FullSimplify a trigonometric expression doesn't work as expected
$begingroup$
I know this kind of question is frequent asked, yet each case has its own particularities. I will show my problem.
I define the following:
f[k_] := Binomial[n, k] (Sin[Φ]^2)^k (Cos[Φ]^2)^(n - k);
g[k_] := 2 (f[n - k - 1] (k + 1) - f[n - k] k);
After that I want to perform the sum:
FullSimplify[Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}],
Element[n, Integers] && n > 0 && Element[Φ, Reals] && 0 < Φ < Pi/2]
And I got the result:
4 n^2 Cos[Φ]^2 HypergeometricPFQ[{-n,1-n/2-1/2 n Cos[2 Φ],1-n/2-
1/2 n Cos[2 Φ]},{-n Cos[Φ]^2,-n Cos[Φ]^2},-Cot[Φ]^2] Sin[Φ]^(-2+2 n)
This as much as I can simplify the expression. However, according to book the result of the sum is just 4n
H[n_]=Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}]]//FullSimplify
Table[H[i],{i,1,10}]//FullSimplify
{4, 8, 12, 16, 20, 24, 28, 32, 36, 40}
If I introduce the Assumptions in the following way at the beginning of the notebook:
$Assumptions = n ∈ Integers && n < 100 &&
n > 0 && Φ ∈ Reals && 0 < Φ < Pi/2
The result of the simplification is yet worse.
The problem is that I have to perform some summations similar to this, but this time I don't have previous knowledge of the solution
simplifying-expressions summation assumptions trigonometry
edited 7 hours ago
user64494
3,61411122
asked 9 hours ago
PopeyePopeye
557
$endgroup$
add a comment |
$begingroup$
I know this kind of question is frequent asked, yet each case has its own particularities. I will show my problem.
I define the following:
f[k_] := Binomial[n, k] (Sin[Φ]^2)^k (Cos[Φ]^2)^(n - k);
g[k_] := 2 (f[n - k - 1] (k + 1) - f[n - k] k);
After that I want to perform the sum:
FullSimplify[Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}],
Element[n, Integers] && n > 0 && Element[Φ, Reals] && 0 < Φ < Pi/2]
And I got the result:
4 n^2 Cos[Φ]^2 HypergeometricPFQ[{-n,1-n/2-1/2 n Cos[2 Φ],1-n/2-
1/2 n Cos[2 Φ]},{-n Cos[Φ]^2,-n Cos[Φ]^2},-Cot[Φ]^2] Sin[Φ]^(-2+2 n)
This as much as I can simplify the expression. However, according to book the result of the sum is just 4n
H[n_]=Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}]]//FullSimplify
Table[H[i],{i,1,10}]//FullSimplify
{4, 8, 12, 16, 20, 24, 28, 32, 36, 40}
If I introduce the Assumptions in the following way at the beginning of the notebook:
$Assumptions = n ∈ Integers && n < 100 &&
n > 0 && Φ ∈ Reals && 0 < Φ < Pi/2
The result of the simplification is yet worse.
The problem is that I have to perform some summations similar to this, but this time I don't have previous knowledge of the solution
simplifying-expressions summation assumptions trigonometry
edited 7 hours ago
user64494
3,61411122
asked 9 hours ago
PopeyePopeye
557
$endgroup$
add a comment |
$begingroup$
I know this kind of question is frequent asked, yet each case has its own particularities. I will show my problem.
I define the following:
f[k_] := Binomial[n, k] (Sin[Φ]^2)^k (Cos[Φ]^2)^(n - k);
g[k_] := 2 (f[n - k - 1] (k + 1) - f[n - k] k);
After that I want to perform the sum:
FullSimplify[Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}],
Element[n, Integers] && n > 0 && Element[Φ, Reals] && 0 < Φ < Pi/2]
And I got the result:
4 n^2 Cos[Φ]^2 HypergeometricPFQ[{-n,1-n/2-1/2 n Cos[2 Φ],1-n/2-
1/2 n Cos[2 Φ]},{-n Cos[Φ]^2,-n Cos[Φ]^2},-Cot[Φ]^2] Sin[Φ]^(-2+2 n)
This as much as I can simplify the expression. However, according to book the result of the sum is just 4n
H[n_]=Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}]]//FullSimplify
Table[H[i],{i,1,10}]//FullSimplify
{4, 8, 12, 16, 20, 24, 28, 32, 36, 40}
If I introduce the Assumptions in the following way at the beginning of the notebook:
$Assumptions = n ∈ Integers && n < 100 &&
n > 0 && Φ ∈ Reals && 0 < Φ < Pi/2
The result of the simplification is yet worse.
The problem is that I have to perform some summations similar to this, but this time I don't have previous knowledge of the solution
simplifying-expressions summation assumptions trigonometry
edited 7 hours ago
user64494
3,61411122
asked 9 hours ago
PopeyePopeye
557
$endgroup$
I know this kind of question is frequent asked, yet each case has its own particularities. I will show my problem.
I define the following:
f[k_] := Binomial[n, k] (Sin[Φ]^2)^k (Cos[Φ]^2)^(n - k);
g[k_] := 2 (f[n - k - 1] (k + 1) - f[n - k] k);
After that I want to perform the sum:
FullSimplify[Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}],
Element[n, Integers] && n > 0 && Element[Φ, Reals] && 0 < Φ < Pi/2]
And I got the result:
4 n^2 Cos[Φ]^2 HypergeometricPFQ[{-n,1-n/2-1/2 n Cos[2 Φ],1-n/2-
1/2 n Cos[2 Φ]},{-n Cos[Φ]^2,-n Cos[Φ]^2},-Cot[Φ]^2] Sin[Φ]^(-2+2 n)
This as much as I can simplify the expression. However, according to book the result of the sum is just 4n
H[n_]=Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}]]//FullSimplify
Table[H[i],{i,1,10}]//FullSimplify
{4, 8, 12, 16, 20, 24, 28, 32, 36, 40}
If I introduce the Assumptions in the following way at the beginning of the notebook:
$Assumptions = n ∈ Integers && n < 100 &&
n > 0 && Φ ∈ Reals && 0 < Φ < Pi/2
The result of the simplification is yet worse.
The problem is that I have to perform some summations similar to this, but this time I don't have previous knowledge of the solution
simplifying-expressions summation assumptions trigonometry
simplifying-expressions summation assumptions trigonometry
edited 7 hours ago
user64494
3,61411122
asked 9 hours ago
PopeyePopeye
557
edited 7 hours ago
user64494
3,61411122
asked 9 hours ago
PopeyePopeye
557
edited 7 hours ago
user64494
3,61411122
edited 7 hours ago
user64494
3,61411122
edited 7 hours ago
user64494
3,61411122
3,61411122
asked 9 hours ago
PopeyePopeye
557
asked 9 hours ago
PopeyePopeye
557
asked 9 hours ago
PopeyePopeye
557
557
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Eliminating the trigonometric terms work in this case:
expr = 4 n^2 Cos[Φ]^2 HypergeometricPFQ[{-n,1-n/2-1/2 n Cos[2 Φ],1-n/2-
1/2 n Cos[2 Φ]},{-n Cos[Φ]^2,-n Cos[Φ]^2},-Cot[Φ]^2] Sin[Φ]^(-2+2 n);
FullSimplify[expr /. Φ -> ArcCos[q], 0 < q < 1]
4n
answered 6 hours ago
Chip HurstChip Hurst
23.8k15995
$endgroup$
add a comment |
$begingroup$
Clear["Global`*"];
$Assumptions =
Element[n, Integers] && n > 0 && 0 < Φ < Pi/2;
f[k_] := Binomial[n, k] (Sin[Φ]^2)^
k (Cos[Φ]^2)^(n - k);
g[k_] := 2 (f[n - k - 1] (k + 1) - f[n - k] k);
seq = Table[
Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}] // Simplify,
{n, 1, 10}]
(* {4, 8, 12, 16, 20, 24, 28, 32, 36, 40} *)
Using FindSequenceFunction,
H[n_] = FindSequenceFunction[seq, n]
(* 4 n *)
Verifying over wider range,
And @@ Table[
H[n] == Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}] // Simplify, {n,
1, 25}]
(* True *)
answered 6 hours ago
Bob HanlonBob Hanlon
61.8k33598
$endgroup$
$begingroup$
Thank, that's very useful answer. That function may be helpful
$endgroup$
– Popeye
6 hours ago
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Eliminating the trigonometric terms work in this case:
expr = 4 n^2 Cos[Φ]^2 HypergeometricPFQ[{-n,1-n/2-1/2 n Cos[2 Φ],1-n/2-
1/2 n Cos[2 Φ]},{-n Cos[Φ]^2,-n Cos[Φ]^2},-Cot[Φ]^2] Sin[Φ]^(-2+2 n);
FullSimplify[expr /. Φ -> ArcCos[q], 0 < q < 1]
4n
answered 6 hours ago
Chip HurstChip Hurst
23.8k15995
$endgroup$
add a comment |
$begingroup$
Eliminating the trigonometric terms work in this case:
expr = 4 n^2 Cos[Φ]^2 HypergeometricPFQ[{-n,1-n/2-1/2 n Cos[2 Φ],1-n/2-
1/2 n Cos[2 Φ]},{-n Cos[Φ]^2,-n Cos[Φ]^2},-Cot[Φ]^2] Sin[Φ]^(-2+2 n);
FullSimplify[expr /. Φ -> ArcCos[q], 0 < q < 1]
4n
answered 6 hours ago
Chip HurstChip Hurst
23.8k15995
$endgroup$
add a comment |
$begingroup$
Eliminating the trigonometric terms work in this case:
expr = 4 n^2 Cos[Φ]^2 HypergeometricPFQ[{-n,1-n/2-1/2 n Cos[2 Φ],1-n/2-
1/2 n Cos[2 Φ]},{-n Cos[Φ]^2,-n Cos[Φ]^2},-Cot[Φ]^2] Sin[Φ]^(-2+2 n);
FullSimplify[expr /. Φ -> ArcCos[q], 0 < q < 1]
4n
answered 6 hours ago
Chip HurstChip Hurst
23.8k15995
$endgroup$
Eliminating the trigonometric terms work in this case:
expr = 4 n^2 Cos[Φ]^2 HypergeometricPFQ[{-n,1-n/2-1/2 n Cos[2 Φ],1-n/2-
1/2 n Cos[2 Φ]},{-n Cos[Φ]^2,-n Cos[Φ]^2},-Cot[Φ]^2] Sin[Φ]^(-2+2 n);
FullSimplify[expr /. Φ -> ArcCos[q], 0 < q < 1]
4n
answered 6 hours ago
Chip HurstChip Hurst
23.8k15995
answered 6 hours ago
Chip HurstChip Hurst
23.8k15995
answered 6 hours ago
Chip HurstChip Hurst
23.8k15995
answered 6 hours ago
Chip HurstChip Hurst
23.8k15995
23.8k15995
add a comment |
add a comment |
$begingroup$
Clear["Global`*"];
$Assumptions =
Element[n, Integers] && n > 0 && 0 < Φ < Pi/2;
f[k_] := Binomial[n, k] (Sin[Φ]^2)^
k (Cos[Φ]^2)^(n - k);
g[k_] := 2 (f[n - k - 1] (k + 1) - f[n - k] k);
seq = Table[
Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}] // Simplify,
{n, 1, 10}]
(* {4, 8, 12, 16, 20, 24, 28, 32, 36, 40} *)
Using FindSequenceFunction,
H[n_] = FindSequenceFunction[seq, n]
(* 4 n *)
Verifying over wider range,
And @@ Table[
H[n] == Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}] // Simplify, {n,
1, 25}]
(* True *)
answered 6 hours ago
Bob HanlonBob Hanlon
61.8k33598
$endgroup$
$begingroup$
Thank, that's very useful answer. That function may be helpful
$endgroup$
– Popeye
6 hours ago
add a comment |
$begingroup$
Clear["Global`*"];
$Assumptions =
Element[n, Integers] && n > 0 && 0 < Φ < Pi/2;
f[k_] := Binomial[n, k] (Sin[Φ]^2)^
k (Cos[Φ]^2)^(n - k);
g[k_] := 2 (f[n - k - 1] (k + 1) - f[n - k] k);
seq = Table[
Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}] // Simplify,
{n, 1, 10}]
(* {4, 8, 12, 16, 20, 24, 28, 32, 36, 40} *)
Using FindSequenceFunction,
H[n_] = FindSequenceFunction[seq, n]
(* 4 n *)
Verifying over wider range,
And @@ Table[
H[n] == Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}] // Simplify, {n,
1, 25}]
(* True *)
answered 6 hours ago
Bob HanlonBob Hanlon
61.8k33598
$endgroup$
$begingroup$
Thank, that's very useful answer. That function may be helpful
$endgroup$
– Popeye
6 hours ago
add a comment |
$begingroup$
Clear["Global`*"];
$Assumptions =
Element[n, Integers] && n > 0 && 0 < Φ < Pi/2;
f[k_] := Binomial[n, k] (Sin[Φ]^2)^
k (Cos[Φ]^2)^(n - k);
g[k_] := 2 (f[n - k - 1] (k + 1) - f[n - k] k);
seq = Table[
Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}] // Simplify,
{n, 1, 10}]
(* {4, 8, 12, 16, 20, 24, 28, 32, 36, 40} *)
Using FindSequenceFunction,
H[n_] = FindSequenceFunction[seq, n]
(* 4 n *)
Verifying over wider range,
And @@ Table[
H[n] == Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}] // Simplify, {n,
1, 25}]
(* True *)
answered 6 hours ago
Bob HanlonBob Hanlon
61.8k33598
$endgroup$
Clear["Global`*"];
$Assumptions =
Element[n, Integers] && n > 0 && 0 < Φ < Pi/2;
f[k_] := Binomial[n, k] (Sin[Φ]^2)^
k (Cos[Φ]^2)^(n - k);
g[k_] := 2 (f[n - k - 1] (k + 1) - f[n - k] k);
seq = Table[
Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}] // Simplify,
{n, 1, 10}]
(* {4, 8, 12, 16, 20, 24, 28, 32, 36, 40} *)
Using FindSequenceFunction,
H[n_] = FindSequenceFunction[seq, n]
(* 4 n *)
Verifying over wider range,
And @@ Table[
H[n] == Tan[Φ]^2*Sum[g[k]^2/f[n - k], {k, 0, n}] // Simplify, {n,
1, 25}]
(* True *)
answered 6 hours ago
Bob HanlonBob Hanlon
61.8k33598
answered 6 hours ago
Bob HanlonBob Hanlon
61.8k33598
answered 6 hours ago
Bob HanlonBob Hanlon
61.8k33598
answered 6 hours ago
Bob HanlonBob Hanlon
61.8k33598
61.8k33598
$begingroup$
Thank, that's very useful answer. That function may be helpful
$endgroup$
– Popeye
6 hours ago
add a comment |
$begingroup$
Thank, that's very useful answer. That function may be helpful
$endgroup$
– Popeye
6 hours ago
$begingroup$
Thank, that's very useful answer. That function may be helpful
$endgroup$
– Popeye
6 hours ago
$begingroup$
Thank, that's very useful answer. That function may be helpful
$endgroup$
– Popeye
6 hours ago
add a comment |
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