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Please help me convert equation to matlab code.
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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 个评论
Walter Roberson
2018-4-1
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?
回答(8 个)
Roger Stafford
2018-4-1
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 个评论
kalai selvi
2020-9-15
pls answer this question ...how to write the equation into code
2 个评论
Walter Roberson
2020-9-15
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
2020-9-16
How to write a code on IOTA filter in fbmc system
2 个评论
Walter Roberson
2020-9-16
编辑:Walter Roberson
2020-9-17
Warning: pudn has questionable security. Take precautions when you access it.
Kunwar Pal Singh
2021-4-26
please answer this....how to write this equation into MATLAB CODE
1 个评论
Walter Roberson
2021-4-26
%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
2021-5-19
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
Lukasz Sarnacki
2022-8-17
Please help
5 个评论
Lukasz Sarnacki
2022-8-17
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
2022-8-18
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)
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