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How to define a function consisting of multiple parts (i.e different f:n at different times) in Matlab using a single equation?

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Harshit Jain
Harshit Jain 2013-4-22
评论： Samuel Silas Ng'habi 2023-4-27
Example
f(x)
= x^2 for 0<x<1;
= x^3 for 1<x<2;
= ......and so on.
Please help me with this.

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采纳的回答

Sally Al Khamees
Sally Al Khamees 2016-12-23
编辑：Sally Al Khamees 2017-2-21
If you have R2016b and the Symbolic Math Toolbox installed, you can just use the piecewise function:
https://www.mathworks.com/help/symbolic/piecewise.html
Here is an example:
syms y(x);
y(x) = piecewise(1<x <2, x^3, 2 <= x <= 3, x^2, 1)
fplot(y)
%You can evaluate it at one point. Example when x = 2
y(2)
%You can evaluate a vector. Example
v = linspace(1,4,4)
y(v)

更多回答（2 个）

Walter Roberson
Walter Roberson 2013-4-22
You can define it symbolically using MuPAD's "piecewise" construct.
In some cases you can define it numerically using logical constructs such as
(x > 1 & x < 2) .* x.^3 + (x > 0 & x < 1) .* x^2
This will not work properly for locations that generate NaN or infinity when evaluated for any part. For example, if f(x) = 1 for x = 0, and f(x) = 1/x for other x, then you cannot use
(x == 0) .* 1 + (x ~= 0) .* 1./x
because the second part will generate NaN when evaluated for x(K) = 0, and the NaN multiplied by the 0 of (x(K) ~= 0) will still be NaN instead of vanishing to 0 as it does for finite values. Similarily, 0 * inf is NaN rather than 0.
John BG
John BG 2016-12-23
y=[1:0.001:2].^3
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Stephen23
Stephen23 2017-2-22
@John BG: how could this be used in a function of x (as the question requests), e.g.:
fun = @(x) ???
Note that both Sally Al Khamees' and Walter Roberson's answers provide this.
Walter Roberson
Walter Roberson 2017-2-22
Consider, for example, if the task is to find the point at which the function equals 3.5,
x0 = 2 * rand(); %range is 0 to 2
fzero(@(x) f(x) - 3.5, x0)
Using a fixed dx is not going to be able to handle this task -- not unless dx = eps(realmin), so that you are testing all 2^62 representable numbers between 0 and 2.
You could, of course, write code that assumes that the input is a scalar:
function y = f(x)
y = 0;
if x > 0 & x < 1
y = x.^2;
elseif x > 1 & x < 2
y = x.^3;
end
end
and you could loop that code for non-scalar x.
You can use logical indexing:
function y = f(x);
y = zeros(size(x));
mask = x > 0 & x < 1;
y(mask) = x(mask).^2;
mask = x > 1 & x < 2;
y(mask) = x(mask).^3;
end
You can define it with an anonymous function,
f = @(x) (x > 1 & x < 2) .* x.^3 + (x > 0 & x < 1) .* x^2;
You can look at the pattern and predict
f = @(x) (x ~= ceil(x)) .* x.^(1 + ceil(x));
And all of those versions are functions that could be used as functions over arbitrary domains such as for fzero() purposes.
But using a fixed dx is not an approach that can be used for this kind of common application.

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