Force a sum to be calculated
$begingroup$
I want to force this sum to be calculated to 1
Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}]
Only DiscretePlot3D gives the correct result showing all point to 1
summation
$endgroup$
add a comment |
$begingroup$
I want to force this sum to be calculated to 1
Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}]
Only DiscretePlot3D gives the correct result showing all point to 1
summation
$endgroup$
$begingroup$
For what values of $n$ and $m$?
$endgroup$
– MarcoB
5 hours ago
add a comment |
$begingroup$
I want to force this sum to be calculated to 1
Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}]
Only DiscretePlot3D gives the correct result showing all point to 1
summation
$endgroup$
I want to force this sum to be calculated to 1
Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}]
Only DiscretePlot3D gives the correct result showing all point to 1
summation
summation
asked 5 hours ago
NitraNitra
676
676
$begingroup$
For what values of $n$ and $m$?
$endgroup$
– MarcoB
5 hours ago
add a comment |
$begingroup$
For what values of $n$ and $m$?
$endgroup$
– MarcoB
5 hours ago
$begingroup$
For what values of $n$ and $m$?
$endgroup$
– MarcoB
5 hours ago
$begingroup$
For what values of $n$ and $m$?
$endgroup$
– MarcoB
5 hours ago
add a comment |
1 Answer
1
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votes
$begingroup$
The following code might do what you want:
s0 = Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}] // Simplify;
s1 = s0 /. {Cos[X_] - Cos[Y_] -> 2 Sin[(X + Y)/2] Sin[(Y - X)/2]};
s2 = s1 /. {Sin[x_] :> Sin[Factor@x],
Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)]};
s3 = Simplify[s2, m [Element] Integers]
which evaluates to 1
. The step to s1
was easy, but I had to do lots of experimentation to find the steps to s2
and s3
. In particular the rule
Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)]
assumes that m
is an integer but I could not get Mathematica to do it automatically. There may be better ways to simplify the sum s0
but perhaps others can produce them.
$endgroup$
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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oldest
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oldest
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$begingroup$
The following code might do what you want:
s0 = Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}] // Simplify;
s1 = s0 /. {Cos[X_] - Cos[Y_] -> 2 Sin[(X + Y)/2] Sin[(Y - X)/2]};
s2 = s1 /. {Sin[x_] :> Sin[Factor@x],
Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)]};
s3 = Simplify[s2, m [Element] Integers]
which evaluates to 1
. The step to s1
was easy, but I had to do lots of experimentation to find the steps to s2
and s3
. In particular the rule
Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)]
assumes that m
is an integer but I could not get Mathematica to do it automatically. There may be better ways to simplify the sum s0
but perhaps others can produce them.
$endgroup$
add a comment |
$begingroup$
The following code might do what you want:
s0 = Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}] // Simplify;
s1 = s0 /. {Cos[X_] - Cos[Y_] -> 2 Sin[(X + Y)/2] Sin[(Y - X)/2]};
s2 = s1 /. {Sin[x_] :> Sin[Factor@x],
Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)]};
s3 = Simplify[s2, m [Element] Integers]
which evaluates to 1
. The step to s1
was easy, but I had to do lots of experimentation to find the steps to s2
and s3
. In particular the rule
Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)]
assumes that m
is an integer but I could not get Mathematica to do it automatically. There may be better ways to simplify the sum s0
but perhaps others can produce them.
$endgroup$
add a comment |
$begingroup$
The following code might do what you want:
s0 = Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}] // Simplify;
s1 = s0 /. {Cos[X_] - Cos[Y_] -> 2 Sin[(X + Y)/2] Sin[(Y - X)/2]};
s2 = s1 /. {Sin[x_] :> Sin[Factor@x],
Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)]};
s3 = Simplify[s2, m [Element] Integers]
which evaluates to 1
. The step to s1
was easy, but I had to do lots of experimentation to find the steps to s2
and s3
. In particular the rule
Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)]
assumes that m
is an integer but I could not get Mathematica to do it automatically. There may be better ways to simplify the sum s0
but perhaps others can produce them.
$endgroup$
The following code might do what you want:
s0 = Sum[Cos[(Pi*l*(2*m + 1))/(n + 1)], {l, 0, n}] // Simplify;
s1 = s0 /. {Cos[X_] - Cos[Y_] -> 2 Sin[(X + Y)/2] Sin[(Y - X)/2]};
s2 = s1 /. {Sin[x_] :> Sin[Factor@x],
Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)]};
s3 = Simplify[s2, m [Element] Integers]
which evaluates to 1
. The step to s1
was easy, but I had to do lots of experimentation to find the steps to s2
and s3
. In particular the rule
Csc[x_] :> -Sin[(m + 1/2) Pi]/Sin[(m Pi - x)]
assumes that m
is an integer but I could not get Mathematica to do it automatically. There may be better ways to simplify the sum s0
but perhaps others can produce them.
edited 36 mins ago
answered 1 hour ago
SomosSomos
1,07019
1,07019
add a comment |
add a comment |
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$begingroup$
For what values of $n$ and $m$?
$endgroup$
– MarcoB
5 hours ago