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Posers and Puzzles

Posers and Puzzles

  1. Subscriber joe shmo On Vacation
    Strange Egg
    28 Mar '09 21:35 / 2 edits
    this is consuming to much of my time, I may be overlooking somthing simple....Ive been trying to use calculus to solve it ( also unsuccessfully)

    ok

    0 = 1-(2x)(arctan(x))

    ive been trying to do sort of an algebriac substituion...trying to integrate to obtain the arclength in terms of x .....which I relly dont think will actually get me anywhere. be interested to see just how much time ive wasted
  2. Subscriber joe shmo On Vacation
    Strange Egg
    31 Mar '09 04:32
    After much work trying to involve integration to determine arclength ( which will be my next step) Ive decided to try and simplify the process by using the chord length as an approximation

    so Ill start by trying to find a right triangle whose tangent is "x"

    this yields a triangle with the following sides

    HYP : sqrt(x^2 + 1)
    OPP : x
    ADJ: 1

    such that the radius of the circle is = to sqrt(x^2 + 1 )

    the arclength in question can be estimated with a chord length

    cord length = sqrt[(sqrt(x^2 +1) - 1)^2 + x^2]= sqrt(2x^2 - 2sqrt(x^2 +1)+2 )

    using the above function as an approximation for "arctanx", and for the variable "x" = approx acrctanx/radius

    i plug it all into the original equation

    y =1-(2x)(arctanx) is approximatley = to y = -4sqrt(x^2+1) +5

    now the approximation function becomes less accurate as x increases/decreases because of the difference between chord length and arclength in large angles, but I got lucky....

    so, can integration of the arclength of the circle in question be done without introducing more trancendental functions....I cant seem to figure a way to do it?
  3. Standard member PBE6
    Bananarama
    31 Mar '09 17:45
    Originally posted by joe shmo
    After much work trying to involve integration to determine arclength ( which will be my next step) Ive decided to try and simplify the process by using the chord length as an approximation

    so Ill start by trying to find a right triangle whose tangent is "x"

    this yields a triangle with the following sides

    HYP : sqrt(x^2 + 1)
    OPP : x
    ADJ: 1

    such ...[text shortened]... without introducing more trancendental functions....I cant seem to figure a way to do it?
    The arclength of the sector swept out is fairly easy to calculate. All you really need to do is read the following equation out loud:

    arctan(x) = 1/2x

    "The angle whose tangent is x is 1/2x."

    From that, it's easy to see that the angle A = 1/2x, and the corresponding right angle triangle has side lengths 1, x and SQRT(1+x^2) as you stated above. From the definition of a radian, the arc length is therefore:

    L = (A/2pi)*(2pi*r) = A*r = (1/2x)*SQRT(1+x^2)

    However, I'm not sure this really helps you find an analytical answer to your equation. I'm not even sure it's possible, given that arctan(x) seems to be defined in terms of an infinite series:

    arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + ...
  4. Standard member Palynka
    Upward Spiral
    31 Mar '09 18:07 / 1 edit
    x = 2.331122368031441.

    Or so Matlab tells me.

    Edit - You have to love Matlab when you want to solve a fixed point problem.
  5. 27 May '09 11:39
    use newton-ramphson method and find an approximation..
  6. Standard member PBE6
    Bananarama
    27 May '09 13:29 / 1 edit
    Originally posted by blacknight1985
    use newton-ramphson method and find an approximation..
    He could get an answer that way, but it wouldn't be an explicit algebraic solution he's looking for.

    EDIT: Oops! He doesn't actually care to find an explicit algebraic solution. An approximation method, like Newton-Raphson, should be sufficient then.
  7. Subscriber joe shmo On Vacation
    Strange Egg
    27 May '09 13:46
    Originally posted by blacknight1985
    use newton-ramphson method and find an approximation..
    Yeah, I had already done tangentline approximation. I just wanted to explore the concept a little more... Anyway, as PBE6 stated, arctanx is defined by an infinite series which seems to draw the line on the journey, or just maybe......... naaaa
  8. Subscriber joe shmo On Vacation
    Strange Egg
    27 May '09 13:48 / 1 edit
    Originally posted by PBE6
    He could get an answer that way, but it wouldn't be an explicit algebraic solution he's looking for.

    EDIT: Oops! He doesn't actually care to find an explicit algebraic solution. An approximation method, like Newton-Raphson, should be sufficient then.
    your were right the first time, I was trying to use appoximations as "stepping stones", i was looking for an analytical result.
  9. Standard member PBE6
    Bananarama
    27 May '09 14:39
    Originally posted by joe shmo
    your were right the first time, I was trying to use appoximations as "stepping stones", i was looking for an analytical result.
    Oh.

    Double-oops then!
  10. 27 May '09 19:48 / 2 edits
    Originally posted by joe shmo
    this is consuming to much of my time, I may be overlooking somthing simple....Ive been trying to use calculus to solve it ( also unsuccessfully)

    ok

    0 = 1-(2x)(arctan(x))

    ive been trying to do sort of an algebriac substituion...trying to integrate to obtain the arclength in terms of x .....which I relly dont think will actually get me anywhere. be interested to see just how much time ive wasted
    Sorry, I was mistaken. So I deleted the whole posting...
  11. 28 May '09 07:12
    Originally posted by Palynka
    x = 2.331122368031441.

    Or so Matlab tells me.

    Edit - You have to love Matlab when you want to solve a fixed point problem.
    but wen u substitute it back, it doesnt give zero on rhs..my answer is 0.764 to 3 decimals gives 0.003 on rhs..i used newtons method..:-)
  12. Standard member Palynka
    Upward Spiral
    28 May '09 18:00
    Originally posted by blacknight1985
    but wen u substitute it back, it doesnt give zero on rhs..my answer is 0.764 to 3 decimals gives 0.003 on rhs..i used newtons method..:-)
    LOL, right you are. Matlab is great, you just have to not make a coding mistake.
  13. 29 May '09 17:38
    Originally posted by Palynka
    LOL, right you are. Matlab is great, you just have to not make a coding mistake.
    yeah u bet..i love MATLAB too.. big fan..
  14. 31 May '09 11:09
    u guys know www.wolframalpha.com?
    it said the answer is: +- 0.765...
  15. 31 May '09 19:35
    Originally posted by crazyblue
    u guys know www.wolframalpha.com?
    it said the answer is: +- 0.765...
    so it seems my manual calculation is right...