フィルターのクリア

Results of Newton-Raphson method to find the root ?

4 ビュー (過去 30 日間)
zayed
zayed 2011 年 12 月 31 日
Hi,
I need to solve the following equation by using Newton-raphson method :
a. f(lambda)=-z'*inv(M)*inv(W)*Q*inv(W)*inv(M)*z,
W=inv(M)+lambda*Q
f'(lambda)=z'*inv(M)*inv(W)*(Q^2*inv(W)+inv(W)*Q^2)*inv(W)*inv(M)*z
Q and W are complex square matrices ,also z and M are metrices .
I wrote a code to find the root by Newton's method ,but the value of the root is complex ,but it must be real.I am not sure a bout the derivative of f(x).
I have another form to the function f(x) ,but I don't know if it's suitable to be solved by Newton's method in matlab,the other form is:
m gamma_k*|x_k|^2
b. f(lambda)=sum -----------------------
k=1 (1+lambda(gamma_k))^2
m (gamma_k)^2*|x_k|^2
f'(lambda)= 2* sum --------------------------
k=1 (1+lambda(gamma_k))^3
where gamma_k is sub indices , gamma is eigenvalue
m is total number of eigenvalues of T=(M)^(1/3)*Q*(M)^(1/2)
x=U'*M^(-1/2)*z ,U is eigen vectors of T.
Is this form in (b) of equation suitable to be solved by Newton Raphson method.
The code for form (a):
delta=1e-12;
epsilon=1e-12;
max1=500;
lambda=-1/(2*gamma);
for k=1:max1
zeta=cos(theta);
I=eye(n,n);
Q=zeta*I-p*p';
W=inv(M)+lambda*Q;
y1=2*z'*inv(M)*inv(W)*Q.^2*inv(W)*inv(M)*z;
y=-z'*inv(M)*inv(W)*Q*inv(W)*inv(M)*z;
p1=lambda-y/y1;
err=abs(p1-lambda);
lambda=p1
if (err<delta)
break
end
k
err
end
  7 件のコメント
Andrew Newell
Andrew Newell 2012 年 1 月 1 日
In (b), the denominator is missing a right parenthesis, so its interpretation is ambiguous.
zayed
zayed 2012 年 1 月 1 日
I fixed the equation of form (b) it.Just I need to represent it in matlab

サインインしてコメントする。

採用された回答

Walter Roberson
Walter Roberson 2012 年 1 月 1 日
Are you able to find the eigenvalues and eigenvectors of T? If so, then (b) should be straight-forward to code (though perhaps tedious.)
plot the real and imaginary parts of solutions on the same graph, but in different colors or symbols. Use "hold on" and plot each complex point as it is generated. Newton's method will move back and forth on x values, so it is probably best not to attempt to connect the solutions with lines until after all the solutions have been generated, after which you would sort based on the x coordinate.
  3 件のコメント
Walter Roberson
Walter Roberson 2012 年 1 月 1 日
X = Something;
plot(X, real(lambda), 'r*', X, imag(lambda), 'bs');
If gamma_k is the list of eigenvalues, then
flambda = 0;
for K = 1 : length(gamma_k)
flambda = flambda + gamma_k(K) * x_k(K) / (1+lambda(gamma_k(K)))^2;
end
and likewise for f' .
However, I cannot tell from what you have posted what x_k is. I also cannot tell whether lambda(gamma_k) is intended to indicate multiplication or something else.
There might be vectorized ways of calculating f(lambda), but that is going to depend on what lambda(gamma_k) and x_k mean.
zayed
zayed 2012 年 1 月 1 日
I don't recognize what X mean.
x_k :z'*inv(M)*U ,U is eigenvectors
lambda*(gamma_k) is muliplication
gamma-k is eigen value

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeSparse Matrices についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by