How do I square an array?

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Danielle Brunig
Danielle Brunig 2022-8-29
回答： Nisarg 2024-5-8
Hi, I'm trying to square an array, but it won't work for some reason, and I cannot figure out why. I'm sure I'm probably overlooking something big though.
x=[a:10:b];
y=x.^2;
gz=-0.2794*(p)*(r^3)*z/((y+z^2)^(3/2))
It gives me multiple errors whenever I try to run this code through, even though I coded all of the variables, it doesn't like trying to square them.
  1 个评论
Torsten
Torsten 2022-8-29
编辑：Torsten 2022-8-29
Ran in:
Please include the full code and write exactly what does not work.
In gz, you will of course have to use .^, .* and ./ instead of ^, * and /
x = 1:10;
y = x.^2
y = 1×10
1 4 9 16 25 36 49 64 81 100

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回答（2 个）

Image Analyst
Image Analyst 2022-8-29
编辑：Image Analyst 2022-8-29
Ran in:
This works fine for me
% Define parameters
a = 10;
b = 40;
x=[a:10:b]
x = 1×4
10 20 30 40
y=x.^2
y = 1×4
100 400 900 1600
% Define parameters
p = 2;
r = 3;
z = 6;
gz=-0.2794*(p)*(r^3)*z ./ ((y+z.^2) .^ (3/2))
gz = 1×4
-0.0571 -0.0099 -0.0032 -0.0014
If you have any more questions, then attach your actual error message (ALL the red text) and data (actual parameters) and code to read it in with the paperclip icon after you read this:
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Nisarg
Nisarg 2024-5-8
class_intervals = [10, 20, 30, 40, 50];
frequencies = [5, 8, 12, 10, 6];
n = sum(frequencies)
mean = sum(class_intervals .* frequencies)/n
y = sum(frequencies.*((class_intervals - mean).^2))
std = sqrt(y/n)

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