How to not use for loop

Hi, I have a function that I am trying to get rid of the for loop and rewrite the function so that it doesnt use any loops. I have looked on various links like the Vector Creation (https://au.mathworks.com/help/matlab/ref/colon.html) and Vectorisation (https://au.mathworks.com/help/matlab/matlab_prog/vectorization.html) but I still cant get it to work. Below I have the function with the for loop.

function dfdx = ddx(f, h)
% Add description, name, date, inputs, outputs
dfdx = nan(size(f));
dfdx(1) = (f(2) - f(1))/h;
for j = 2:length(f)-1;
    dfdx(j) = 0.5*(f(j+1) - f(j-1))/h;
end
dfdx(end) = (f(end) - f(end-1))/h;

And here is the code to call the function

format compact
a = randn(2, 1)
x = linspace(-1, 1, 20)    % equispaced x
f = a(1) + a(2)*x          % function values
dfdx = ddx(f, x(2)-x(1))   % derivatives should be exact for linear
computeError = a(2) - dfdx % should be zeros to 1e-15

Answer 1

[Star Strider]

Try something like this —

format compact
a = randn(2, 1)

a = 2×1

0.4175 1.4768

x = linspace(-1, 1, 20)    % equispaced x

x = 1×20

-1.0000 -0.8947 -0.7895 -0.6842 -0.5789 -0.4737 -0.3684 -0.2632 -0.1579 -0.0526 0.0526 0.1579 0.2632 0.3684 0.4737 0.5789 0.6842 0.7895 0.8947 1.0000

f = a(1) + a(2)*x          % function values

f = 1×20

-1.0593 -0.9038 -0.7484 -0.5929 -0.4374 -0.2820 -0.1265 0.0289 0.1844 0.3398 0.4953 0.6507 0.8062 0.9616 1.1171 1.2725 1.4280 1.5835 1.7389 1.8944

dfdx = ddx(f, x(2)-x(1))   % derivatives should be exact for linear

dfdx = 1×20

1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768 1.4768

computeError = a(2) - dfdx % should be zeros to 1e-15

computeError = 1×20

1.0e-14 * -0.0444 -0.0444 0.3775 -0.0444 0.3331 0.1776 0.0666 0.2220 0.1332 0.1332 0.1776 0.1110 0.1776 0.1776 0.1776 0.1776 0.1776 -0.0444 0.3775 -0.0444

function dfdx = ddx(f,h)
dfdx(1) = (f(2) - f(1))/h;
dfdx(2:numel(f)) = (f(2:end) - f(1:end-1))/h;
end

EDIT — The gradient function already exists to do this, however I’m assuming here that you want to write your own function to do the numerical derivative.

.

[Declan]

Thanks for the answer!

[Torsten]

@Star Strider

But

for j = 2:length(f)-1;
    dfdx(j) = 0.5*(f(j+1) - f(j-1))/h;
end
dfdx(end) = (f(end) - f(end-1))/h;

doesn't translate to

dfdx(2:numel(f)) = (f(2:end) - f(1:end-1))/h;

[Star Strider]

@Declan — As always, my pleasure!

@Torsten —

I checked it against the gradient function and both gave the same result.

That was my criterion —

format compact
a = randn(2, 1)

a = 2×1

0.6877 1.4736

x = linspace(-1, 1, 20)    % equispaced x

x = 1×20

-1.0000 -0.8947 -0.7895 -0.6842 -0.5789 -0.4737 -0.3684 -0.2632 -0.1579 -0.0526 0.0526 0.1579 0.2632 0.3684 0.4737 0.5789 0.6842 0.7895 0.8947 1.0000

f = a(1) + a(2)*x          % function values

f = 1×20

-0.7859 -0.6307 -0.4756 -0.3205 -0.1654 -0.0103 0.1448 0.2999 0.4550 0.6102 0.7653 0.9204 1.0755 1.2306 1.3857 1.5408 1.6959 1.8510 2.0062 2.1613

dfdx = ddx(f, x(2)-x(1))   % derivatives should be exact for linear

dfdx = 1×20

1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736 1.4736

computeError = a(2) - dfdx % should be zeros to 1e-15

computeError = 1×20

1.0e-14 * 0.0666 0.0666 0.2887 -0.1332 0.3775 0.0666 0.1776 0.1776 0.1776 0.1776 0.0666 0.1776 0.2887 0.0666 0.0666 0.2887 0.2887 -0.1332 0.2887 -0.1332

CompareResults = ["gradient" gradient(f, x(2)-x(1)); "ddx" dfdx]

CompareResults = 2×21 string array
    "gradient"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"
    "ddx"         "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"    "1.4736"

function dfdx = ddx(f,h)
dfdx(1) = (f(2) - f(1))/h;
dfdx(2:numel(f)) = (f(2:end) - f(1:end-1))/h;
end

.

[Torsten]

@Star Strider

Yes, for linear functions, centered and forward differencing to approximate the derivative give the same result.

Answer 2

[Torsten]

function dfdx = ddx(f, h)
  dfdx = gradient(f,h);
end

[Declan]

Oh, I didnt realise that there was a gradient function inbuilt. Thanks!

[Stephen23]

+1 very neat.