Solving a system of equation with two different series.

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Shaibal Ahmed
Shaibal Ahmed 2016-8-12
评论： Torsten 2016-8-15
Hi, could someone help me writing the following equation? The two different series (j & i) is confusing me.
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Shaibal Ahmed
Shaibal Ahmed 2016-8-12
编辑：Shaibal Ahmed 2016-8-12
a1 & a2 were calculated to make the equation simpler, they are fixed value and i is the index, the sum of all h's (36 of them) should be equal to 1.
John D'Errico
John D'Errico 2016-8-12
编辑：John D'Errico 2016-8-12
When did you tell s about the sum constraint on h? lol. If you tell us everything in the beginning, it would help.
Without that constraint on h, this is a simple loop. With it, we have a set of 36 LINEAR equations in 35 unknowns, since h(1) is known. As such, there will generally be no exact solution.

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回答（2 个）

Torsten
Torsten 2016-8-12
beta=...;
a1=...;
a2=...;
h(1) = 0.026;
for i=2:36
h(i)=1-sum(h(1:i-1))-(beta*exp(-a1*i)*(a1*i+1)+(1-beta)*exp(-a2*i)*(a2*i+1);
end
Best wishes
Torsten.
John D'Errico
John D'Errico 2016-8-12
编辑：John D'Errico 2016-8-12
Given your comments, assuming that a1 and a2 are not problematic, this is a simple loop. Not sure what the issue is. Start at the beginning.
h = zeros(1,36);
h(1) = 0.026.
for i = 2:36
h(i) = 1 - sum(h(1:(i-1)) - (beta*exp(-a1*i)*(a1*i+1) + (1-beta)*exp(-a2*i)*(a2*i+1)));
end
Note that I preallocated the vector h. Growing a vector over time is a slow operation, that will get slower as your vector gets larger. So preallocate vectors when they will be grown.
Edit: given the extra information about the sum of the h vector to be 1...
Without that constraint on h, this is a simple loop. With it, we have a set of 36 LINEAR equations in 35 unknowns, since h(1) is known. As such, there will generally be no exact solution.
So you need to explain what you really need to do. Tell us what we need to know, or else we are just chasing a moving target.
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Shaibal Ahmed
Shaibal Ahmed 2016-8-13
编辑：Shaibal Ahmed 2016-8-14
Hi, this is how I did it.
beta=...;
a_1=...;
a_2=...;
x = zeros (1,36)';
for i = 1:36
x(i) = beta * (exp (-a_1 * i)) * ( a_1 * i + 1 );
end
y = zeros (1,36)';
for i = 1:36
y(i) = (1 - beta) * (exp (-a_2 * i)) * ( a_2 * i + 1 );
end
z = zeros (1,36);
for i = 1:36
z(i) = (x(i) + y(i));
end
h = [0.026 -diff(z)]';
s = sum(h);
Torsten
Torsten 2016-8-15
This is not what your formula says.
Also, the sum s of the h_i's from i=1 to i=35 is not 1, but
s=35*h(1)+(z(1)-z(36))
Best wishes
Torsten.

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