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square root of pi

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nathan blanc
nathan blanc on 13 Jan 2021
Edited: nathan blanc on 27 Jan 2021
I was banging my head against the wall looking for the error in my code, that kept returning "false" when it should have returned true. I finally realised that my error was because in MATLAB (or at least in my copy) pi^0.5^2==pi returns "false". The same goes for sqrt(pi)^2==pi. I assume this is because the machine precision of pi is more exact then the machine precison of the square root function, even though they are identical up to 16 digits.
I have some unelegant solutions to this. e.g writing down pi=3.141592 at the begining of my code. is there a better solution? a more percise calculaion of the square root?
many thanks

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Adam Danz
Adam Danz on 13 Jan 2021
pi - (pi^0.5^2)
ans =
4.44089209850063e-16
NASA only uses around 15 digits of pi in its calculations for sending rockets into space (source).
dpb
dpb on 13 Jan 2021
Your initialization code has warnings of one property making reference to another and explicit instructions to not leave code that has such an issue.
I don't do objects so not positive of the manner in which this should be resolved, but redesign appears in your future... :)
nathan blanc
nathan blanc on 13 Jan 2021
dpb i actualy couldn't see a way around this that allows setting both properties. in any case this is a seperate issue from my question

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Answers (1)

Steven Lord
Steven Lord on 13 Jan 2021
I would have both the Area and Diameter properties depend upon a third (maybe public, maybe internal) property. Setting either Area or Diameter computes what the value of that internal property should be and getting them computes the value using that internal property.
classdef Circle715788
properties(Dependent)
Area
Diameter
end
properties
Radius
end
methods % Constructor
function obj = Circle715788(r)
obj.Radius = r;
end
end
methods % Property accessors
function obj = set.Area(obj, newArea)
R = sqrt(newArea./pi);
obj.Radius = R;
end
function obj = set.Diameter(obj, newDiameter)
obj.Radius = newDiameter./2;
end
function A = get.Area(obj)
A = pi*obj.Radius.^2;
end
function D = get.Diameter(obj)
D = 2*obj.Radius;
end
end
end
Use as:
>> Q = Circle715788(1)
Q =
Circle715788 with properties:
Area: 3.1416
Diameter: 2
Radius: 1
>> Q = Circle715788(4)
Q =
Circle715788 with properties:
Area: 50.2655
Diameter: 8
Radius: 4
>> Q.Area = pi
Q =
Circle715788 with properties:
Area: 3.1416
Diameter: 2
Radius: 1

  2 Comments

nathan blanc
nathan blanc on 26 Jan 2021
Thank you Steven and sorry for the late reply. I thought of doing this but I am worried that this might slow down my code. I am calling these properties about 10^6 times during a run. What do you think?
Steven Lord
Steven Lord on 26 Jan 2021
On my Windows 10 machine running release R2020b:
>> C = Circle715788(7);
>> tic; for k = 1:1e6, A = C.Area; end, toc
Elapsed time is 0.132099 seconds.
>> tic; for k = 1:1e6, A = C.Area; end, toc
Elapsed time is 0.112353 seconds.
>> tic; for k = 1:1e6, A = C.Area; end, toc
Elapsed time is 0.108369 seconds.
>> tic; for k = 1:1e6, A = C.Area; end, toc
Elapsed time is 0.105188 seconds.
>> tic; for k = 1:1e6, A = C.Area; end, toc
Elapsed time is 0.102445 seconds.
So about a tenth of a second for 1e6 queries.
>> tic; for k = 1:1e6, C.Radius = k; A = C.Area; end, toc
Elapsed time is 0.231171 seconds.
>> tic; for k = 1:1e6, C.Radius = k; A = C.Area; end, toc
Elapsed time is 0.210652 seconds.
>> tic; for k = 1:1e6, C.Radius = k; A = C.Area; end, toc
Elapsed time is 0.208293 seconds.
Two-tenths for 1e6 property sets and gets.
You might be able to speed this up, particularly if the calculations of Area / Diameter are expensive, using a memoize object inside your Area / Diameter calculations to cache the values for repeated Radius values.

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