how can i convert from one value to multi values

19 Jun. 2022
2 Respuestas
Respuesta aceptada
Actualizado a las 23 Jun. 2022
6 Visualizaciones (30 días)

0 votos

how can i replace value of
alpha = 0.5
by multi values as
alpha =[0.1 0.3 0.5 0.6 0.66 0.9 1]
in the following code:
alpha = 0.5;
u0 = 0;
a_k = @(k) (k + 1)^(1 - alpha) - (k)^(1 - alpha);
n = 100;
a = 0;
b = 1;
h = (b - a) / n;
t = a:h:b;
f = @(t,u) (-u.^4) + (gamma(2*alpha+1) ./ gamma(alpha+1) ) .* (t.^alpha) - ...
(2./gamma(3 - alpha) ) .* (t.^(2 - alpha)) + (t.^(2*alpha) - t.^2).^4;
up = zeros(1, n);
uc = zeros(1, n);
zp = zeros(1, n);
uc = zeros(1, n); % ??? is this your "u"?
C = gamma(2 - alpha) * h ^ alpha;
for ni = 1:n
up(ni) = a_k(ni - 1) * u0;
for k = 1:ni - 1
up(ni) = up(ni) + (a_k(ni - 1 - k) - a_k(ni - k)) * uc(k);
end
zp(ni) = C * f(t(ni), up(ni));
uc(ni) = up(ni) + C * f(t(ni), up(ni) + zp(ni));
end
fprintf('%g\n', up(1:20))

Iniciar sesión para responder a esta pregunta.

 Respuesta aceptada

Image Analyst
Image Analyst el 20 de Jun. de 2022

1 voto

Try this:
% Define all the alphas that we want to use.
allAlpha =[0.1 0.3 0.5 0.6 0.66 0.9 1]
% Iterate the code for each value of alpha.
for kk = 1 : length(allAlpha)
alpha = allAlpha(kk);
% Existing code below:
u0 = 0;
a_k = @(k) (k + 1)^(1 - alpha) - (k)^(1 - alpha);
n = 100;
a = 0;
b = 1;
h = (b - a) / n;
t = a:h:b;
f = @(t,u) (-u.^4) + (gamma(2*alpha+1) ./ gamma(alpha+1) ) .* (t.^alpha) - ...
(2./gamma(3 - alpha) ) .* (t.^(2 - alpha)) + (t.^(2*alpha) - t.^2).^4;
up = zeros(1, n);
uc = zeros(1, n);
zp = zeros(1, n);
uc = zeros(1, n); % ??? is this your "u"?
C = gamma(2 - alpha) * h ^ alpha;
for ni = 1:n
up(ni) = a_k(ni - 1) * u0;
for k = 1:ni - 1
up(ni) = up(ni) + (a_k(ni - 1 - k) - a_k(ni - k)) * uc(k);
end % of k loop
zp(ni) = C * f(t(ni), up(ni));
uc(ni) = up(ni) + C * f(t(ni), up(ni) + zp(ni));
end % of ni loop
fprintf('%g\n', up(1:20))
end % of kk loop

6 comentarios
Mostrar 4 comentarios más antiguos Ocultar 4 comentarios más antiguos

work wolf
work wolf el 20 de Jun. de 2022
Editada: work wolf el 20 de Jun. de 2022
@Image Analyst Thank you so much. Please, how can i store every
up at allAlpha(kk) separated,
meaning
output=[ up(1:20) at (allallAlpha(1))' up(1:20) at (allAlpha(2))' up(1:20) at (allAlpha(3))' up(1:20) at (allAlpha(4))' ...
up(1:20) at (allAlpha(5))' up(1:20) at (allAlpha(6))' up(1:20) at (allAlpha(7))'];
This is useful for comparison between result.
Image Analyst
Image Analyst el 20 de Jun. de 2022
Is that a matrix, row vector, or column vector? I don't understand what you put. Like (allAlpha(5))' is the same as allAlpha(5) since allAlpha(5) is just a scalar and there is no need to transpose it with '.
work wolf
work wolf el 20 de Jun. de 2022
@Image Analyst i want result as matrix including every column at evey iteration,meaning up at alpha=0.1 then alpha=0.3 and so on as follows
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0.0497389463654563 0.0213889960049408 0.00577975952460106 0.00286678264632385 0.00185596757939561 0.000217889077015351 0
0.0756032841848556 0.0394787481302252 0.0123999751730029 0.00655408489750118 0.00438988350200604 0.000568453154565168 0
0.0937806925446395 0.0556290379398444 0.0193798935286129 0.0107113922486972 0.00734884592892711 0.00102120965764830 0
0.107860035746299 0.0704308289266767 0.0265305609236315 0.0151729898408803 0.0106037797699722 0.00155716732200875 0
0.119276626823768 0.0841960303939159 0.0337537642308159 0.0198414190139364 0.0140747131135280 0.00216271878416734 0
0.128782977112920 0.0971145894937620 0.0409904050975548 0.0246522821978120 0.0177066747337709 0.00282736971807349 0
0.136836519308874 0.109313451384330 0.0482014917128138 0.0295597555016475 0.0214592905356894 0.00354264272057228 0
0.143742977513616 0.120882992733044 0.0553596049729811 0.0345295051483806 0.0253015035929395 0.00430146626493324 0
0.149720693652571 0.131890546165088 0.0624445380331634 0.0395348080402283 0.0292085925801107 0.00509780051317483 0
0.154933371449271 0.142388021211973 0.0694408532510186 0.0445542468881334 0.0331603538578347 0.00592639232345276 0
0.159508321299395 0.152416520942504 0.0763364150931947 0.0495702537425901 0.0371399266969435 0.00678260656294221 0
0.163547351524805 0.162009299532333 0.0831214599873308 0.0545681439491122 0.0411329985829613 0.00766230529256888 0
0.167133636814165 0.171193743128752 0.0897879802538048 0.0595354498177205 0.0451272473771931 0.00856175844777340 0
0.170336250166931 0.179992745424877 0.0963293011057279 0.0644614459321516 0.0491119375908007 0.00947757605512695 0
0.173213269808024 0.188425691974366 0.102739781309983 0.0693368016829291 0.0530776205939369 0.0104066556516299 0
0.175813980719025 0.196509182520684 0.109014595844465 0.0741533209796437 0.0570159070743530 0.0113461407282561 0
0.178180480530615 0.204257572561355 0.115149574557707 0.0789037433441743 0.0609192910387687 0.0122933873534719 0
0.180348881531626 0.211683386918866 0.121141080069081 0.0835815892474123 0.0647810114200892 0.0132459369880677 0
Image Analyst
Image Analyst el 20 de Jun. de 2022
You can just index the matrices with kk
zp(kk, ni) = C * f(t(ni), up(ni));
uc(kk, ni) = up(ni) + C * f(t(ni), up(ni) + zp(ni));
assuming zp and uc are row vectors in your original code. Using kk as the first argument would load them into the kk'th row of a matrix.
work wolf
work wolf el 21 de Jun. de 2022
i did it, but give me all output, zores !!
% Define all the alphas that we want to use.
allAlpha =[0.1 0.3 0.5 0.6 0.66 0.9 1];
% Iterate the code for each value of alpha.
for kk = 1 : length(allAlpha)
alpha = allAlpha(kk);
% Existing code below:
u0 = 0;
a_k = @(k) (k + 1)^(1 - alpha) - (k)^(1 - alpha);
n = 100;
a = 0;
b = 1;
h = (b - a) / n;
t = a:h:b;
f = @(t,u) (-u.^4) + (gamma(2*alpha+1) ./ gamma(alpha+1) ) .* (t.^alpha) - ...
(2./gamma(3 - alpha) ) .* (t.^(2 - alpha)) + (t.^(2*alpha) - t.^2).^4;
up = zeros(kk, n+1);
uc = zeros(kk, n+1);
zp = zeros(kk, n+1);
% uc = zeros(kk, n+1); % ??? is this your "u"?
C = gamma(2 - alpha) * h ^ alpha;
for ni = 1:n+1 %length(t)
% up(ni) = a_k(ni - 1) * u0;
up(kk,ni) = a_k(ni - 1) * u0; %modfied
for k = 1:ni - 1
% up(ni) = up(ni) + (a_k(ni - 1 - k) - a_k(ni - k)) * uc(k);
up(kk,ni) = up(kk,ni) + (a_k(ni - 1 - k) - a_k(ni - k)) * uc(kk,k); % modfied
end % of k loop
% zp(ni) = C * f(t(ni), up(ni));
% uc(ni) = up(ni) + C * f(t(ni), up(ni) + zp(ni));
zp(kk, ni) = C * f(t(ni), up(kk,ni)); % modfied
uc(kk, ni) = up(kk,ni) + C * f(t(ni), up(kk,ni) + zp(kk,ni)); % modfied
end % of ni loop
% fprintf('%g\n', up(1:20))
end % of kk loop
up
gives,
up =
Columns 1 through 13
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
Columns 14 through 26
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
Columns 27 through 39
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
Columns 40 through 52
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
Columns 53 through 65
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
Columns 66 through 78
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
Columns 79 through 91
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
Columns 92 through 101
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
work wolf
work wolf el 23 de Jun. de 2022
@Image Analyst thank you so much again. it's done with edite
up = zeros(kk, n+1);
uc = zeros(kk, n+1);
zp = zeros(kk, n+1);
before loop (for kk = 1 : length(allAlpha) ) and replace kk by length(allAlpha).
Best regards

Iniciar sesión para comentar.

Más respuestas (1)

Ayush Kumar Jaiswal
Ayush Kumar Jaiswal el 19 de Jun. de 2022
Editada: Ayush Kumar Jaiswal el 19 de Jun. de 2022

1 voto

You want to calculate that function at different values of alpha, it can done using
arrayfun (func, arr);
Read the documentation here.

Categorías

Más información sobre Modify Image Colors en Centro de ayuda y File Exchange.

Etiquetas

Preguntada:

work wolf
work wolf
el 19 de Jun. de 2022

Comentada:

work wolf
work wolf
el 23 de Jun. de 2022

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by