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We ended Page 3 with an initial value problem to be solved: Find ** ** *v* = *v*(*t*)** ** so that

More generally, our problem is to solve any initial value problem of the form

We will calculate approximate values for the velocity ** ** *v* at ** ** *n* equally spaced points in some fixed time interval. Our procedure is simple: We repeatedly calculate a *rise* in ** ** *v* as *slope* x *run*. Then we add the *rise* to the current value of ** ** *v* to get the next value of ** ** *v*.

Our goal is to estimate the velocity *v*(*t*)** ** at times

Our estimated velocity values at these times will be denoted by

Our method for estimating the velocity values will be recursive, i.e., ** ** *v*_{k}** ** will be calculated from the preceding ** ** *v*_{k--1}** ** for each ** ** *k* = 1, 2, 3, ... .

How do we obtain ** ** *v*_{1}** ** from ** ** *v*_{0}, the initial velocity? We will answer this in a geometric fashion. We will look at the graph of velocity versus time on the ** ** (*t*,*v*)-plane. The following figure shows a graph of the starting situation: the initial velocity ** ** *v*_{0 } is shown as a vertical line segment of length ** ** *v*_{0 } at the starting time ** ** *t*_{0 } = 0.

We now include the graph of ** ** *v* versus ** ** *t*. Our next velocity value, ** ** *v*_{1}, is shown as the length of a vertical line segment at time ** ** *t*_{1}.

However, the value of ** ** *v*_{1 } is not known to us, and hence we will estimate its value. We do this by drawing the tangent line to the graph at ** ** *t*_{ } = *t*_{0}. Follow this tangent line to the point ** ** *P*, the top of a vertical line segment that approximates ** ** *v*_{1}.

We can compute the length of this new line segment: We separate the line segment into two pieces -- the bottom piece having length ** ** *v*_{0}, and the top piece being the rise of a right triangle with Using

we see that rise** ** equals slope ** ** times . Hence,

(We use the symbol to mean "almost equal to".)

This is the key to **Euler's Method** for approximating the solution of an initial value problem. It's valuable because the slope (of the tangent line) equals the derivative *dv/dt*, which is given by our original differential equation when ** ** *t*_{ } = *t*_{0 } and ** ** *v*_{ } = *v*_{0}:

Substituting this value of the slope into the preceding equation, we find

Great! This gives us a method for going from ** ** *v*_{0 } to ** ** *v*_{1}. But how do we go from ** ** *v*_{1 } to ** ** *v*_{2}? Easy -- we use the same equation, only with ** ** *v*_{0 } and** ** *v*_{1 } replaced by ** ** *v*_{1 } and ** ** *v*_{2}:

In general, to go from ** ** *v*_{k - 1 } to ** ** *v _{k }* we have

This equation, along with the initial value ** ** *v*_{0 } = 0** ** and the assignment of a value to the step size , plays the central role in our computations.

David A. Smith and Lawrence C. Moore, "Raindrops - Euler's Method," *Loci* (December 2004)

Journal of Online Mathematics and its Applications