How to access the equation above the while loop

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Stark Volt
Stark Volt am 17 Sep. 2021
Kommentiert: Stark Volt am 20 Sep. 2021
How to access the y and limit equation above the while loop so that i do not need to type the equation inside the while loop?
y = 0;
limit= (exp(y)-(1+y+y^2/2+y^3/6))/exp(y)
while y>=0
y=y+0.01;
limit=(exp(y)-(1+y+y^2/2+y^3/6))/exp(y);
if limit>0.01
break
end
end
y=y

Akzeptierte Antwort

the cyclist
the cyclist am 17 Sep. 2021
Bearbeitet: the cyclist am 17 Sep. 2021
One straightforward way to do this is by using an anonymous function:
% Define an anonymous function for the limit
f_limit = @(y) (exp(y)-(1+y+y^2/2+y^3/6))/exp(y);
y = 0;
limit = f_limit(y)
while y>=0
y=y+0.01;
limit = f_limit(y);
if limit>0.01
break
end
end
y = y
  3 Kommentare
the cyclist
the cyclist am 18 Sep. 2021
I don't understand why you need another way.
Can you please explain in more detail what you need, and why your solution (with my modification) is not ideal?
Stark Volt
Stark Volt am 18 Sep. 2021
Ah sorry for that, your solution is good, I just want to see other types of solution and somehow learn from them just like the (anonymous function) I did not know this until you shown it me. Thanks

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Weitere Antworten (1)

Image Analyst
Image Analyst am 18 Sep. 2021
Try this:
% Demo by Image Analyst
clc; % Clear the command window.
close all; % Close all figures (except those of imtool.)
clear; % Erase all existing variables. Or clearvars if you want.
workspace; % Make sure the workspace panel is showing.
format short g;
format compact;
fontSize = 20;
y = 0;
limit = CustomFunction(y)
loopCounter = 1; % Failsafe to prevent infinite loop
maxIterations = 1000; % Failsafe - way more than you think you'll ever need.
while y >= 0 && loopCounter < maxIterations
fprintf('On iteration #%d, y = %f and limit = %f.\n', loopCounter, y, limit);
y = y + 0.01;
limit = CustomFunction(y);
if limit > 0.01
break
end
loopCounter = loopCounter + 1;
end
% Define a function for the limit
function f_limit = CustomFunction(y)
f_limit = (exp(y)-(1+y+y^2/2+y^3/6))/exp(y);
end
Put it all into one m-file, like test.m or whatever (don't call it limit.m or f_limit.m). You'll see
On iteration #1, y = 0.000000 and limit = 0.000000.
On iteration #2, y = 0.010000 and limit = 0.000000.
On iteration #3, y = 0.020000 and limit = 0.000000.
....
On iteration #80, y = 0.790000 and limit = 0.008702.
On iteration #81, y = 0.800000 and limit = 0.009080.
On iteration #82, y = 0.810000 and limit = 0.009469.
On iteration #83, y = 0.820000 and limit = 0.009868.
  1 Kommentar
Stark Volt
Stark Volt am 20 Sep. 2021
I also like your way, but (the cyclist) was the first to comment. Thats why I voted for your answer, Thanks

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