absolute value of a vector

how can i determine the absolute value of a vector?

so for example

if i had x=[ -0.5 .3 .5 -0.5]

how would i determine the absolute value of each individual term in the vector?

i've tried using "abs" but it spits back one value.

Answer 1

[Roger Stafford]

That isn't true of matlab's 'abs' function. For every value in x it will return its absolute value. For every individual complex number it will return the square root of the sum of the squares of its real and imaginary parts.

I suspect you have a false 'abs' function somewhere on your system that is being called instead of the true 'abs'. You had better check on that.

[Matt J]

Use

 >>which abs(x)

to find the version of abs() that is being called on x.

[Kyle]

here is my code. i'm trying to do abs(((yexact-y)/yexact)) i'm getting 1 value and i don't know why.

h=0.1; t(1)=0; y(1)=1; t=0:h:5;

for i=1:(numel(t)-1);
    k=2-exp(-4*t(i))-2*y(i) ;
    y(i+1)=y(i)+h*k;

end

plot(t,y) 
xlabel('t')
ylabel('y') 
title('Euler Method: Problem 1 ') 
hold on 

h2=0.001;
t2(1)=0;
y2(1)=1;
t2=0:h2:5;

for i=1:(numel(t2))-1;
    k2=2-exp(-4*t2(i))-2*y2(i) ;
    y2(i+1)=y2(i)+h2*k2;
end 
plot(t2,y2,'g')

t3=[0:0.1:5];
yexact=1+0.5*exp(-4*t3)-0.5*exp(-2*t3);

plot(t3,yexact,'r')

[Matt J]

Note the missing '.'

abs(((yexact-y)./yexact))

[Kyle]

thank you!

[Roger Stafford]

I'll expand Matt's answer. You have used matrix division in yexact-y)/yexact and for this case the result would be a single 1 x 1 value, not a vector.