square root of pi

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nathan blanc 2021-1-13

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https://ww2.mathworks.cn/matlabcentral/answers/715788-square-root-of-pi

编辑： nathan blanc 2021-1-27

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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dpb 2021-1-13

编辑：dpb 2021-1-13

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 2021-1-13

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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Answer 1

Steven Lord 2021-1-13

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https://ww2.mathworks.cn/matlabcentral/answers/715788-square-root-of-pi#answer_597168

在 MATLAB Online 中打开

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       

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nathan blanc 2021-1-26

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 2021-1-26

在 MATLAB Online 中打开

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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