An expression involving trig functions and binomial coefficients
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Prove that
begineqnarraysum_i=0^k5^k-i-1binomkileft(frac(37+62cosfrac2pi5+56cosfrac4pi5)^ileft(2sinfracpi5right)^2(k-i-1)+frac(37+62cosfrac4pi5+56cosfrac8pi5)^ileft(2sinfrac2pi5right)^2(k-i-1)right)endeqnarray
is an integer. Can we further simplify the above expression? Thanks.
trigonometry binomial-coefficients
asked Aug 30 at 7:29
Alex Fok
3,836617
add a comment |Â
up vote
1
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Prove that
begineqnarraysum_i=0^k5^k-i-1binomkileft(frac(37+62cosfrac2pi5+56cosfrac4pi5)^ileft(2sinfracpi5right)^2(k-i-1)+frac(37+62cosfrac4pi5+56cosfrac8pi5)^ileft(2sinfrac2pi5right)^2(k-i-1)right)endeqnarray
is an integer. Can we further simplify the above expression? Thanks.
trigonometry binomial-coefficients
asked Aug 30 at 7:29
Alex Fok
3,836617
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Prove that
begineqnarraysum_i=0^k5^k-i-1binomkileft(frac(37+62cosfrac2pi5+56cosfrac4pi5)^ileft(2sinfracpi5right)^2(k-i-1)+frac(37+62cosfrac4pi5+56cosfrac8pi5)^ileft(2sinfrac2pi5right)^2(k-i-1)right)endeqnarray
is an integer. Can we further simplify the above expression? Thanks.
trigonometry binomial-coefficients
asked Aug 30 at 7:29
Alex Fok
3,836617
Prove that
begineqnarraysum_i=0^k5^k-i-1binomkileft(frac(37+62cosfrac2pi5+56cosfrac4pi5)^ileft(2sinfracpi5right)^2(k-i-1)+frac(37+62cosfrac4pi5+56cosfrac8pi5)^ileft(2sinfrac2pi5right)^2(k-i-1)right)endeqnarray
is an integer. Can we further simplify the above expression? Thanks.
trigonometry binomial-coefficients
trigonometry binomial-coefficients
asked Aug 30 at 7:29
Alex Fok
3,836617
asked Aug 30 at 7:29
Alex Fok
3,836617
asked Aug 30 at 7:29
Alex Fok
3,836617
asked Aug 30 at 7:29
Alex Fok
3,836617
asked Aug 30 at 7:29
Alex Fok
3,836617
3,836617
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
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accepted
Using the same approach as @drhab in his answer, I arrived to a slightly different result
$$colorbluet_k=fracb5left(a+frac 5bright)^k+fracd5left(c+frac5dright)^k$$ Now, using the values of the trigonometric terms, we have
$$a=frac32 left(5+sqrt5right)qquad b=frac12 left(5-sqrt5right)qquad c=frac32 left(5-sqrt5right)qquad d=frac12 left(5+sqrt5right)$$
$$a+frac 5b=2 left(5+sqrt5right)qquad c+frac5d=2 left(5-sqrt5right)$$ leading the the sequence
$$1,8,80,960,12800,179200,2560000,36864000,532480000,7700480000,111411200000$$
We could make $t_k$ more fancy in terms of $phi$, the golden ratio.
answered Aug 31 at 4:29
Claude Leibovici
113k1155127
+1 Nice answer! Btw I am a "he".
– drhab
Aug 31 at 6:22
add a comment |Â
up vote
1
down vote
You could start with something like:
$$sum_i=0^k5^k-i-1binomkileft[fraca^ib^k-i-1+fracc^id^k-i-1right]=fracb5sum_i=0^kbinomkia^ileft(frac5bright)^k-i+fracd5sum_i=0^kbinomkic^ileft(frac5dright)^k-i=$$$$fracb5left(a+frac5bright)^k+fracd5left(c+frac5dright)^k$$
This to get rid of the sum.
I hope it helps, but too risky to give any guarantees :-).
answered Aug 30 at 7:40
drhab
88.8k541121
Using the values of $a,b,c,d$, this leads to an interesting sequence.
– Claude Leibovici
Aug 30 at 8:28
@ClaudeLeibovici Why is it interesting?
– Alex Fok
Aug 30 at 13:14
@AlexFok. Replace $a,b,c,d$ by their values (they are simple). Now, using the expression given in drhab's answer, compute for $k=1,2,3,4,cdots,n$. Look at the numbers (all integers) and try to see a pattern (if any).
– Claude Leibovici
Aug 30 at 14:54
I suppose that there are typo's in the answer.
– Claude Leibovici
Aug 31 at 4:30
@ClaudeLeibovici thank you. Repaired.
– drhab
Aug 31 at 6:19
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Using the same approach as @drhab in his answer, I arrived to a slightly different result
$$colorbluet_k=fracb5left(a+frac 5bright)^k+fracd5left(c+frac5dright)^k$$ Now, using the values of the trigonometric terms, we have
$$a=frac32 left(5+sqrt5right)qquad b=frac12 left(5-sqrt5right)qquad c=frac32 left(5-sqrt5right)qquad d=frac12 left(5+sqrt5right)$$
$$a+frac 5b=2 left(5+sqrt5right)qquad c+frac5d=2 left(5-sqrt5right)$$ leading the the sequence
$$1,8,80,960,12800,179200,2560000,36864000,532480000,7700480000,111411200000$$
We could make $t_k$ more fancy in terms of $phi$, the golden ratio.
answered Aug 31 at 4:29
Claude Leibovici
113k1155127
+1 Nice answer! Btw I am a "he".
– drhab
Aug 31 at 6:22
add a comment |Â
up vote
2
down vote
accepted
Using the same approach as @drhab in his answer, I arrived to a slightly different result
$$colorbluet_k=fracb5left(a+frac 5bright)^k+fracd5left(c+frac5dright)^k$$ Now, using the values of the trigonometric terms, we have
$$a=frac32 left(5+sqrt5right)qquad b=frac12 left(5-sqrt5right)qquad c=frac32 left(5-sqrt5right)qquad d=frac12 left(5+sqrt5right)$$
$$a+frac 5b=2 left(5+sqrt5right)qquad c+frac5d=2 left(5-sqrt5right)$$ leading the the sequence
$$1,8,80,960,12800,179200,2560000,36864000,532480000,7700480000,111411200000$$
We could make $t_k$ more fancy in terms of $phi$, the golden ratio.
answered Aug 31 at 4:29
Claude Leibovici
113k1155127
+1 Nice answer! Btw I am a "he".
– drhab
Aug 31 at 6:22
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Using the same approach as @drhab in his answer, I arrived to a slightly different result
$$colorbluet_k=fracb5left(a+frac 5bright)^k+fracd5left(c+frac5dright)^k$$ Now, using the values of the trigonometric terms, we have
$$a=frac32 left(5+sqrt5right)qquad b=frac12 left(5-sqrt5right)qquad c=frac32 left(5-sqrt5right)qquad d=frac12 left(5+sqrt5right)$$
$$a+frac 5b=2 left(5+sqrt5right)qquad c+frac5d=2 left(5-sqrt5right)$$ leading the the sequence
$$1,8,80,960,12800,179200,2560000,36864000,532480000,7700480000,111411200000$$
We could make $t_k$ more fancy in terms of $phi$, the golden ratio.
answered Aug 31 at 4:29
Claude Leibovici
113k1155127
Using the same approach as @drhab in his answer, I arrived to a slightly different result
$$colorbluet_k=fracb5left(a+frac 5bright)^k+fracd5left(c+frac5dright)^k$$ Now, using the values of the trigonometric terms, we have
$$a=frac32 left(5+sqrt5right)qquad b=frac12 left(5-sqrt5right)qquad c=frac32 left(5-sqrt5right)qquad d=frac12 left(5+sqrt5right)$$
$$a+frac 5b=2 left(5+sqrt5right)qquad c+frac5d=2 left(5-sqrt5right)$$ leading the the sequence
$$1,8,80,960,12800,179200,2560000,36864000,532480000,7700480000,111411200000$$
We could make $t_k$ more fancy in terms of $phi$, the golden ratio.
answered Aug 31 at 4:29
Claude Leibovici
113k1155127
edited Aug 31 at 6:47
answered Aug 31 at 4:29
Claude Leibovici
113k1155127
answered Aug 31 at 4:29
Claude Leibovici
113k1155127
answered Aug 31 at 4:29
Claude Leibovici
113k1155127
113k1155127
+1 Nice answer! Btw I am a "he".
– drhab
Aug 31 at 6:22
add a comment |Â
+1 Nice answer! Btw I am a "he".
– drhab
Aug 31 at 6:22
+1 Nice answer! Btw I am a "he".
– drhab
Aug 31 at 6:22
+1 Nice answer! Btw I am a "he".
– drhab
Aug 31 at 6:22
add a comment |Â
up vote
1
down vote
You could start with something like:
$$sum_i=0^k5^k-i-1binomkileft[fraca^ib^k-i-1+fracc^id^k-i-1right]=fracb5sum_i=0^kbinomkia^ileft(frac5bright)^k-i+fracd5sum_i=0^kbinomkic^ileft(frac5dright)^k-i=$$$$fracb5left(a+frac5bright)^k+fracd5left(c+frac5dright)^k$$
This to get rid of the sum.
I hope it helps, but too risky to give any guarantees :-).
answered Aug 30 at 7:40
drhab
88.8k541121
Using the values of $a,b,c,d$, this leads to an interesting sequence.
– Claude Leibovici
Aug 30 at 8:28
@ClaudeLeibovici Why is it interesting?
– Alex Fok
Aug 30 at 13:14
@AlexFok. Replace $a,b,c,d$ by their values (they are simple). Now, using the expression given in drhab's answer, compute for $k=1,2,3,4,cdots,n$. Look at the numbers (all integers) and try to see a pattern (if any).
– Claude Leibovici
Aug 30 at 14:54
I suppose that there are typo's in the answer.
– Claude Leibovici
Aug 31 at 4:30
@ClaudeLeibovici thank you. Repaired.
– drhab
Aug 31 at 6:19
add a comment |Â
up vote
1
down vote
You could start with something like:
$$sum_i=0^k5^k-i-1binomkileft[fraca^ib^k-i-1+fracc^id^k-i-1right]=fracb5sum_i=0^kbinomkia^ileft(frac5bright)^k-i+fracd5sum_i=0^kbinomkic^ileft(frac5dright)^k-i=$$$$fracb5left(a+frac5bright)^k+fracd5left(c+frac5dright)^k$$
This to get rid of the sum.
I hope it helps, but too risky to give any guarantees :-).
answered Aug 30 at 7:40
drhab
88.8k541121
Using the values of $a,b,c,d$, this leads to an interesting sequence.
– Claude Leibovici
Aug 30 at 8:28
@ClaudeLeibovici Why is it interesting?
– Alex Fok
Aug 30 at 13:14
@AlexFok. Replace $a,b,c,d$ by their values (they are simple). Now, using the expression given in drhab's answer, compute for $k=1,2,3,4,cdots,n$. Look at the numbers (all integers) and try to see a pattern (if any).
– Claude Leibovici
Aug 30 at 14:54
I suppose that there are typo's in the answer.
– Claude Leibovici
Aug 31 at 4:30
@ClaudeLeibovici thank you. Repaired.
– drhab
Aug 31 at 6:19
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You could start with something like:
$$sum_i=0^k5^k-i-1binomkileft[fraca^ib^k-i-1+fracc^id^k-i-1right]=fracb5sum_i=0^kbinomkia^ileft(frac5bright)^k-i+fracd5sum_i=0^kbinomkic^ileft(frac5dright)^k-i=$$$$fracb5left(a+frac5bright)^k+fracd5left(c+frac5dright)^k$$
This to get rid of the sum.
I hope it helps, but too risky to give any guarantees :-).
answered Aug 30 at 7:40
drhab
88.8k541121
You could start with something like:
$$sum_i=0^k5^k-i-1binomkileft[fraca^ib^k-i-1+fracc^id^k-i-1right]=fracb5sum_i=0^kbinomkia^ileft(frac5bright)^k-i+fracd5sum_i=0^kbinomkic^ileft(frac5dright)^k-i=$$$$fracb5left(a+frac5bright)^k+fracd5left(c+frac5dright)^k$$
This to get rid of the sum.
I hope it helps, but too risky to give any guarantees :-).
answered Aug 30 at 7:40
drhab
88.8k541121
edited Aug 31 at 6:18
answered Aug 30 at 7:40
drhab
88.8k541121
answered Aug 30 at 7:40
drhab
88.8k541121
answered Aug 30 at 7:40
drhab
88.8k541121
88.8k541121
Using the values of $a,b,c,d$, this leads to an interesting sequence.
– Claude Leibovici
Aug 30 at 8:28
@ClaudeLeibovici Why is it interesting?
– Alex Fok
Aug 30 at 13:14
@AlexFok. Replace $a,b,c,d$ by their values (they are simple). Now, using the expression given in drhab's answer, compute for $k=1,2,3,4,cdots,n$. Look at the numbers (all integers) and try to see a pattern (if any).
– Claude Leibovici
Aug 30 at 14:54
I suppose that there are typo's in the answer.
– Claude Leibovici
Aug 31 at 4:30
@ClaudeLeibovici thank you. Repaired.
– drhab
Aug 31 at 6:19
add a comment |Â
Using the values of $a,b,c,d$, this leads to an interesting sequence.
– Claude Leibovici
Aug 30 at 8:28
@ClaudeLeibovici Why is it interesting?
– Alex Fok
Aug 30 at 13:14
@AlexFok. Replace $a,b,c,d$ by their values (they are simple). Now, using the expression given in drhab's answer, compute for $k=1,2,3,4,cdots,n$. Look at the numbers (all integers) and try to see a pattern (if any).
– Claude Leibovici
Aug 30 at 14:54
I suppose that there are typo's in the answer.
– Claude Leibovici
Aug 31 at 4:30
@ClaudeLeibovici thank you. Repaired.
– drhab
Aug 31 at 6:19
Using the values of $a,b,c,d$, this leads to an interesting sequence.
– Claude Leibovici
Aug 30 at 8:28
Using the values of $a,b,c,d$, this leads to an interesting sequence.
– Claude Leibovici
Aug 30 at 8:28
@ClaudeLeibovici Why is it interesting?
– Alex Fok
Aug 30 at 13:14
@ClaudeLeibovici Why is it interesting?
– Alex Fok
Aug 30 at 13:14
@AlexFok. Replace $a,b,c,d$ by their values (they are simple). Now, using the expression given in drhab's answer, compute for $k=1,2,3,4,cdots,n$. Look at the numbers (all integers) and try to see a pattern (if any).
– Claude Leibovici
Aug 30 at 14:54
@AlexFok. Replace $a,b,c,d$ by their values (they are simple). Now, using the expression given in drhab's answer, compute for $k=1,2,3,4,cdots,n$. Look at the numbers (all integers) and try to see a pattern (if any).
– Claude Leibovici
Aug 30 at 14:54
I suppose that there are typo's in the answer.
– Claude Leibovici
Aug 31 at 4:30
I suppose that there are typo's in the answer.
– Claude Leibovici
Aug 31 at 4:30
@ClaudeLeibovici thank you. Repaired.
– drhab
Aug 31 at 6:19
@ClaudeLeibovici thank you. Repaired.
– drhab
Aug 31 at 6:19
add a comment |Â
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