Numerical
Approximation of Zeroes: The Midpoint Method (page
1 of 2)
You can always find the exact zeroes of a quadratic
equation, because you have a formula: The Quadratic Formula. There are formulas for cubic and quartic
equations, but they're so complicated that you'll probably never see them, let alone use them.
And there are no general formulas for polynomial equations of degrees higher than 4.
Most of the time, when they ask you to find the
zeroes (x-intercepts,
solutions, roots) of a polynomial equation, they'll have given you a polynomial that is solvable
in some straightforward manner. But most polynomials (in "real life") aren't solvable.
How do you find their zeroes? You don't, exactly; instead, you find a numerical approximation to
some degree of accuracy. (This is the sort of answer you get when you punch buttons on your graphing
calculator to find the root of a graphed equation.) How does this work?
The secret to approximating zeroes is to use the "continuity
property" of polynomials. This property says that, if your polynomial equals, say, 5
at some value of x
and equals, say, 10
at some other value of x,
then the polynomial takes on every value between 5
and 10
because polynomials are continuous (connected) lines. They don't have gaps or jumps or angles or
elbows. They don't do anything hinky. They're just nice smooth solid curvy lines.
In particular, if your polynomial is negative (below
the x-axis)
at one spot and positive (above the x-axis)
at another spot, then it has to be zero somewhere in between, because zero is between the negatives
and the positives.
For instance, if you've got the polynomial
x5 + x3
– 3x – 2, you can quickly see from
a graph that it is negative (below the axis) at x
= 1 and positive (above the axis) at x
= 2. For x
= 1, y
= –3; for x
= 2, y
= 32. The polynomial is zero (it crosses
the axis) somewhere in between.
(In practice, you'll probably be given x-values
to use as your starting points, rather than having to find them from a graph. One of the x-values
will give a negative value for the polynomial and the other will give a positive value. This is
similar to when you use your calculator to find zeroes on a graph, and the calculator asks you
to pick left- and right-hand bounds for the zero.)
There are two methods for narrowing in on a polynomial's
zero: using midpoints, and picking clever values for x.
I'll show midpoints first.
Find the root of the polynomial x5
+ x3 – 3x – 2,
accurate to three decimal places.
Right now, I only know that the zero is somewhere
between x = 1 and
x = 2.
To get closer to the zero by using midpoints, I will split the difference, literally: I'll try x
= 1.5.
At x
= 1.5, I get y
= 4.46875. The actual y-value
isn't as important as the sign: since the polynomial is positive here, the graph is above
the axis. Since the polynomial is positive at x
= 1.5 and x
= 2, but is negative at x
= 1, I know that the zero has to be
between x = 1 and
x = 1.5.
(The new interval endpoints, bracketing
the zero, are marked in red on the graph.)
So I'll split the difference again. The
halfway point between x
= 1 and x
= 1.5 is x
= 1.25. At this point, y
= –0.745117, approximately. The important
thing to note is that this is a negative y-value.
The polynomial is positive at the x
= 1.5 endpoint, so the new bounds
on my interval are x
= 1.25 and x
= 1.5.
This interval is halved
by x = 1.375.
(Just take the average of the two x-values
to find your midpoint.) This gives me y
= 1.38950, approximately, which
is positive. So I'll keep x
= 1.25 as the lower endpoint and
use x = 1.375
as the new upper boundary.
The midpoint between x
= 1.25 and x
= 1.375 is x
= 1.3125, which gives y
= 0.218389, approximately. Since the polynomial is
positive here, I'll keep x
= 1.25 as the lower boundary and use x
= 1.3125 as the new upper boundary.
The midpoint between x
= 1.25 and x
= 1.3125 is x
= 1.28125, which gives y
= –0.287664,
approximately. Since the polynomial is negative here, I'll keep x
= 1.3125 as the upper boundary and use x
= 1.28125 as the new lower boundary.
The midpoint between x
= 1.28125 and x
= 1.3125 is x
= 1.296875, which gives y
= –0.40913,
approximately. Since the polynomial is negative here, I'll keep x
= 1.3125 as the upper boundary and use x
= 1.296875 as the new lower boundary.
As you can see, I'm getting a lot closer to where the
zero is.
The midpoint between x
= 1.296875 and x
= 1.3125 is x
= 1.3046875, which gives y
= 0.87143, approximately. Since the polynomial is
positive here, I'll keep x
= 1.296875 as the lower boundary and use x
= 1.3046875 as the new upper boundary.
The midpoint between x
= 1.296875 and x
= 1.3046875 is x
= 1.30078125, which gives y
= 0.0227196, approximately. Since the polynomial is
positive here, I'll keep x
= 1.296875 as the lower boundary and use x
= 1.30078125 as the new upper boundary.
You've probably got the idea by now, so I'll just show the rest
of my computations:
The last midpoint becomes the new lower boundary
of the interval, and I now have bracketed the polynomial's zero between x
= 1.299316407 and
x = 1.299804688.
Since these two x-values
agree in the first three decimal places, I'm done.
The polynomial's zero, to three decimal places
of accuracy, is at x
= 1.299.
Stapel, Elizabeth. "Numerical Approximation
of Zeroes: The Midpoint Method." Purplemath. Available
from http://www.purplemath.com/modules/numeric.htm.
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