Date: 08/10/99 at 23:16:01 From: Anonymous Subject: Arcsin and Taylor Series In my last question, I asked about how to find arcsin from only sin, cos, and tan. But the page you directed me to reads: "The basis of the calculation is a Taylor series: arcsin(x) = x + 1/2 (x^3/3) + (1/2)(3/4)(x^5/5) + [1/2)(3/4)(5/6)(x^7/7) + ... " I noticed a pattern, but when does it stop? How do you use this formula? In addition, for example, does "x^3/3" reduce to "x^1" or does the order of operations straighten that out? Can you help me; I only want to find the angle from its trigonometric ratio.
Date: 08/11/99 at 08:08:22 From: Doctor Jerry Subject: Re: Arcsin and Taylor Series The Taylor series for arcsin(x) continues indefinitely; it is an infinite series. It would not be an efficient method for calculating arcsin(x), except, perhaps, for small x. The catch about using series approximations is to know how far out to go in the series sum to achieve a certain accuracy. There are advanced methods for finding polynomials that approximate arcsin(x). For example, this approximation formula came from page 81 of the Handbook of Mathematical Functions, by Milton Abramowitz and Irene Stegun: arcsin(x) = pi/2 - sqrt(1 - x)(a0 + a1*x + a2*x^2 + a3*x^3), where a0 = 1.5707288 a1 = -0.2121144 a2 = 0.0742610 a3 = -0.0187293 Alternately, you could interpolate, using the sine function and arcsin for a few known values. This could be programmed without great difficulty. Here is an example. Suppose you want, working in radians, arcsin(0.7). Using a brief sine table, precalculated and stored, determine arcsin(0.7) bounded below and above by sin(0.75) = 0.681638760023 sin(0.80) = 0.717356090900 The sine curve is approximately the line joining (0.75, 0.681...) and (0.80, 0.717...). Write the equation of this line, solve it for x in terms of y, and then substitute 0.7 for y: x = 1.39988(y - 0.145879) Replacing y by 0.7, we find x = 0.775704, In fact, arcsin(0.7) = 0.775397, so by this interpolation we have achieved arcsin(0.7) to three significant digits, or within an accuracy of about 0.04%. We could get closer: determine that sin(0.77) = 0.696135238627 sin(0.78) = 0.703279419200 and continue interpolating until you achieve the accuracy you desire. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/
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