ltrinang

https://imgur.com/a/Pl1x0D0
négyzetesen konvergáló módszer az newton
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function x = LTriangSolve(a,b)
n = length(b);
x(1,1) = b(1)/a(1,1);
for i = 2:n
    x(i,1)=(b(i)-a(i,1:i-1)*x(1:i-1,1))./a(i,i);
end
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function [invMtx] = I3(A)

    %simetrikus, hatekony modszer a Cholesky
    L = Cholesky(A);

    b = [1;0;0];
    y = LTriangSolve(L,b);
    a1 = UTriangSolve(L',y);

    b = [0;1;0];
    y = LTriangSolve(L,b);
    a2 = UTriangSolve(L',y);

    b = [0;0;1];
    y = LTriangSolve(L,b);
    a3 = UTriangSolve(L',y);

    Mtx = A
    invMtx = [a1;a2;a3]
end
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function [sqrt] = II9 (a,maxit)
    tic
    %sqrt(a) = x |^2
    f = @(x) x^2 - a;
    fd = @(x) 2*x;
    fplot(f,[-a/2,a/2]);
    hold on;

    %Olyan kezdoerteket kell valasztanunk, aminek a negyzete nagyobb a
    %szamunknal (gy�k nem lehet negativ)

    [koz,n] = Newton_Raphson(f,fd,a/2,1/10000,maxit);
    sqrt = koz(n);
    toc
end
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% Adott A es B matrix. Megoldjuk A , b (B oszlop) egyenleteket Gaussal es LU-val.
% Futasi idok tesztelese, melyik gyorsabb?

function M_1()
  A = [1,0,3; 2,4,5;9,1,3];
  B = [1,1,1; 0,2,3; 4,5,2];

  n = length(B(1,:));

  for i= 1:n
    tic
    GaussElimSolve(A,B(:,i))
    toc
    tic
    [L,U,P] = L4_lup(A);
    y = UTriangSolve(U,B(:,i))
    toc

  endfor
 % Az LU felbontas gyorsabb.

end
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function M_1()
  f = @(x) sin(x).^2;
  a = 0;
  b = pi;

  integral = L10_romberg(f,a,b,0.001,15);
  V = pi*integral
end

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% Szamitsuk ki Gauss eliminacioval egy matrix inverzet.
function [I] = M_1(A)
    b = [1;0;0];
    [U1,c1] = GaussElim(A,b);
    x1 = UTriangSolve(U1,c1);

    b = [0;1;0];
    [U2,c2] = GaussElim(A,b);
    x2 = UTriangSolve(U2,c2);

    b = [0;0;1];
    [U3,c3] = GaussElim(A,b);
    x3 = UTriangSolve(U3,c3);


    I = [x1;x2;x3];
    I = I'

end