Info
This question is locked. Vuélvala a abrir para editarla o responderla.
Please help me convert equation to matlab code.
16 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Deal all.
I need you help to convert this equation to matlab code
I spend a lot of time to write it but it doesn't work. Thank you.
1 comentario
Walter Roberson
el 1 de Abr. de 2018
Are you permitted to use the symbolic toolbox?
Is the question about providing some kind of symbolic proof, or is it about calculation of the formula using finite precision and a particular numeric input?
Respuestas (8)
Roger Stafford
el 1 de Abr. de 2018
N = 100; % <-- Choose some large number
s = x;
for n = 2*N-1:-2:1
s = x - s*x^2/((n+2)*(n+1));
end
(I think you meant to take the limit as N approaches infinity, not x.)
0 comentarios
kalai selvi
el 15 de Sept. de 2020
pls answer this question ...how to write the equation into code
2 comentarios
Walter Roberson
el 15 de Sept. de 2020
π is written as pi in MATLAB.
exp of an expression is written as exp(expression) in MATLAB.
is written as sqrt(expression) in MATLAB.
kalai selvi
el 16 de Sept. de 2020
How to write a code on IOTA filter in fbmc system
2 comentarios
Walter Roberson
el 16 de Sept. de 2020
Editada: Walter Roberson
el 17 de Sept. de 2020
Warning: pudn has questionable security. Take precautions when you access it.
Kunwar Pal Singh
el 26 de Abr. de 2021
please answer this....how to write this equation into MATLAB CODE
1 comentario
Walter Roberson
el 26 de Abr. de 2021
%these variables must be defined in a way appropriate for your situation
S_N = rand() * 10
theta = randn() * 2 * pi
l = randi([2 10])
b_1 = rand()
c_11 = rand()
t_year = randi([1950 2049])
d_11 = rand()
t_1 = rand()
t_x = t_1 + rand()
lambda_a = randi([500 579])
LOTF_a = rand()
P = rand()
K_l = rand()
k_0 = rand()
t_tau = randi(10)
overhaulcost_a = 1000 + rand()*100
%the work
syms t
part1 = int(S_N .* cos(theta) .* l .* b_1 .* t_year .* d_11, t, t_1, t_x);
part2 = int(lambda_a .* LOTF_a, t, t_1, t_x);
part3 = int(P*K_l .* t_year + P .* k_0 .* l .* l .* t_tau, t, t_1, t_x);
part4 = overhaulcost_a ;
result = part1 - part2 - part3 - part4;
Jakub Laznovsky
el 19 de Mayo de 2021
Hi guys, can you please help me with conversion this piece of code to the mathematical equation?
It i a simple 3D mask proceeding the image, and searching for adjoining number one and number two. Thank you in advance.
Code:
m1=[0 0 0; 0 1 0; 0 0 0];
m2=[0 1 0; 1 1 1; 0 1 0];
mask=zeros(3,3,3);
mask(:,:,1)=m1;mask(:,:,2)=m2;mask(:,:,3)=m1;
for i=2:size(image,1)-1
for j=2:size(image,2)-1
for k=2:size(image,3)-1
help_var=image(i-1:i+1,j-1:j+1,k-1:k+1);
if sum(unique(help_var(mask==1)))==3
new_image(i,j,k)=3; %marks adjoining pixel with number 3
end
end
end
end
4 comentarios
Adhin Abhi
el 4 de En. de 2022
(λlog vmax−log vmin) /(vmax−vmin )
1 comentario
Walter Roberson
el 4 de En. de 2022
(lambda .* log(vmax) - log(vmin)) ./ (vmax - vmin)
Lukasz Sarnacki
el 17 de Ag. de 2022
Please help
5 comentarios
Lukasz Sarnacki
el 17 de Ag. de 2022
Thay all are arrays
A(x; y) is the average intensity of the fringe image
B(x; y) is the so-called intensity modulation.
ϕ is the corresponding wrapped phase
Walter Roberson
el 18 de Ag. de 2022
syms n N integer
syms A(x,y) B(x,y) phi(x,y)
Pi = sym(pi)
I(n,x,y) = A(x, y) + B(x,y) * cos(phi(x,y) - 2*Pi*n/N)
numerator = simplify(symsum(I(n, x, y) .* sin(2*Pi*n/N), n, 0, N-1))
denominator = simplify(symsum(I(n, x, y) .* cos(2*Pi*n/N), n, 0, N-1))
eqn = phi(x,y) == atan(numerator ./ denominator)
simplify(eqn)
This question is locked.
Ver también
Categorías
Más información sobre Linear Algebra en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!