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Please help me convert equation to matlab code.

16 ビュー (過去 30 日間)
adi putra
adi putra 2018 年 4 月 1 日
Locked: Rik 2024 年 7 月 9 日 4:20
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 件のコメント
Walter Roberson
Walter Roberson 2018 年 4 月 1 日
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 件)

Birdman
Birdman 2018 年 4 月 1 日
編集済み: Birdman 2018 年 4 月 1 日
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 件のコメント
Roger Stafford
Roger Stafford 2018 年 4 月 1 日
@Birdman: I think you meant f(pi/2)
Birdman
Birdman 2018 年 4 月 1 日
Yes, I just now edited it Roger.

Roger Stafford
Roger Stafford 2018 年 4 月 1 日
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 2020 年 9 月 15 日
pls answer this question ...how to write the equation into code
  2 件のコメント
John D'Errico
John D'Errico 2020 年 9 月 15 日
Please don't post a completely distinct question as an answer.
Walter Roberson
Walter Roberson 2020 年 9 月 15 日
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 2020 年 9 月 16 日
How to write a code on IOTA filter in fbmc system
  2 件のコメント
Walter Roberson
Walter Roberson 2020 年 9 月 16 日
編集済み: Walter Roberson 2020 年 9 月 17 日
Warning: pudn has questionable security. Take precautions when you access it.
kalai selvi
kalai selvi 2020 年 9 月 23 日
thank you

Kunwar Pal Singh
Kunwar Pal Singh 2021 年 4 月 26 日
please answer this....how to write this equation into MATLAB CODE
  1 件のコメント
Walter Roberson
Walter Roberson 2021 年 4 月 26 日
%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 2021 年 5 月 19 日
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 件のコメント
Walter Roberson
Walter Roberson 2021 年 5 月 22 日
Jakub
Jakub 2021 年 5 月 22 日
Thanks a lot!

Adhin Abhi
Adhin Abhi 2022 年 1 月 4 日
(λlog vmax−log vmin) /(vmax−vmin )
  1 件のコメント
Walter Roberson
Walter Roberson 2022 年 1 月 4 日
(lambda .* log(vmax) - log(vmin)) ./ (vmax - vmin)

Lukasz Sarnacki
Lukasz Sarnacki 2022 年 8 月 17 日
Please help
  5 件のコメント
Lukasz Sarnacki
Lukasz Sarnacki 2022 年 8 月 17 日
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 2022 年 8 月 18 日
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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