Creating a Smooth Acceleration in AnimationDate: 05/09/2001 at 21:30:56 From: William Fex Subject: 1=X-(X^(119/120)) I need to solve for X I have a formula for the exponential increase of a number from 0 to the target number: ((x+1)^y)-1 where y = {0, 1/120, 2/120, 3/120, ..., 120/120} The first answer to this is ((x+1)^y)-1 = 0 then ((x+1)^y)-1 = x, ... This is not difficult to solve for at all. The problem arises when I am trying to figure out a way to set the value of X to achieve a specific difference between the last two sets in this problem. I am using this for an animation, and I want the object in the animation to accelerate smoothly over the course of 120 frames, and then in the next 120 frames to move at a constant rate of speed. I am using X to represent the location of the object. In the next set of frames I will simply move at x*y, where y = {0, 1/120, 2/120, etc.} What I need to know is how to determine how far I need to move in those first 120 frames to obtain an exact speed. For this example I will seek a movement rate of exactly 1 unit per frame. I have figured out that the number I am seeking can be simplified to 1 = X-(X^(119/120)). This if solved will tell me what X must equal in the formula ((x+1)^y)-1 for the difference in those last two frames where y = 119/120 and then y = 120/120 to be exactly 1. HELP! I lack the math skills to understand how to solve the equation: 1 = X-(X^(119/120)). I figured out a slew of other problems on my own and I am largely self-taught, but my high school math never went beyond Algebra I and Geometry I, and my college math was minimal. This is for a hobby, but it is making my brain bleed! Help me Doctor Math! Tell me how I go about solving a monster like this. Several of my friends have run screaming from this problem. Date: 05/11/2001 at 13:18:06 From: Doctor Douglas Subject: Re: 1=X-(X^(119/120)) I need to solve for X Hi Bill, and thanks for writing to Ask Dr. Math. This appears to be a transcendental equation, which means that there is no solution that is easily expressible in terms of the functions that we usually encounter (say, on a calculator). However, that does not mean that no methods exist for *computing* the solution. For example, as long as we have access to a calculator that can compute powers (y^x), we can do the following simple procedure: we graph the function f(x) = X - X^(119/120) - 1 and see where it crosses zero. We can show that for x = 1, f(x) = -1 and for x = 50, f(50) = 50 - 50^(119/120) - 1 = +0.603. I used a calculator for this last computation. Since f(1) and f(50) have opposite signs, and since the function f is "continuous," meaning that its graph can be drawn as a single line without lifting the pencil from the paper, I know that f must equal zero somewhere for some x between 1 and 50. This is known as "bracketing the root." By trying a few values of x on a calculator, we can focus in on a solution for x by making the bracket tighter and tighter. Once we have a bracket, we just pick a point in the middle of the interval and evaluate the value of f there. Either f = 0 (we're done but this is unlikely to happen) or f < 0 or f > 0. Whichever one of these latter cases occurs, it will have a sign opposite to one of the two function values that we started with - in other words, we will still have the root bracketed, but with an interval that is smaller. So we can do this process repeatedly to focus in on the root. f(25) = -0.3383 pick something in the middle; bisecting works well f(40) = 0.2109 ah, a solution is bracketed by 25 and 40 f(35) = 0.0218 between 25 and 35, and probably much closer to 35 so make our next guess 33, not 30 f(33) = -0.0523 solution is bracketed by 33 and 35 f(34) = -0.0154 bracketed by 34 and 35 . . . . . . If I keep doing this process, I obtain x = 34.415159... I verified this solution by plugging it back into the equation for f(x). Note how it isn't necessary to choose the interior point to be the midpoint (25 is not exactly (50+1)/2, for example) but we were free to choose numbers that were relatively convenient. We also took a chance by guessing 33 rather then 30 - this kind of procedure helps save on the total number of function evaluations, so that we don't do too much work. I hope that this shows you how we can obtain a numerical solution to f(x) = 0 when f is continuous. The method is not much more than trial and error, guessing a point and noting whether f is positive or negative at that point, guided by an intelligent way of determining the next guess. Incidentally, many sophisticated root-finding methods are used in mathematical computing. Some of them involve various enhancements for how the guessing procedure can be made more efficient if you know something about the function. I hope this helps. Please write back if you have more questions about this. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ |
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