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In this segment, we are going to
derive secant method of finding roots
of a nonlinear equation.
So, if somebody gives you a nonlinear equation like this, f of x
equal to 0, which basically implies that if I draw it graphically that you're going to
plot the function f of x of that equation f of x equal to 0, and you're going
to draw it like this, then what you are trying to do is you're trying to find out
where does the function become 0, and that becomes the root of
f of x equal to 0, or some people might say it
is the zero of the function f of x, which is the same thing.|So that's what we want
to be able to do, and let's go ahead and see how we can derive secant method and what it is based on.
So, let me go ahead and draw . . . draw this . . . this
plot again, so that we can show what secant method is all about.
So, let's suppose somebody gives you a function like this, and says . . .
of the function . . . of the equation f of x equal to 0, and wants
you to find out the root of this equation which is right here.|So, what secant method does is
that it says, hey, let me go ahead and take some initial guess, let's suppose xi,
let me go ahead and take another initial guess, let's suppose
x-sub-i-minus-1 here, let's suppose, and what I'm going to do is I'm going
to draw a secant line, I'm going to draw a secant line between the
two points, so this is one point and this is another point.|So this point here is nothing
but xi, and the value of y is the value of the function at that particular point.
This here, again, is x-sub-i-minus-1 and the value of the function at
x-sub-i-minus-1.|So what you want to do is you take the secant line and you continue this secant
line now, and see where it crosses the x-axis.|So that's what
you are doing, you're taking this secant line which you have, so you're taking this secant line which you have,
it's not looking exactly straight, but it's a one secant straight line, and you
find out where it crosses the x-axis, so that's where it's crossing the x-axis, so it's at x-sub-i-plus-1,
and that's going to become your estimate of the root of the equation.
So this becomes the estimate of the second . . . next estimate of the root of the equation, starting with this estimate
and this estimate here, and then you're going to do the same thing, you're going to draw a secant line between this
point and this point here, and, again, see where the secant crosses the x-axis, and
that becomes your root.|So, let's suppose if I call this to be point A,
I call this to be B, I call this to be C, I call
this to be D, and I call this to be E.|Now we can very well see
that triangle ABC
and triangle ADE
are similar.|Now what does that mean is
that you can very well see that they are similar, this triangle ABC
and triangle ADE are similar, because all the angles
are the same.|So, in similar triangles, what you can very well say
is that then this height here divided by this height will be equal to
this width divided by this width right here.|So what I'm trying to say is that
DE, so if I take this height, DE, and I divide it by this height,
which is BC, I'll put it equal to then this length of
AD, and then divided by this length, which is AB.
So we've got, from similar triangles, that's what we get from similar
triangles, that the two heights are proportional to the two widths
which we are seeing right here.|So, but what is . . . what is DE?|DE is
nothing but the value, or the length of DE is nothing but the value of the function at xi.
And what is BC?|BC is nothing but the value of the function
at x-sub-i-minus-1.|Now, what is AD,
what is the length of AD?|The length of AD is nothing but xi minus
x-sub-i-plus-1, because it is this point here, D, and A is
x-sub-i-plus-1, so you've xi and x-sub-i-plus-1, so it's the difference between the two,
and same thing, AB, which is this length right here, is the difference between
this point and this point, which is, again, x-sub-i-minus-1
minus x-sub-i-plus-1, so that's what we're getting
from there.|So you can very well see that, now, in this case,
if I am choosing x-sub-i-minus-1 to be one of the initial guesses,
xi to be the other initial guess, then the only unknown which I have is x-sub-i-plus-1.
So, bottom line is to be able to solve this simple linear equation for x-sub-i-plus-1,
and this is what I'm going to get, if I take x-sub-i-plus-1, I'm going to get x-sub-i-plus-1,
after a little bit of algebra.
.
So that's what I get as the simplification
by writing down the unknown, x-sub-i-plus-1, in terms of the
other two . . . the two guesses, xi and x-sub-i-minus-1, and the values of the function
at those particular points.|And this is nothing but the formula for the secant method.
So, secant method has been derived here by
taking two initial guesses, x-sub-i-minus-1 and xi, and you are able to get x-sub-i-plus-1,
so you have to start with two initial guesses.|Again, keep in mind that although you're going to start with two
initial guesses, you still . . . this stall falls under the category of
the open method, just like other methods, such as Newton-Raphson method, and
the reason why that is so is because these two initial guesses, or any of the estimates,
two estimates which you're going to use for the roots or root of the equation f of x equal to 0,
they don't necessarily have to bracket the root, they can be outside of the bracket, they can be . . .
they can be bracketing the root, so that's why it's still called the open method.
And that's the end of this segment.
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