18.03 Class 2
, Feb 3, 2010
Numerical Methods
[1] How do you know the value of e?
[2] Euler's method
[3] Sources of error
[4] Higher order methods
[1] The study of differential equations has three parts:
. Analytic, exact, symbolic methods
. Quantitative methods (direction fields, isoclines
....
)
. Numerical methods
Even if we can solve symbolically, the question of computing values
remains. The number
e
is the value
y(1)
of the solution to
y' = y
with
y(0) = 1. But how do you find that in fact
e = 2.718282828459045
....
? The answer is: numerical methods.
(Euler already computed
e
to at least 18 decimal places. It is
now known to some 200 billion places, but I won't write them all out here.)
As an example, take the first order ODE
y' = y^2  x = F(x,y)
with initial condition
y(0) = 1
 an Initial Value Problem, IVP.
For example:
what is
y(1) ?
I invoked the Euler's Method Mathlet, and selected
F(x,y) = y^2  x .
The slope field appears. I selected the initial condition
(0,1) ,
and then invoked "actual."
This solution is one of those trapped in the funnel,
so for large
x ,
the graph of
y(x)
is close to
the graph of
 \sqrt(x) :
y(100) is very close to
10 .
But what about
y(1) ?
The listing says
y(1) = .83 .
How did the computer know this?
[2] The tangent line approximation gives one approach.
Recall:
We have a function
y(x), and a point
(a,y0) on its graph
(so y(a) = y0 ). The tangent line to the graph at this point has slope
y'(a), which I will denote
m_0 .
The tangent line approximation is
y(a+h) ~ y0 + m0 h
The right hand side is the
ycoordinate of a point on the tangent line
directly above or below (a+h , y(a+h)).
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In our case, we do know the derivative of
y
at
x = 0 :
y'(0) = F(0,1) = (1)^2 + 0 = 1
So the tangent line at
(0,1)
has slope
1 , and we find the tangent line
approximation
y(1) ~ 1 + 1 = 0. I showed this on the applet.
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 Spring '10
 Prof.HaynesMiller
 Differential Equations, Numerical Analysis, Equations, Euler

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