Why is expression not evaluated completely?
$begingroup$
I have a simple sum of complex numbers. In my example their sum is zero.
sum = Total[{1 , E^(2*I/3*Pi) , E^((4*I)/3*Pi)}]
Print[sum]
Try it online!
When computing this sum, I just get back this "symbolic" expression:
1 + E^((-2*I)/3*Pi) + E^((2*I)/3*Pi)
(I already learned that this can be reduced to a single number using Simplify to get the result I expect.)
But when I replace this list of complex numbers with e.g. a list of integers {1,2,3} it does get evaluated to a single number.
I didn't understand why these two cases behave differently, so can you explain why I get an expression back for the first case and a fully simplified number in the second case?
simplifying-expressions evaluation
asked 4 hours ago
flawrflawr
1105
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add a comment |
$begingroup$
I have a simple sum of complex numbers. In my example their sum is zero.
sum = Total[{1 , E^(2*I/3*Pi) , E^((4*I)/3*Pi)}]
Print[sum]
Try it online!
When computing this sum, I just get back this "symbolic" expression:
1 + E^((-2*I)/3*Pi) + E^((2*I)/3*Pi)
(I already learned that this can be reduced to a single number using Simplify to get the result I expect.)
But when I replace this list of complex numbers with e.g. a list of integers {1,2,3} it does get evaluated to a single number.
I didn't understand why these two cases behave differently, so can you explain why I get an expression back for the first case and a fully simplified number in the second case?
simplifying-expressions evaluation
asked 4 hours ago
flawrflawr
1105
$endgroup$
add a comment |
$begingroup$
I have a simple sum of complex numbers. In my example their sum is zero.
sum = Total[{1 , E^(2*I/3*Pi) , E^((4*I)/3*Pi)}]
Print[sum]
Try it online!
When computing this sum, I just get back this "symbolic" expression:
1 + E^((-2*I)/3*Pi) + E^((2*I)/3*Pi)
(I already learned that this can be reduced to a single number using Simplify to get the result I expect.)
But when I replace this list of complex numbers with e.g. a list of integers {1,2,3} it does get evaluated to a single number.
I didn't understand why these two cases behave differently, so can you explain why I get an expression back for the first case and a fully simplified number in the second case?
simplifying-expressions evaluation
asked 4 hours ago
flawrflawr
1105
$endgroup$
I have a simple sum of complex numbers. In my example their sum is zero.
sum = Total[{1 , E^(2*I/3*Pi) , E^((4*I)/3*Pi)}]
Print[sum]
Try it online!
When computing this sum, I just get back this "symbolic" expression:
1 + E^((-2*I)/3*Pi) + E^((2*I)/3*Pi)
(I already learned that this can be reduced to a single number using Simplify to get the result I expect.)
But when I replace this list of complex numbers with e.g. a list of integers {1,2,3} it does get evaluated to a single number.
I didn't understand why these two cases behave differently, so can you explain why I get an expression back for the first case and a fully simplified number in the second case?
simplifying-expressions evaluation
simplifying-expressions evaluation
asked 4 hours ago
flawrflawr
1105
asked 4 hours ago
flawrflawr
1105
asked 4 hours ago
flawrflawr
1105
asked 4 hours ago
flawrflawr
1105
asked 4 hours ago
flawrflawr
1105
1105
add a comment |
add a comment |
2 Answers
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$begingroup$
Exact numeric expressions are treated symbolically, and there are a very limited number of transformations that will be applied automatically. Adding "actual numbers" (see NumberQ) is one such transformation that occurs. But E^((-2*I)/3*Pi) is not converted to a number, unless such a transformation is explicitly requested (as is done by Simplify, ComplexExpand, and so forth). A similar thing happens with the following:
Total@ArcTan@Range@3
(* π/4 + ArcTan[2] + ArcTan[3] *)
FullSimplify[%]
(* π *)
answered 3 hours ago
Michael E2Michael E2
147k12197469
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$begingroup$
sum = Total[{1, E^(2*I/3*Pi), E^((4*I)/3*Pi)}]
(* 1+E^(-((2 I π)/3))+E^((2 I π)/3) *)
Using machine precision, the sum is not identically zero
sum // N
(* 4.44089*10^-16+0. I *)
In fact, basic symbolic and numerical methods used internally do not show that sum has value zero
sum // PossibleZeroQ
(* PossibleZeroQ::ztest1: Unable to decide whether numeric quantity
1-(-1)^(1/3)+(-1)^(2/3) is equal to zero. Assuming it is.
True *)
Consequently, sum does not automatically evaluate to zero. More robust methods are required such as
#@sum& /@ {Simplify, ComplexExpand, RootReduce}
(* {0,0,0} *)
answered 1 hour ago
Bob HanlonBob Hanlon
59.6k33596
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Exact numeric expressions are treated symbolically, and there are a very limited number of transformations that will be applied automatically. Adding "actual numbers" (see NumberQ) is one such transformation that occurs. But E^((-2*I)/3*Pi) is not converted to a number, unless such a transformation is explicitly requested (as is done by Simplify, ComplexExpand, and so forth). A similar thing happens with the following:
Total@ArcTan@Range@3
(* π/4 + ArcTan[2] + ArcTan[3] *)
FullSimplify[%]
(* π *)
answered 3 hours ago
Michael E2Michael E2
147k12197469
$endgroup$
add a comment |
$begingroup$
Exact numeric expressions are treated symbolically, and there are a very limited number of transformations that will be applied automatically. Adding "actual numbers" (see NumberQ) is one such transformation that occurs. But E^((-2*I)/3*Pi) is not converted to a number, unless such a transformation is explicitly requested (as is done by Simplify, ComplexExpand, and so forth). A similar thing happens with the following:
Total@ArcTan@Range@3
(* π/4 + ArcTan[2] + ArcTan[3] *)
FullSimplify[%]
(* π *)
answered 3 hours ago
Michael E2Michael E2
147k12197469
$endgroup$
add a comment |
$begingroup$
Exact numeric expressions are treated symbolically, and there are a very limited number of transformations that will be applied automatically. Adding "actual numbers" (see NumberQ) is one such transformation that occurs. But E^((-2*I)/3*Pi) is not converted to a number, unless such a transformation is explicitly requested (as is done by Simplify, ComplexExpand, and so forth). A similar thing happens with the following:
Total@ArcTan@Range@3
(* π/4 + ArcTan[2] + ArcTan[3] *)
FullSimplify[%]
(* π *)
answered 3 hours ago
Michael E2Michael E2
147k12197469
$endgroup$
Exact numeric expressions are treated symbolically, and there are a very limited number of transformations that will be applied automatically. Adding "actual numbers" (see NumberQ) is one such transformation that occurs. But E^((-2*I)/3*Pi) is not converted to a number, unless such a transformation is explicitly requested (as is done by Simplify, ComplexExpand, and so forth). A similar thing happens with the following:
Total@ArcTan@Range@3
(* π/4 + ArcTan[2] + ArcTan[3] *)
FullSimplify[%]
(* π *)
answered 3 hours ago
Michael E2Michael E2
147k12197469
edited 3 hours ago
answered 3 hours ago
Michael E2Michael E2
147k12197469
answered 3 hours ago
Michael E2Michael E2
147k12197469
answered 3 hours ago
Michael E2Michael E2
147k12197469
147k12197469
add a comment |
add a comment |
$begingroup$
sum = Total[{1, E^(2*I/3*Pi), E^((4*I)/3*Pi)}]
(* 1+E^(-((2 I π)/3))+E^((2 I π)/3) *)
Using machine precision, the sum is not identically zero
sum // N
(* 4.44089*10^-16+0. I *)
In fact, basic symbolic and numerical methods used internally do not show that sum has value zero
sum // PossibleZeroQ
(* PossibleZeroQ::ztest1: Unable to decide whether numeric quantity
1-(-1)^(1/3)+(-1)^(2/3) is equal to zero. Assuming it is.
True *)
Consequently, sum does not automatically evaluate to zero. More robust methods are required such as
#@sum& /@ {Simplify, ComplexExpand, RootReduce}
(* {0,0,0} *)
answered 1 hour ago
Bob HanlonBob Hanlon
59.6k33596
$endgroup$
add a comment |
$begingroup$
sum = Total[{1, E^(2*I/3*Pi), E^((4*I)/3*Pi)}]
(* 1+E^(-((2 I π)/3))+E^((2 I π)/3) *)
Using machine precision, the sum is not identically zero
sum // N
(* 4.44089*10^-16+0. I *)
In fact, basic symbolic and numerical methods used internally do not show that sum has value zero
sum // PossibleZeroQ
(* PossibleZeroQ::ztest1: Unable to decide whether numeric quantity
1-(-1)^(1/3)+(-1)^(2/3) is equal to zero. Assuming it is.
True *)
Consequently, sum does not automatically evaluate to zero. More robust methods are required such as
#@sum& /@ {Simplify, ComplexExpand, RootReduce}
(* {0,0,0} *)
answered 1 hour ago
Bob HanlonBob Hanlon
59.6k33596
$endgroup$
add a comment |
$begingroup$
sum = Total[{1, E^(2*I/3*Pi), E^((4*I)/3*Pi)}]
(* 1+E^(-((2 I π)/3))+E^((2 I π)/3) *)
Using machine precision, the sum is not identically zero
sum // N
(* 4.44089*10^-16+0. I *)
In fact, basic symbolic and numerical methods used internally do not show that sum has value zero
sum // PossibleZeroQ
(* PossibleZeroQ::ztest1: Unable to decide whether numeric quantity
1-(-1)^(1/3)+(-1)^(2/3) is equal to zero. Assuming it is.
True *)
Consequently, sum does not automatically evaluate to zero. More robust methods are required such as
#@sum& /@ {Simplify, ComplexExpand, RootReduce}
(* {0,0,0} *)
answered 1 hour ago
Bob HanlonBob Hanlon
59.6k33596
$endgroup$
sum = Total[{1, E^(2*I/3*Pi), E^((4*I)/3*Pi)}]
(* 1+E^(-((2 I π)/3))+E^((2 I π)/3) *)
Using machine precision, the sum is not identically zero
sum // N
(* 4.44089*10^-16+0. I *)
In fact, basic symbolic and numerical methods used internally do not show that sum has value zero
sum // PossibleZeroQ
(* PossibleZeroQ::ztest1: Unable to decide whether numeric quantity
1-(-1)^(1/3)+(-1)^(2/3) is equal to zero. Assuming it is.
True *)
Consequently, sum does not automatically evaluate to zero. More robust methods are required such as
#@sum& /@ {Simplify, ComplexExpand, RootReduce}
(* {0,0,0} *)
answered 1 hour ago
Bob HanlonBob Hanlon
59.6k33596
answered 1 hour ago
Bob HanlonBob Hanlon
59.6k33596
answered 1 hour ago
Bob HanlonBob Hanlon
59.6k33596
answered 1 hour ago
Bob HanlonBob Hanlon
59.6k33596
59.6k33596
add a comment |
add a comment |
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