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### Creating a Smooth Acceleration in Animation

Date: 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/

Associated Topics:
High School Analysis
High School Equations, Graphs, Translations

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