Input argument "alpha" is undefined.

i get this error. how to solve this error?????

""""??? Input argument "alpha" is undefined.

Error in ==> newvod3RL at 6 alphaINT = (real(alpha)>0) .* ceil(real(alpha)) + ... """

 function out = vod3RL(t, in, alpha)
% function out = vod3RL(t, in, alpha)
% Variable order derivative, Riemann-Liouville definition, variant 3.
% Duarte Valrio, 2009.

 alphaINT = (real(alpha)>0) .* ceil(real(alpha)) + ...
    ( (real(alpha)==0) & (imag(alpha)~=0) ); % number of differentiations to perform
alphaRES = alpha - alphaINT; % residue of alphaINT with negative real part only

 % negative differentiation (i.e. integration) performed
out = zeros(size(t));
for k = 2 : length(t) % the first integral is always 0, since no time elapsed, hence no need to make k = 1...
    tau = t(1:k);
    integrand = (t(k) - tau).^(-alphaRES(k - (1:k) + 1) - 1) ./ gamma(-alphaRES(k - (1:k) + 1)) .* in(1:k);
    if any(~isfinite(integrand)) % eliminate eventual inf and nan points
        integrand(find(isinf(integrand))) = 0;
        integrand(find(isnan(integrand))) = 0;
    end
    integrand(1) = 2*integrand(1); % this is to handle steps properly with Caputo's   definition
    % (otherwise the first point only has half the influence and the result comes halved all the way to the end)
    out(k) = sum( (integrand(1:end-1)+integrand(2:end))/2 .* (tau(2:end)-tau(1:end-1)) );
end

 % necessary differentiations performed
while 1
    if sum(alphaINT) == 0, break, end % exit condition
    temp = diff([0; out]) ./ [t(2)-t(1); diff(t)]; % this is to handle steps properly
    out = out.*(alphaINT==0) + temp.*(alphaINT~=0);
    alphaINT = (alphaINT-1) .* (alphaINT~=0); % take away one integration
end

alphaINT = (real(alpha)>0) .* ceil(real(alpha)) + ...

[Torsten]

The error addresses a function called "newvod3RL", but you posted "vod3RL".

I suspect that a variable "alpha" does not appear in the list of parameters for function "newvod3RL".

Best wishes

Torsten.

Answer 1

[Stalin Samuel]

before using a variable alpha in function you must use that variable alpha.It may be initialization or some other value

[naeem]

i am supplying three inputs as

alpha=0.9 in=0.5 t=1 and now i am getting an error of

 ??? Attempted to access t(2); index out of bounds because numel(t)=1.
Error in ==> vod1RL at 30
  temp = diff([0; out]) ./ [t(2)-t(1); diff(t)]; % this is to handle steps properly

[Torsten]

But you write yourself that you supply t=1 to vod3RL.

This is a scalar value - consequently t(2) (which you address to in the above expression) does not exist.

Best wishes

Torsten.