How do is code this last summation? I was able to do the first one but have no clue how to finish it?

Question

Adam Schlei 2019-3-23

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/451824-how-do-is-code-this-last-summation-i-was-able-to-do-the-first-one-but-have-no-clue-how-to-finish-i

评论： Adam Schlei 2019-3-23

function Senior_Project_code
%input values
y = 4;
n = 1;
L = 3;
%equations
%z=17;
a=L/(2*n);
x = 0:1:L;
E = 808/(y^2);
%c = atan(abs((x-a(2*i-1)))/y)*(180/3.14);
%s = (-0.0000000667*atan(abs((x-a(2*i-1)))/y)*(180/3.14).^4 ...
%    + 0.0000157209*atan(abs((x-a(2*i-1)))/y)*(180/3.14).^3 ...
%    - 0.0010081511*atan(abs((x-a(2*i-1)))/y)*(180/3.14).^2 ...
%    + 0.0020817855*atan(abs((x-a(2*i-1)))/y)*(180/3.14)    ...
%    + 0.9991821678);
    Axx(x+1) = 0;
        for k = 1:n
            Axx(x+1) = Axx(x+1) + E* ...
            (-0.0000000667*(atan(abs((x-a*(2*k-1)))/y)*(180/3.14)).^4   ...
            + 0.0000157209*(atan(abs((x-a*(2*k-1)))/y)*(180/3.14)).^3   ...
            - 0.0010081511*(atan(abs((x-a*(2*k-1)))/y)*(180/3.14)).^2   ...
            + 0.0020817855*(atan(abs((x-a*(2*k-1)))/y)*(180/3.14))      ...
            + 0.9991821678);
        end
    Ax = sum(Axx)/(L+1);   %can I redo this to match my other equation? 
display (Axx,'Brightness Values');
display (Ax,'Average Brightness');
end

Any help redoing Ax to match my above summation with i=1, to z-2... would be much appreciated.

Coding should be in for,if, then, and, or....type statements if possible.

4 个评论
显示 2更早的评论隐藏 2更早的评论

Adam Schlei 2019-3-23

Yes that is correct. I'm trying to code my MathCad formulas from the picture above. The problem I'm having is non-integer values. What I have in the picture is correct. The coding in Matlab is the problem I'm having. When I have non integer values I get array size errors. Is their ways around this, cause I originally had x = 0 : 0.1 : L so their were more test points being used, but the array sizes don't match then. And yes I do agree with all of what you saying (efficincy and rcalculating of atan four times), I'm not great at coding and any help for matching my MathCad program shown in the picture to the Matlab code would be great. Thanks a bunch.

Adam Schlei 2019-3-23

In other words I need the matlab equation of Ax, to be this summation equation with z=17: Matlab question3.GIF

or if not possible I need the matlab equation of Ax, to be this summation equation with z=17:

Any help would be great!!

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

darova 2019-3-23

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/451824-how-do-is-code-this-last-summation-i-was-able-to-do-the-first-one-but-have-no-clue-how-to-finish-i#answer_366979

在 MATLAB Online 中打开

Maybe a little bit clearer version:

y= 7.5, n=16, L=16 result: 118.22

y= 4, n=1, L=3 result: 46.31

function main 
    % input values
    y = 4;
    L = 3;
    z = 17;
    n = 1;
    av = 0;
    for i = 1:z-2
        x = L/(z-1)*i;
        av = av + s(x,y,n,L);       
    end
    x = L/(2*z-2)*i;
    av = av + s(x,y,n,L) + s(L-x,y,n,L);
    av = av/z;
    display('Average Brightness: '), av
end
function result = s(x,y,n,L)
    % equations
    E = 808.01/y^2.008;
    a = L/(2*n);
    c = @(i) atand( abs(x-a*[2*i-1])/y );
    coeff = [
        -0.0000000667
         0.0000157209
        -0.0010081511
         0.0020817855
         0.9991821678];
    S = 0;
    for i = 1:n
        for j = 1:5
            S = S + coeff(j)*c(i)^(5-j);
        end
    end
    
    result = E*S;
end

2 个评论
显示无隐藏无

Adam Schlei 2019-3-23

Thats awesome. Thanks.

Adam Schlei 2019-3-23

found the error. when x = L/(2*z-2)*i get rid of the i and it comes out prefect!

请先登录，再进行评论。

Answer 2

Walter Roberson 2019-3-23

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/451824-how-do-is-code-this-last-summation-i-was-able-to-do-the-first-one-but-have-no-clue-how-to-finish-i#answer_366931

在 MATLAB Online 中打开

%input values
y = 4;
n = 1;
L = 3;
z = 17;
%equations
c = @(x, y, n, l, i) atand(abs((x-l./(2.*n).*(2*i-1)))./y);
p = @(X) -0.0000000667*X.^4 + 0.0000157209*X.^3 - 0.0010081511*X.^2 + 0.0020817855*X + 0.9991821678;
s = @(x, y, n, l, c) sum(p(c(x(:), y, n, l, 1:n)),2);
zs = @(z, y, n, l, c) sum( s(l./(z-1).*(1:z-2), y, n, l, c), 1);
result = @(z, y, n, l, c) (zs(z, y, n, l, c) + s(l./(2*z-2), y, n, l, c) + s(l - l./(2*z-2), y, n, l, c)) ./ z;
result(z, y, n, L, c)

However the result is nowhere near 111

3 个评论
显示 1更早的评论隐藏 1更早的评论

Walter Roberson 2019-3-23

在 MATLAB Online 中打开

p = @(TEMPORARY_VARIABLE) -0.0000000667*TEMPORARY_VARIABLE.^4 + 0.0000157209*TEMPORARY_VARIABLE.^3 - 0.0010081511*TEMPORARY_VARIABLE.^2 + 0.0020817855*TEMPORARY_VARIABLE + 0.9991821678;

not sure why you say L=3 then use a lower case l in your equation. Doesn't this screw things up?

No. You need to learn about dummy parameters. For example

s = @(x, y, n, l, c) sum(p(c(x(:), y, n, l, 1:n)),2);

means the same thing as

s = @temporary_function_s;
function result = temporary_function_s(x, y, n, l, c)
    p = evalin('caller', 'p');
    result = sum(p(c(x(:), y, n, l, 1:n)),2);
end

which in turn means the same as

s = @temporary_function_s;
function result = temporary_function_s(First_Parameter, Second_Parameter, Third_Parameter, Fourth_Parameter, Fifth_Parameter)
  p = evalin('caller', 'p');
  result = sum(p(Fifth_Parameter(First_Parameter(:), Second_Parameter, Third_Parameter, Fourth_Parameter, 1:Third_Parameter)),2);
end