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Find sum

  1. Jan 8, 2008 #1
    [tex]\tan^2{(1^\circ)}+\tan^2{(3^\circ)}+\tan^2{(5^\circ)}+\ldots+\tan^2{(89^\circ)}[/tex]
     
  2. jcsd
  3. Jan 8, 2008 #2

    berkeman

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    Moved to Homework Help forums. santa, you need to show us your own work, before we can offer tutorial assistance.
     
  4. Jan 8, 2008 #3

    Gib Z

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    The Gib-Z Constant.
     
  5. Jan 8, 2008 #4

    berkeman

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    Okay, I was informed via friendly PM that santa is offering this kind of question as a puzzle, not as a homework or coursework question. I'm moving santa's threads back to General Math for now.

    santa, maybe preface your puzzle posts with a brief note, so that other Mentors who are stopping by don't move your puzzles to Homework Help like I did. Thanks!
     
  6. Jan 8, 2008 #5

    morphism

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    4005? (Otherwise known as The Gib-Z Constant.)
     
  7. Jan 8, 2008 #6

    rock.freak667

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    Hm...this seems difficult with my limited knowledge as I only know AP and GP...


    So far all I have simplified that to is

    [tex]\sum_{n=0} ^{44} tan^2(2n+1)[/tex]


    and that really didn't help...
     
  8. Jan 8, 2008 #7

    Gib Z

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    The number is probably transcendental, since the the sine or cosine (and hence tangent) of x is transcendental if x is not a rational multiple of pi. My conjecture is that the squares and sum of these transcendental numbers gives another ugly transcendental number, which until further investigation remains to be named after me =]
     
  9. Jan 8, 2008 #8

    morphism

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    Here's an outline of what I did:

    Start with De Moivre and the binomial formula:
    [tex]\cos nx + i \sin nx = (\cos x + i \sin x)^n = \sum_{k=0}^{n} i^k {n \choose k} \cos^{n-k} x \sin^k x[/tex]

    In particular,
    [tex]\cos nx = \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k {n \choose 2k} \cos^{n-2k} x \sin^{2k} x[/tex]

    [tex]\Rightarrow \frac{\cos nx}{\cos^n x} = \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k {n \choose 2k} \left(\tan^2 x\right)^k[/tex]

    Now is where it gets tricky. Notice that if we let n=90, we get:
    [tex]\sum_{k=0}^{45} (-1)^k {90 \choose 2k} \left(\tan^2 x\right)^k = 0[/tex]

    Plug x=1,3,...,89 into this, and note that the 45 numbers tan^2(1), tan^2(3), ..., tan^2(89) are all distinct. Thus they are precisely the roots of the degree 45 polynomial
    [tex]\sum_{k=0}^{45} (-1)^k {90 \choose 2k} t^k[/tex]

    So from this we get a bunch of equalities, like

    [tex]\sum_{n=0}^{44} \tan^2(2n+1) = {90 \choose 2\cdot44}= 4005[/itex]

    and

    [tex]\prod_{n=0}^{44} \tan^2(2n+1) = {90 \choose 0} = 1[/itex]
     
    Last edited: Jan 8, 2008
  10. Jan 8, 2008 #9

    Gib Z

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    Did you do that all in radians or degrees :( ?
     
  11. Jan 8, 2008 #10

    morphism

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    Degrees. I'm actually worried that I made a degree-radian slip somewhere. Maple seems to agree with my formula for the product, but not the sum.

    Edit:
    Nevermind - Maple does agree - I just forgot to square my tans! I got 4004.999892 for the sum and 0.9999999699 for the product, which is good enough.
     
    Last edited: Jan 8, 2008
  12. Jan 8, 2008 #11

    Gib Z

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    Taking the last 5 terms manually yields about 4015 :( Already over, but you can't be b far off since the terms are decreasing pretty fast...
     
  13. Jan 8, 2008 #12

    morphism

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    I get ~3883 for the first five terms!? You probably added in an even number somewhere. :tongue2:
     
  14. Jan 8, 2008 #13

    morphism

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    I'm not really satisfied with my solution, but haven't been able to find anything better. I did notice some curious symmetries though:

    S = tan2(1) + tan2(3) + ... + tan2(87) + tan2(89) = tan2(90-89) + tan2(90 - 87) + ... + tan2(90-3) + tan2(90-1) = cot2(1) + ... + cot2(89)

    2S = (tan2(1) + tan2(3) + ... + tan2(87) + tan2(89)) + (cot2(1) + ... + cot2(89))
    = (tan2(1) + cot2(1)) + ... + (tan2(89) + cot2(89))

    Now everything on the left of the centre (that is: (tan2(45) + cot2(45)) = 2) is equal to the corresponding thing on the right of it, e.g. (tan2(1) + cot2(1)) = (tan2(89) + cot2(89)).

    But I haven't been able to get very far with this, and I'm too sleepy to continue trying. Maybe someone who's a trig magician can take it from here!
     
  15. Jan 9, 2008 #14

    Gib Z

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    I readded the last 5 terms and I got around 3909.5, which still isn't what you have but I'll just take your calculation. What does Maple give for the sum?
     
  16. Jan 9, 2008 #15

    mda

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    Looks to me like morphism has it nailed. Maple gives me 4005 to more than 1000sf, and the sum for the largest 5 terms is 3883.069... even manually on a calculator.
     
  17. Jan 9, 2008 #16

    Gib Z

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    Looks like problem is solved then =] By the incredible work of morphism !
     
  18. Jan 9, 2008 #17

    CRGreathouse

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    Likewise: on Pari, to 10,000 digits, I also get 4005. That took almost 30 seconds!
     
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