Facing problem while solving this.

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AVINASH SAHU
AVINASH SAHU 2022-7-31
评论： Walter Roberson 2022-7-31
p1 = @(x) (1./F(x)) .* ((0.5 .* hbar(x)) - Q);
P = @(x) integral(@(x) p1(x),0,x,'ArrayValued', true);
xi = linspace(0,1) ;
P1 = zeros(size(xi)) ;
for j = 1:length(xi)
P1(j) = P(xi(j));
end
Using the abpve approach I am getting datasets of P1 but not able to figure out how to approach in the following method given below
for i = 1:100
p1(i) =(((0.5*hbar(i))-Q)/F(i));
% P = @(x) integral(@(x) p1(x),0,x,'ArrayValued', true); %%%% how
% to write this line here?
end

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Walter Roberson
Walter Roberson 2022-7-31
p1{1} = @(x) (1./F(x)) .* ((0.5 .* hbar(x)) - Q);
P{1} = @(x) integral(p1{1}, 0, x, 'ArrayValued', true);
xi = linspace(0,1) ;
P1 = zeros(size(xi)) ;
for j = 1:length(xi)
P1(j) = P{1}(xi(j));
end
for i = 1:100
p1{i} = @(x) (((0.5*hbar(i))-Q)/F(i));
P{i} = @(x) integral(p1{i}, 0, x, 'ArrayValued', true);
end
Note that this would overwrite the original p1{1} and P{1}
Also note that your p1{i} does not involve x, so the body would effectively be constant, and the result of the integral would be x times the constant minus 0 times the constant, which is predictable ahead of time.
  2 个评论
AVINASH SAHU
AVINASH SAHU 2022-7-31
Appreciating your help. But I want to say that I am getting the results (i.e. datasets of P1) using the following approach
p1 = @(x) (1./F(x)) .* ((0.5 .* hbar(x)) - Q);
P = @(x) integral(@(x) p1(x),0,x,'ArrayValued', true);
xi = linspace(0,1) ;
P1 = zeros(size(xi)) ;
for j = 1:length(xi)
P1(j) = P(xi(j));
end
and I want to get the same results using the below approach but don't know how to proceed
p1(i) =(((0.5*hbar(i))-Q)/F(i)); % same as above p1
% P = ?????
% P = @(x) integral(@(x) p1(x),0,x,'ArrayValued', true); %%%% how
% to write this line here in the form of indices and the obtain the
% datasets of uppercase "P1"
xi = linspace(0,1) ;
P1 = zeros(size(xi)) ;
for j = 1:length(xi)
P1(j) = P(xi(j));
end
Thank you very much for your help. Hope I am able to clarify my problem.
Walter Roberson
Walter Roberson 2022-7-31
You cannot do that.
Your original code defines p1 as an anonymous function handle. Your new version defines p1(i) as a numeric constant, not as a function handle. It is not useful to integrate a numeric constant.

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