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Topic: Find sum (Read 1418 times) |
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tony123
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tan^2(1°) + tan^2(3°) + ... + tan^2(87°) + tan^2(89°)
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| « Last Edit: Jan 9th, 2008, 1:46am by tony123 » |
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Aryabhatta
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Re: Find sum
« Reply #1 on: Jan 10th, 2008, 5:13pm » |
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Seems like following might work (though there might be mistakes...)
Let x be any angle in {1,2 ,..., 89, 90} (all in degrees)
Let z = cos x + i sin x
Let c = cos x and s = sin x
Consider z90.
Expanding using binomial theorem, we get
z90 = Sum C(90,k) ck(i*s)90-k
Sum C(90,2m) c2m* (-1)m s90-2m + Imaginary part.
Real part is zero, so dividing by s90 we get
0 = Sum_{m=0 to 45} C(90,2m) (-1)mcot(x)2m
This is a 90 degree polynomial in cot(x) whose roots are cot(1), cot(2), ..., cot(89) and cot(90) = 0
We need the sum of squares of the roots.
So it is C(90,2) = 4005.
[edit]
Hey! last time I saw this it had all of them, now only the odd degrees are there! The working I did is for Sum tan2x where x is from 1,2,3, ..., 89.
[/edit]
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| « Last Edit: Jan 10th, 2008, 5:16pm by Aryabhatta » |
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towr
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Re: Find sum
« Reply #2 on: Jan 11th, 2008, 12:25am » |
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on Jan 10th, 2008, 5:13pm, Aryabhatta wrote:[edit]
Hey! last time I saw this it had all of them, now only the odd degrees are there! The working I did is for Sum tan2x where x is from 1,2,3, ..., 89.
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The problem does seem to have suddenly changed; but for the original the answer is around 5310 1/3 anyway; and I doubt computer imprecision accounts for a difference of over a thousand.
However, the answer you got for the old problem seems to fit the current one rather well.
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| « Last Edit: Jan 11th, 2008, 12:26am by towr » |
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Aryabhatta
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Re: Find sum
« Reply #3 on: Jan 11th, 2008, 1:29am » |
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on Jan 11th, 2008, 12:25am, towr wrote:
However, the answer you got for the old problem seems to fit the current one rather well. |
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You are right, I did make a mistake as I was afraid.
The claim that real part of z90 is zero is false for even degrees.
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Grimbal
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Re: Find sum
« Reply #4 on: Jan 11th, 2008, 1:30am » |
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5310 1/3 can also be written 179*89/3 or (2n-1)*(n-1)/3 with n=90.
on Jan 11th, 2008, 1:29am, Aryabhatta wrote:| The claim that real part of z90 is zero is false for even degrees. |
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I thought you meant the imaginary part of z90 is zero ...
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| « Last Edit: Jan 11th, 2008, 1:33am by Grimbal » |
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Eigenray
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Re: Find sum
« Reply #5 on: Jan 11th, 2008, 4:33am » |
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Setting the real (or imaginary) part to 0 gives the sum over the odd (or even) degrees to be C(90,2))/C(90,0) (or C(90,3)/C(90,1)).
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temporary
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Re: Find sum
« Reply #6 on: Jan 23rd, 2008, 6:47pm » |
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Can that be written as a geometric series? I doubt it can be answered if it goes on forever. If it stops at 89, the answer can be defined by calculator.
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Icarus
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Re: Find sum
« Reply #7 on: Jan 23rd, 2008, 6:55pm » |
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The series is not geometric.
By the periodicity of tan, it repeats itself after 180o. Since the terms are all non-negative, it cannot converge if continued to  . (Besides which, you would have to skip 90o, 270o, etc., where tan is undefined.)
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| « Last Edit: Jan 23rd, 2008, 6:55pm by Icarus » |
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temporary
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Re: Find sum
« Reply #8 on: Jan 23rd, 2008, 8:11pm » |
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Why undefined? Is the denominator zero?
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mikedagr8
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Re: Find sum
« Reply #9 on: Jan 23rd, 2008, 11:52pm » |
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on Jan 23rd, 2008, 8:11pm, temporary wrote:| Why undefined? Is the denominator zero? |
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No. The tangent is not touched.
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Icarus
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Re: Find sum
« Reply #10 on: Jan 24th, 2008, 3:41pm » |
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on Jan 23rd, 2008, 8:11pm, temporary wrote:| Why undefined? Is the denominator zero? |
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yes. cos 90o = 0.
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mikedagr8
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Re: Find sum
« Reply #11 on: Jan 24th, 2008, 10:44pm » |
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on Jan 24th, 2008, 3:41pm, Icarus wrote:
LOL 
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