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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
2013 年 4 月 22 日
コメント済み: Samuel Silas Ng'habi
2023 年 4 月 27 日
Example
f(x)
= x^2 for 0<x<1;
= x^3 for 1<x<2;
= ......and so on.
Please help me with this.
0 件のコメント
採用された回答
Sally Al Khamees
2016 年 12 月 23 日
編集済み: Sally Al Khamees
2017 年 2 月 21 日
If you have R2016b and the Symbolic Math Toolbox installed, you can just use the piecewise function:
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)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/176226/image.png)
1 件のコメント
Samuel Silas Ng'habi
2023 年 4 月 27 日
What if the value and the conditions are expressed as nth -term. How will you plot the function See pictures attached
その他の回答 (2 件)
Walter Roberson
2013 年 4 月 22 日
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
2016 年 12 月 23 日
y=[1:0.001:2].^3
4 件のコメント
Stephen23
2017 年 2 月 22 日
@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
2017 年 2 月 22 日
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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