Please help me convert equation to matlab code.

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]

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?

Answer 1

[Birdman]

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)

[adi putra]

so, how i can get sin(x)? ex sin(90)?

[Birdman]

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]

@Birdman: I think you meant f(pi/2)

[Birdman]

Yes, I just now edited it Roger.

Answer 2

[Roger Stafford]

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.)

Answer 3

[kalai selvi]

pls answer this question ...how to write the equation into code

[John D'Errico]

Please don't post a completely distinct question as an answer.

[Walter Roberson]

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.

Answer 4

[kalai selvi]

How to write a code on IOTA filter in fbmc system

[Walter Roberson]

http://en.pudn.com/Download/item/id/2208316.html

Warning: pudn has questionable security. Take precautions when you access it.

[kalai selvi]

thank you

Answer 5

[Kunwar Pal Singh]

please answer this....how to write this equation into MATLAB CODE

[Walter Roberson]

%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;

Answer 6

[Jakub Laznovsky]

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

[Walter Roberson]

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]

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]

∨

[Jakub]

Thanks a lot!

Answer 7

[Adhin Abhi]

(λlog vmax−log vmin) /(vmax−vmin )

[Walter Roberson]

(lambda .* log(vmax) - log(vmin)) ./ (vmax - vmin)

Answer 8

[Lukasz Sarnacki]

Please help

[Walter Roberson]

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 .

[Lukasz Sarnacki]

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]

Are A and B and ϕ functions, or are they arrays?

[Lukasz Sarnacki]

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]

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 =