how can i convert from one value to multi values

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work wolf
work wolf 2022-6-19
评论： work wolf 2022-6-23
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))

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采纳的回答

Image Analyst
Image Analyst 2022-6-20
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
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work wolf
work wolf 2022-6-21
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 2022-6-23
@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

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更多回答（1 个）

Ayush Kumar Jaiswal
编辑：Ayush Kumar Jaiswal 2022-6-19
You want to calculate that function at different values of alpha, it can done using
arrayfun (func, arr);
Read the documentation here.

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