Constructing piecewise symbolic function in Matlab

旧城冷巷雨未停 提交于 2019-11-29 01:57:13
gnovice

One option is to use the heaviside function to make each equation equal zero outside of its given range, then add them all together into one equation:

syms x;
f = (heaviside(x)-heaviside(x-1))*x^3/6 + ...
    (heaviside(x-1)-heaviside(x-2))*(1/6)*(-3*x^3+12*x^2-12*x+4) + ...
    (heaviside(x-2)-heaviside(x-3))*(1/6)*(3*x^3-24*x^2+60*x-44) + ...
    (heaviside(x-3)-heaviside(x-4))*(1/6)*(4-x)^3;
double(int(diff(f, 1)^2, x, 0, 4))

ans =

    0.6667

Another alternative is to perform your integration for each function over each subrange then add the results:

syms x;
eq1 = x^3/6;
eq2 = (1/6)*(-3*x^3+12*x^2-12*x+4);
eq3 = (1/6)*(3*x^3-24*x^2+60*x-44);
eq4 = (1/6)*(4-x)^3;
total = int(diff(eq1, 1)^2, x, 0, 1) + ...
        int(diff(eq2, 1)^2, x, 1, 2) + ...
        int(diff(eq3, 1)^2, x, 2, 3) + ...
        int(diff(eq4, 1)^2, x, 3, 4)

total =

2/3

UPDATE:

Although it's mentioned in the question that the piecewise function didn't work, Karan's answer suggests it does, at least in newer versions. The documentation for piecewise currently says it was introduced in R2016b, but it was clearly present much earlier. I found it in the documentation for the Symbolic Math Toolbox as far back as R2012b, but the calling syntax was different than it is now. I couldn't find it in earlier documentation for the Symbolic Math Toolbox, but it did show up as a function in other toolboxes (such as the Statistics and Spline Toolboxes), which explains its mention in the question (and why it didn't work for symbolic equations at the time).

Starting R2016b, use the piecewise function

syms x
y = piecewise(x<0, -1, x>0, 1)

y =
piecewise(x < 0, -1, 0 < x, 1)

For this case:

syms x
f = piecewise( ...
0< x <=1, x^3/6, ...
1 < x <= 2, (1/6)*(-3*x^3+12*x^2-12*x+4), ...
2 < x <= 3, (1/6)*(3*x^3-24*x^2+60*x-44), ...
3 < x <= 4, (1/6)*(4-x)^3, ...
0)

f =
piecewise(x in Dom::Interval(0, [1]), x^3/6, x in Dom::Interval(1, [2]), - x^3/2 + 2*x^2 - 2*x + 2/3, x in Dom::Interval(2, [3]), x^3/2 - 4*x^2 + 10*x - 22/3, x in Dom::Interval(3, [4]), -(x - 4)^3/6, 0)

int(diff(f, 1)^2, x, 0, 4)
ans =
2/3
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