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）
Basically, Symbolic Toolbox will help you:
syms y(x) n
f(x)=symsum((-1).^n*(x.^(2*n+1))/factorial(2*n+1),n,0,Inf)
4 个评论
adi putra
2018-4-1
Now that you have f(x)=sin(x), simply write
f(90)
but remember that sin function takes input arguments in radians, you need to write
f(pi/2)
to get a numerical result.
Roger Stafford
2018-4-1
@Birdman: I think you meant f(pi/2)
Birdman
2018-4-1
Yes, I just now edited it Roger.
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
0 个投票
pls answer this question ...how to write the equation into code
2 个评论
John D'Errico
2020-9-15
Please don't post a completely distinct question as an answer.
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
0 个投票
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.
kalai selvi
2020-9-23
thank you
Kunwar Pal Singh
2021-4-26
0 个投票
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
4 个评论
Walter Roberson
2021-5-20
What are the permitted values inside image ? The code logic posted would work if the array is all 1's and 2's, but it would also work if 0's could also be present. The logic could potentially also locate regions that were exclusively 3's.
The logic looks at the "adjacent faces" of each pixel (no diagonals), and it believes it is detecting that if the center pixel is 1 then there is a 2 among the 6 faces, and it believes it is detecting that if the center pixel is 2 then there is a 1 among the 6 faces. However, that is not what is actually happening if 0's or 3's are possible in the matrix.
The code can written without loops, and more accurately.
se1 = [0 0 0; 0 1 0; 0 0 0]; se2 = [0 1 0; 1 0 1; 0 1 0]; %do not set center
mask = cat(3, se1, se2, se1);
m1 = image == 1;
m2 = image == 2;
r1 = m1 & imdilate(m2, mask);
r2 = m2 & imdilate(m1, mask);
new_image = zeros(size(image));
new_image(r1 | r2) = 3;
Jakub
2021-5-21
Dear Walter,
thanks for the detailed reply. The permitted values in the image variable are 0's,1's and 2's. Location of the regions that were exclusively 3's is not desired for my purposes.
Thanks for your code, it works really well and much faster.
However, I'm still wondering, how to express this logic by a mathematical equation, do you have any idea?
Thanks, best regards
Walter Roberson
2021-5-22
∨
Jakub
2021-5-22
Thanks a lot!
Adhin Abhi
2022-1-4
0 个投票
(λlog vmax−log vmin) /(vmax−vmin )
1 个评论
Walter Roberson
2022-1-4
(lambda .* log(vmax) - log(vmin)) ./ (vmax - vmin)
Lukasz Sarnacki
2022-8-17
0 个投票
Please help
5 个评论
Walter Roberson
2022-8-17
What is 
? The 
looks like the Modified Bessel Function of the First Kind, MATLAB besseli(n, Z) -- but that function only accepts one parameter beyond the ν .
? The looks like the Modified Bessel Function of the First Kind, MATLAB besseli(n, Z) -- but that function only accepts one parameter beyond the ν .
Lukasz Sarnacki
2022-8-17
编辑：Lukasz Sarnacki
2022-8-17
In distorted fringe distribution, denoted as In(x; y) captured by the camera.
n represents the phase-shift index n = 0; 1; 2; :::;N - 1.
This equation is Standard N-step phase shifting.
Walter Roberson
2022-8-17
编辑：Walter Roberson
2022-8-17
Are A and B and ϕ functions, or are they arrays?
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
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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