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        riddles >> medium >> Find sum
        
(Message started by: tony123 on Jan 9th, 2008, 12:21am)
Title: Find sum
Post by tony123 on Jan 9th, 2008, 12:21am
tan^2(1°) + tan^2(3°) + ... + tan^2(87°) + tan^2(89°)
Title: Re: Find sum
Post by Aryabhatta on Jan 10th, 2008, 5:13pm
Seems like following might work (though there might be mistakes...)
[hide]

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.

[/hide]

[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]
Title: Re: Find sum
Post by towr on Jan 11th, 2008, 12:25am

on 01/10/08 at 17:13:19, 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.
[/edit]
The problem does seem to have suddenly changed; but for the original the answer is around [hide]5310 1/3[/hide] 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.
Title: Re: Find sum
Post by Aryabhatta on Jan 11th, 2008, 1:29am

on 01/11/08 at 00:25:31, towr wrote:
However, the answer you got for the old problem seems to fit the current one rather well.



You are right, I did make a mistake as I was afraid.

The claim that [hide] real part of z90 is zero [/hide] is false for even degrees.
Title: Re: Find sum
Post by Grimbal on Jan 11th, 2008, 1:30am
5310 1/3 can also be written 179*89/3 or (2n-1)*(n-1)/3 with n=90.


on 01/11/08 at 01:29:33, Aryabhatta wrote:
The claim that [hide] real part of z90 is zero [/hide] is false for even degrees.

I thought you meant the [hide]imaginary part of z90 is zero [/hide]...
Title: Re: Find sum
Post by Eigenray on Jan 11th, 2008, 4:33am
[hide]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)).[/hide]
Title: Re: Find sum
Post by temporary on Jan 23rd, 2008, 6:47pm
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.
Title: Re: Find sum
Post by Icarus on Jan 23rd, 2008, 6:55pm
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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif. (Besides which, you would have to skip 90o, 270o, etc., where tan is undefined.)
Title: Re: Find sum
Post by temporary on Jan 23rd, 2008, 8:11pm
Why  undefined? Is the denominator zero?
Title: Re: Find sum
Post by mikedagr8 on Jan 23rd, 2008, 11:52pm

on 01/23/08 at 20:11:05, temporary wrote:
Why  undefined? Is the denominator zero?

No. The tangent is not touched.
Title: Re: Find sum
Post by Icarus on Jan 24th, 2008, 3:41pm

on 01/23/08 at 20:11:05, temporary wrote:
Why  undefined? Is the denominator zero?

yes. cos 90o = 0.
Title: Re: Find sum
Post by mikedagr8 on Jan 24th, 2008, 10:44pm

on 01/24/08 at 15:41:21, Icarus wrote:
yes. cos 90o = 0.

LOL  ;D


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