Mathematics |
Finding Zeros of Functions
The fzero
function attempts to find a zero of one equation with one variable. You can call this function with either a one-element starting point or a two-element vector that designates a starting interval. If you give fzero
a starting point x0
, fzero
first searches for an interval around this point where the function changes sign. If the interval is found, fzero
returns a value near where the function changes sign. If no such interval is found, fzero
returns NaN
. Alternatively, if you know two points where the function value differs in sign, you can specify this starting interval using a two-element vector; fzero
is guaranteed to narrow down the interval and return a value near a sign change.
The following sections contain two examples that illustrate how to find a zero of a function using a starting interval and a starting point. The examples use the function humps
, which is provided with MATLAB. The following figure shows the graph of humps
.
Using a Starting Interval
The graph of humps
indicates that the function is negative at x = -1
and positive at x = 1
. You can confirm this by calculating humps
at these two points.
Consequently, you can use [-1 1]
as a starting interval for fzero
.
The iterative algorithm for fzero
finds smaller and smaller subintervals of [-1 1]
. For each subinterval, the sign of humps
differs at the two endpoints. As the endpoints of the subintervals get closer and closer, they converge to a zero for humps
.
To show the progress of fzero
at each iteration, set the Display
option to iter
using the function optimset
.
This returns the following iterative output:
a = fzero(@humps,[-1 1],options) Func-count x f(x) Procedure 2 -1 -5.13779 initial 3 -0.513876 -4.02235 interpolation 4 -0.513876 -4.02235 bisection 5 -0.473635 -3.83767 interpolation 6 -0.115287 0.414441 bisection 7 -0.115287 0.414441 interpolation 8 -0.132562 -0.0226907 interpolation 9 -0.131666 -0.0011492 interpolation 10 -0.131618 1.88371e-007 interpolation 11 -0.131618 -2.7935e-011 interpolation 12 -0.131618 8.88178e-016 interpolation 13 -0.131618 8.88178e-016 interpolation Zero found in the interval [-1, 1] a = -0.1316
Each value x
represents the best endpoint so far. The Procedure
column tells you whether each step of the algorithm uses bisection or interpolation.
You can verify that the function value at a
is close to zero by entering
Using a Starting Point
Suppose you do not know two points at which the function values of humps
differ in sign. In that case, you can choose a scalar x0
as the starting point for fzero
. fzero
first searches for an interval around this point on which the function changes sign. If fzero
finds such an interval, it proceeds with the algorithm described in the previous section. If no such interval is found, fzero
returns NaN
.
For example, if you set the starting point to -0.2
, the Display
option to Iter
, and call fzero
by
fzero
returns the following output:
Search for an interval around -0.2 containing a sign change: Func-count a f(a) b f(b) Procedure 1 -0.2 -1.35385 -0.2 -1.35385 initial interval 3 -0.194343 -1.26077 -0.205657 -1.44411 search 5 -0.192 -1.22137 -0.208 -1.4807 search 7 -0.188686 -1.16477 -0.211314 -1.53167 search 9 -0.184 -1.08293 -0.216 -1.60224 search 11 -0.177373 -0.963455 -0.222627 -1.69911 search 13 -0.168 -0.786636 -0.232 -1.83055 search 15 -0.154745 -0.51962 -0.245255 -2.00602 search 17 -0.136 -0.104165 -0.264 -2.23521 search 18 -0.10949 0.572246 -0.264 -2.23521 search Search for a zero in the interval [-0.10949, -0.264]: Func-count x f(x) Procedure 18 -0.10949 0.572246 initial 19 -0.140984 -0.219277 interpolation 20 -0.132259 -0.0154224 interpolation 21 -0.131617 3.40729e-005 interpolation 22 -0.131618 -6.79505e-008 interpolation 23 -0.131618 -2.98428e-013 interpolation 24 -0.131618 8.88178e-016 interpolation 25 -0.131618 8.88178e-016 interpolation Zero found in the interval [-0.10949, -0.264] a = -0.1316
The endpoints of the current subinterval at each iteration are listed under the headings a
and b
, while the corresponding values of humps
at the endpoints are listed under f(a)
and f(b)
, respectively.
Note
The endpoints a and b are not listed in any specific order: a can be greater than b or less than b .
|
For the first nine steps, the sign of humps
is negative at both endpoints of the current subinterval, which are listed under in the output. At the tenth step, the sign of humps
is positive at the endpoint, -0.10949
, but negative at the endpoint, -0.264
. From this point on, the algorithm continues to narrow down the interval [-0.10949 -0.264]
, as described in the previous section, until it reaches the value -0.1316
.
Output Functions | Tips |
© 1994-2005 The MathWorks, Inc.