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Another easy method to apply is to pick two arguments, x0
and x0' that you hope are near the solution you
want, evaluate the function at them and assume that the straight
line defined by the two values and arguments is reasonable
approximation to f. You can then find where that straight
line meets the x axis, and use it to replace one of the two
initial values.
In doing this there are two cases; one in which the value
of f has the same sign at both points. This means that their
straight line will not meet the x axis between your guesses,
and then you might as well use the new point to replace the
one of the two others furthest from it.
If, on the other hand, the values of f at your two arguments
have the opposite, sign, you should replace the old argument
at which f has the same sign as it has at the new one, so
that the values of f will have the opposite sign on the next
turn as well.
This method has the advantage over Newton’s that once you
find arguments whose values of f have the opposite sign, you
must home in on a solution between those arguments. (a solution
being a point at which f changes sign; for a continuous f
this is a point at which f is 0; functions like tan x change
signs at arguments at which they are infinite (or more properly,
undefined) as well as those at which they are 0.
It has the disadvantage that there are functions for which
the true derivative between your guesses is far from a straight
line, and you gain very little in each iteration. It can then
be painfully slow to converge.
The equation used for finding xi +1 given the
two values xi and say xi' that you use
to compute it is:
If you compare this with the previous method, this corresponds
to approximating the derivative in Newton’s method by using
values at the previous two guesses, rather than by the last
guess and it plus d.
This equation can be solved and we can find a new xi
easily as before. With a small amount of effort we can figure
out which of the old arguments to discard; the following setup
does it:
In s2 and t2 put your initial guesses. In u2 put "= f(s2)"’
and copy that into v2.
Now put "= s2 -(s2-t2)*u2/(u2-v2)"in s3, and "=
if(abs(s3-t2)<abs(s3-s2),t2,s2)" in w2.(this puts
the old argument closer to the new one in w2).
Copy s3 and u2 down, and now put "= if(u2*v2>0,w2,if(u2*u3>0,t2,s2))"
in t3 and copy t3 w2, and v2 down.
This last horrible instruction makes t3 hold the nearer old
argument to s3 if u2 and v2 had the same sign, and otherwise
puts the old argument whose value of f has the opposite sign
to u3 in t3.
Exercise 13.8 How does this scheme compare to the others
on your examples?
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