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john le 20 Nov 2014
Commenté : Walter Roberson le 10 Juin 2021
f(x)=2*sqrt(x) for 0<=x<=1 and f(x)=3-(x) for 1<=x<=3
How would I plot this function on the range -9<=x<=9? The questions states to make use of the "floor function".
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Azzi Abdelmalek le 20 Nov 2014
Your function is defined from 0 to 3, what about the other ranges?
john le 20 Nov 2014
Modifié(e) : john le 20 Nov 2014
It is a periodic function with period 3, so it would just repeat itself for x<0 and x>3.

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### Réponse acceptée

Star Strider le 20 Nov 2014
Modifié(e) : Star Strider le 20 Nov 2014
I don’t understand where the floor function comes into it. This is how I would code it:
f = @(x) [2*sqrt(x).*(0<=x & x<1) + (3-x).*(1<=x & x<=3)];
x = linspace(-9,9);
figure(1)
plot(x, f(x))
grid
Note that your function does not exist for x<0 and x>3, so while the plot defaults to zero outside that region, I would only plot it for [0,3].
EDIT —
‘Periodic function’ ... this works:
f = @(x) [2*sqrt(x).*(0<=x & x<1) + (3-x).*(1<=x & x<=3)];
x = linspace(0,3);
intvl = [-6 6];
pfx = repmat(f(x),1,diff(intvl)/3);
px = linspace(intvl(1),intvl(2),length(pfx));
figure(1)
plot(px, pfx)
grid
I had to revise my code every time you updated your Question. This (EDIT) code reproduces the plot you posted. I did the function in a one-line anonymous function for simplicity and efficiency. It shouldn’t be difficult for you to follow its logic and write a function .m-file to do the same calculation.
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Diptanshu De le 10 Juin 2021
Modifié(e) : Diptanshu De le 10 Juin 2021
How to change my code if my function is
f(x) = -x ; -5<=x<0
x^2+1; 0<=x<5
Walter Roberson le 10 Juin 2021
x = linspace(-6,6);
f = zeros(size(x));
mask = -5 <= x & x < 0;
mask = 0 <= x & x < 5;
plot(x, f)

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### Plus de réponses (1)

Sally Al Khamees le 21 Fév 2017
If you have R2016b and the Symbolic Math Toolbox installed, you can use the piecewise function to recreate this example:
syms y(x) a(x) b(x);
y(x) = piecewise(0<=x <1, 2*sqrt(x), 1 <= x <= 3, 3-x);
interval = [-6 6];
pw=y;
for i=1:diff(interval/6)
a(x)= piecewise(i*3<=x<1+(i*3),2*sqrt(x-3*i),1+(i*3)<=x<=3+(i*3),3-(x-3*i));
b(x)= piecewise(i*-3<=x<1+(i*-3),2*sqrt(x+3*i),1+(i*-3)<=x<=3+(i*-3),3-(x+3*i));
pw = [pw a b];
end
pw
fplot(pw,interval)
You can read more about the piecewise function in Symbolic Math Toolbox here https://www.mathworks.com/help/symbolic/piecewise.html
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Puech gabriel le 2 Août 2017
Hi,
Thanks for giving this code. Here are some parameters added to your code and using just one function
syms a(x);
T = pi; % period value
i = -2; % number of periods, must be integer!
interval = [i*T -i*T];
pw = [];
while i<=diff(interval/(2*T))
a(x)= piecewise(i*T<=x<1+(i*T),2*sqrt(x-T*i),1+(i*T)<=x<=3+(i*T),3-(x-T*i),3+(i*T)<=x<=T+(i*T),0); %+diff(interval/6)-floor(diff(interval/6))
i = i +1;
pw = [pw a ]; % concatenation des periodes
end
pw
fplot(pw,interval)
Would it have a way of preallocating the pw symfun matrix before the loop?
Kind regards, Gabriel

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