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😵

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?

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?

x = 2.331122368031441.

Or so Matlab tells me.

Edit - You have to love Matlab when you want to solve a fixed point problem.

use newton-ramphson method and find an approximation..😀

Originally posted by blacknight1985
use newton-ramphson method and find an approximation..😀

Originally posted by blacknight1985
use newton-ramphson method and find an approximation..😀

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.

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.

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😵

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.

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..:-)

Originally posted by Palynka
LOL, right you are. Matlab is great, you just have to not make a coding mistake. 😵

u guys know www.wolframalpha.com?
it said the answer is: +- 0.765...
😀

Originally posted by crazyblue
u guys know www.wolframalpha.com?
it said the answer is: +- 0.765...
😀