# Array indices must be positive integers or logical values.

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Mark Prentice on 10 Jul 2021
Commented: Mark Prentice on 10 Jul 2021
h = 0.2;
T = 4:h:8
X_s=[-5.87 -4.23 -2.55 -0.89 0.67 2.09 3.31 4.31 5.06 5.55 5.78 5.77 5.52 5.08 4.46 3.72 2.88 2.00 1.10 .23 -0.59]
for X=4:0.2:8 %%%need to take the first and last point off since central difffrencing doesnt work on them
k=[X];
x = round(k/0.2 - 4/0.2 + 1);
V = ((X_s((x)+1))-(X_s((x)-1)))/(2*h);
if X==4
k=[X];
x = round(k/0.2 - 4/0.2 + 1);
V = ((X_s((x)+1))-(X_s((x))))/(2*h);
end
if X==8
k=[X];
x = round(k/0.2 - 4/0.2 + 1);
V = ((X_s((x)))-(X_s((x-1))))/(2*h);
end
Velcoity=V
end
Hey there,
Im now trying to get the array to work with two end points where it can take them and redirect them to equations suited to solving them with the given data. However I find myself getting the error "Array indices must be positive integers or logical values". I've tried reorginizing my if statments to come first and that doesnt seem to work. If i change the start to 4.2 and the end to 7.8 it works like a charm but cant figure out why it wont accept the start as 4 and end as 8 and use the forward and backward diffrencing equations. Any help or knowledge helps and thank you for looking.

Walter Roberson on 10 Jul 2021
h = 0.2;
X_s = [-5.87 -4.23 -2.55 -0.89 0.67 2.09 3.31 4.31 5.06 5.55 5.78 5.77 5.52 5.08 4.46 3.72 2.88 2.00 1.10 .23 -0.59]
X_s = 1×21
-5.8700 -4.2300 -2.5500 -0.8900 0.6700 2.0900 3.3100 4.3100 5.0600 5.5500 5.7800 5.7700 5.5200 5.0800 4.4600 3.7200 2.8800 2.0000 1.1000 0.2300 -0.5900
nX = length(X_s);
for x = 1 : nX
if x == 1
V = ((X_s((x)+1))-(X_s((x))))/(2*h);
elseif x == nX
V = ((X_s((x)))-(X_s((x-1))))/(2*h);
else
V = ((X_s((x)+1))-(X_s((x)-1)))/(2*h);
end
Velocity(x) = V;
end
plot(1:nX, X_s, 'k-', 1:nX, Velocity, 'b--') Mark Prentice on 10 Jul 2021
thank you, this actually made a good bit of sense.

G A on 10 Jul 2021
When x=1, then X_s(x-1) = X_s(0), which is not allowed in Matlab
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Mark Prentice on 10 Jul 2021
Yes, I've been rearranging the whole system of 3 equations in every and all configurations. Such as putting it last however I'm still getting the error. I'm wondering if theres another variable im missing or a function that makes the 4 value on go into one equation ( if im useing the if functions right). But yes you are correct V = ((X_s((x)+1))-(X_s((x)-1)))/(2*h); should not be first.

R2021a

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