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Please help me convert equation to matlab code.

8 Ansichten (letzte 30 Tage)
adi putra
adi putra am 1 Apr. 2018
Locked: Rik am 9 Jul. 2024
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 Kommentar
Walter Roberson
Walter Roberson am 1 Apr. 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?

Antworten (8)

Birdman
Birdman am 1 Apr. 2018
Bearbeitet: Birdman am 1 Apr. 2018
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 Kommentare
Roger Stafford
Roger Stafford am 1 Apr. 2018
@Birdman: I think you meant f(pi/2)
Birdman
Birdman am 1 Apr. 2018
Yes, I just now edited it Roger.

Roger Stafford
Roger Stafford am 1 Apr. 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.)

kalai selvi
kalai selvi am 15 Sep. 2020
pls answer this question ...how to write the equation into code
  2 Kommentare
John D'Errico
John D'Errico am 15 Sep. 2020
Please don't post a completely distinct question as an answer.
Walter Roberson
Walter Roberson am 15 Sep. 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
kalai selvi am 16 Sep. 2020
How to write a code on IOTA filter in fbmc system
  2 Kommentare
Walter Roberson
Walter Roberson am 16 Sep. 2020
Bearbeitet: Walter Roberson am 17 Sep. 2020
Warning: pudn has questionable security. Take precautions when you access it.
kalai selvi
kalai selvi am 23 Sep. 2020
thank you

Kunwar Pal Singh
Kunwar Pal Singh am 26 Apr. 2021
please answer this....how to write this equation into MATLAB CODE
  1 Kommentar
Walter Roberson
Walter Roberson am 26 Apr. 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
Jakub Laznovsky am 19 Mai 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 Kommentare
Jakub
Jakub am 22 Mai 2021
Thanks a lot!

Adhin Abhi
Adhin Abhi am 4 Jan. 2022
(λlog vmax−log vmin) /(vmax−vmin )
  1 Kommentar
Walter Roberson
Walter Roberson am 4 Jan. 2022
(lambda .* log(vmax) - log(vmin)) ./ (vmax - vmin)

Lukasz Sarnacki
Lukasz Sarnacki am 17 Aug. 2022
Please help
  5 Kommentare
Lukasz Sarnacki
Lukasz Sarnacki am 17 Aug. 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
Walter Roberson am 18 Aug. 2022
syms n N integer
syms A(x,y) B(x,y) phi(x,y)
Pi = sym(pi)
Pi = 
π
I(n,x,y) = A(x, y) + B(x,y) * cos(phi(x,y) - 2*Pi*n/N)
I(n, x, y) = 
numerator = simplify(symsum(I(n, x, y) .* sin(2*Pi*n/N), n, 0, N-1))
numerator = 
denominator = simplify(symsum(I(n, x, y) .* cos(2*Pi*n/N), n, 0, N-1))
denominator = 
eqn = phi(x,y) == atan(numerator ./ denominator)
eqn = 
simplify(eqn)
ans = 

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