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adi putra on 1 Apr 2018
Commented: Walter Roberson on 18 Aug 2022
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.
Walter Roberson on 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?

Birdman on 1 Apr 2018
Edited: Birdman on 1 Apr 2018
syms y(x) n
f(x)=symsum((-1).^n*(x.^(2*n+1))/factorial(2*n+1),n,0,Inf)
Birdman on 1 Apr 2018
Yes, I just now edited it Roger.

Roger Stafford on 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 on 15 Sep 2020
pls answer this question ...how to write the equation into code ##### 2 CommentsShowHide 1 older comment
Walter Roberson on 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 on 16 Sep 2020
How to write a code on IOTA filter in fbmc system
##### 2 CommentsShowHide 1 older comment
kalai selvi on 23 Sep 2020
thank you

Kunwar Pal Singh on 26 Apr 2021 Walter Roberson on 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 on 19 May 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];
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);
new_image(i,j,k)=3; %marks adjoining pixel with number 3
end
end
end
end
Jakub on 22 May 2021
Thanks a lot!

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

Lukasz Sarnacki on 17 Aug 2022 Walter Roberson on 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 = 