Is this correct?
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Use loops to compute and plot (not animated)the following piecewise function for -15<=x<=15 .
F(x)={5x, x<0
{x^2, 0<=x<2
{ 2lnx, x>=2
My code:
x=-15:15;
if x<0;
5*x;
elseif x>=0 & x<2
y=x^2;
else x>=2;
y=2*log(x);
end
plot(x,y)
4 Comments
Chunru
on 18 Jul 2022
Although the code is not doing what it intends to, the code can be run and MATLAB only gives a warning message (orange color).
Accepted Answer
Chunru
on 18 Jul 2022
x=-15:15;
y = zeros(size(x));
% This is more MATLAB way
idx = x<0;
y(idx) = 5 * x(idx);
idx = x>=0 & x<2;
y(idx) =x(idx).^2;
idx = x>=2;
y(idx) = 2*log(x(idx));
plot(x,y)
% This is more conventional way (some othter programming language)
x=-15:15;
y = zeros(size(x));
for i=1:length(x)
if x(i)<0
y(i) = 5 * x(i);
elseif x(i)>=0 & x(i)<2
y(i) =x(i).^2;
elseif x(i)>=2;
y(i) = 2*log(x(i));
end
end
figure
plot(x,y)
More Answers (1)
M.B
on 18 Jul 2022
x = -15:1:15;% init x
y = nan*x;% init y to be of same size as x
for index = -15:15;
if index<0;
y(index+16) = 5*index;
elseif index>=0 & index<2
y(index+16) = index^2;
else index>=2;
y(index+16) = 2*log(index);
end
end
plot(x,y)
2 Comments
Chunru
on 18 Jul 2022
The "index" is the loop variable the code uses and it need to be added with 16 in order to become array index (which is 1,2,...31). It's working code, but it may be further improved.
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