:16
add a comment |
$begingroup$
Use RootReduce
Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // RootReduce
(* 45 *)
answered Feb 8 at 0:16
Bob HanlonBob Hanlon
60.7k33597
$endgroup$
1
$begingroup$
+1. It is unclear to me how it works.
$endgroup$
– user64494
Feb 8 at 3:50
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
Feb 8 at 4:16
add a comment |
$begingroup$
Use RootReduce
Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // RootReduce
(* 45 *)
answered Feb 8 at 0:16
Bob HanlonBob Hanlon
60.7k33597
$endgroup$
Use RootReduce
Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}] // RootReduce
(* 45 *)
answered Feb 8 at 0:16
Bob HanlonBob Hanlon
60.7k33597
answered Feb 8 at 0:16
Bob HanlonBob Hanlon
60.7k33597
answered Feb 8 at 0:16
Bob HanlonBob Hanlon
60.7k33597
answered Feb 8 at 0:16
Bob HanlonBob Hanlon
60.7k33597
60.7k33597
1
$begingroup$
+1. It is unclear to me how it works.
$endgroup$
– user64494
Feb 8 at 3:50
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
Feb 8 at 4:16
add a comment |
1
$begingroup$
+1. It is unclear to me how it works.
$endgroup$
– user64494
Feb 8 at 3:50
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
Feb 8 at 4:16
1
1
$begingroup$
+1. It is unclear to me how it works.
$endgroup$
– user64494
Feb 8 at 3:50
$begingroup$
+1. It is unclear to me how it works.
$endgroup$
– user64494
Feb 8 at 3:50
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
Feb 8 at 4:16
$begingroup$
@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
Feb 8 at 4:16
add a comment |
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$begingroup$
N[Sum[Tan[(4*j - 3)*Pi/180], {j, 1, 45}]]gives 45.$endgroup$
– Carl Lange
Feb 7 at 20:02
$begingroup$
@Carl Lange: Up to a certain precision, is not so? Did you carefully read the question and its tags?
$endgroup$
– user64494
Feb 7 at 20:04
$begingroup$
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
Feb 7 at 20:11
$begingroup$
@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
Feb 7 at 20:14
1
$begingroup$
@CarlLange I guess, it is only about challenging Mathematica's symbolic capabilities.
$endgroup$
– Henrik Schumacher
Feb 7 at 20:27