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adi putra il 1 Apr 2018
Commentato: Walter Roberson il 18 Ago 2022
Questo/a domanda è stato/a segnalato/a da 2 collaboratori
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 CommentoMostra NessunoNascondi Nessuno
Walter Roberson il 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?

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### Risposte (8)

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

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Roger Stafford il 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.)
##### 0 CommentiMostra -1 commenti meno recentiNascondi -1 commenti meno recenti

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kalai selvi il 15 Set 2020
pls answer this question ...how to write the equation into code ##### 2 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
Walter Roberson il 15 Set 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.

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kalai selvi il 16 Set 2020
How to write a code on IOTA filter in fbmc system
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kalai selvi il 23 Set 2020
thank you

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Kunwar Pal Singh il 26 Apr 2021 ##### 1 CommentoMostra NessunoNascondi Nessuno
Walter Roberson il 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;

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Jakub Laznovsky il 19 Mag 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
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Jakub il 22 Mag 2021
Thanks a lot!

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

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Lukasz Sarnacki il 17 Ago 2022 ##### 5 CommentiMostra 4 commenti meno recentiNascondi 4 commenti meno recenti
Walter Roberson il 18 Ago 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 = Accedi per commentare.

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