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Unformatted text preview: e it explicitly.
We have
0 = f (c) = f (a) + m(c − a).
This implies that
−f (a) = m(c − a).
If m = 0, we can divide both sides of the
equation by m to get
−f (a)
= c − a.
m
Finally, adding a to both sides of the equation
yields
f (a)
m
f (a)
= a−
f (a) c = a− Therefore, if f (x) = f (a) + m(x − a) and m = 0, we can easily solve the equation
f (x) = 0.
Note: If m = 0, the graph is a horizontal line, so if f (a) = 0, it does NOT cross the xaxis and NO such c exists.
What do we do if the graph of f (x) is not a line? In this case we can use tangent line approximation (this is formally called
linear approximation ). The following steps outline a general method for using tangent lines to approximate the point c for
f (c) = 0.
Step 1: Pick a point x1 that is reasonably close to a point c with f (c) = 0. (The IVT or a rough plot might be helpful in
ﬁnding such an x1 .)
Step 2: If f (x) is diﬀerentiable at x = x1 , then we can approximate f (x) by the function
Lx1 (x) = f (x1 ) + f (x1 )(x − x1 )
The graph of Lx1 (x) is the tangent line to f (x) at x = x1 . Since, for x values near x1 ,
f (x) ∼ Lx1 (x)
=
it would make sense that the graphs of f (x) and Lx1 (x) cross the xaxis at roughly the same place.
vii Copyrighted by B.A. Forrest (bforrest@sympatico.ca) Newton’s Method in Maple Recall from above that Lx1 (x) is a linear
function. Therefore, if f (x1 ) = 0, the graph of
Lx1 (x) crosses the xaxis at the point x2 where
x2 = x1 − f (x1 )
.
f (x1 ) Step 3: We now repeat the above procedure,
replacing x1 by x2 and using the linear
approximation at x2 , to get a new approximation
x3 = x2 − f (x2 )
f (x2 ) for c. The diagram shows that in this example, x3 is very close to c.
Continuing in this manner, we get a recursively deﬁned sequence
xn+1 = xn − f (xn )
f (xn ) where xn+1 is simply the point at which the tangent line to the graph of f (x) through (xn , f (xn )) crosses the xaxis.
It can be shown that for most nice functions and reasonable choices of x1 , the sequence {xn } converges very
rapidly to a number c with f (c) = 0. Still, it makes sense for us to ask: How do we know how good our approximation
is?
If we want to approximate c to k decimal places of accuracy, at least k decimal places must be carried throughout our
calculations.
The procedure stops when two successive terms, xn and xn+1 , agree to k decimal places. In most case...
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 Winter '09
 Prof.Smith
 Calculus, Equations

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