Incomplete Elliptic Integrals of the second kind -- evaluation

Discussion in 'Differentiation and Integration' started by Sequoia, Apr 6, 2020.

  1. Sequoia

    Sequoia

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    I am programming as a hobby and trying something with elliptical orbits of satellites.
    I need to understand how to measure the length of an arc on an ellipse, e.g. from 0 to pi/3.
    Now I've come upon Elliptic Integrals and an associated brick wall.

    For the "Complete Elliptic Integral of the second kind" the length of an arc from 0 to pi/2, I've found a series the program can use to approximate the value of this kind of arc.

    However, for the INCOMPLETE sort, there is nowhere a description of how the value is calculated, no series, no formula. Nothing.

    They tell you in most places either :
    1. Look it up
    or
    2. use a website to calculate it, such as https://keisan.casio.com/

    The casio.com website can indeed calculate the value where k = 0.6 and the limits are 0 to pi/3 bringing the value:
    0.98922159...... so there must be an infinite series one utilizes to calculate this.

    It is most certainly not an elementary function. So, what is the series?
     
    Sequoia, Apr 6, 2020
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  2. Sequoia

    Country Boy

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    I am not clear why you say "there must be an infinite series one utilizes to calculate this." I would be inclined to assume that they used a numerical integration.

    The "incomplete elliptic integral of the second kind" for k= 0.6 and limits 0 and [math]\pi/3[/math] is
     
    Country Boy, Jan 19, 2022
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  3. Sequoia

    Country Boy

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    I accidently terminated my previous post:

    I am not clear why you say "there must be an infinite series one utilizes to calculate this." I would be inclined to assume that they used a numerical integration.
    You want to find int_0^(pi/3) sqrt(1- 0.36 sin^2(x)) dx.
    The "incomplete elliptic integral of the second kind" for k= 0.6 and limits 0 and pi/3 is int_(0 to pi/3) sqrt(1- 0.6 sin^2(x))dx. The "trapezoid rule" divides the interval into n equal sections, raises vertical lines at each endpoint from the x-axis to the curve, then replaces the curve above each interval by the straight line between the two points. So the area under the curve is approximated by n trapezoids. Here, the ith trapezoid has "height", on the x-axis, from (pi/3n)(i-1) to (pi/3n)i, a length of pi/3n. the "base" on the left is sqrt(1- 0.36 sin^2((pi/3n)(i-1))) and the "base" on the right is sqrt(1- 0.36 sin^2((pi/3n)i).

    The area of each such trapezoid is the average of the two "bases", [sqrt(1- 0.36 sin^2((pi/3n)(i-1)))+ sqrt(1- 0.36 sin^2((pi/3n)i)]/2, times the "height", pi/3n, so (pi/3n)[sqrt(1- 0.36 sin^2((pi/3n)(i-1)))+ sqrt(1- 0.36 sin^2((pi/3n)i)]/2.

    The integral is the sum of those for i= 1 to n.
     
    Country Boy, Jan 19, 2022
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