Solving a system of equation with two different series.

Hi, could someone help me writing the following equation? The two different series (j & i) is confusing me.

[John D'Errico]

Be clear. (So far, this is very incomplete.)

1. What is known? What is unknown? What are you trying to solve for? a? h? beta?

2. In some places you use i where it might be the imaginary unit, sqrt(-1). In other places, i is clearly an index vector. Are these all real variables? Or are they complex? Use of i as a variable can often be dangerous to those who might want to read what you are doing.

Assuming h is a vector, how large is it? Or does this series go on forever?

What is the goal here? BE CLEAR! Are you trying to write an equation, or solve one? You cannot get an intelligent answer unless you provide a clear question.

[Shaibal Ahmed]

Hi, I'm really sorry, I made it unclear.

I know beta, a1, a2, I want to solve it for h, its a vector, the series goes until 36 units.

These are all real variables.

The goal is to solve the equation by writing it, maybe inside a loop (get values of all 36 units).

[Torsten]

What is h_0 ?

Best wishes

Torsten.

[Shaibal Ahmed]

Hi, you mean the value?

There's no h_0, it starts from h_1 (0.026).

These h's are ordinates of a unit hydrograph.

[John D'Errico]

What is a1 and a2? Note that since i is APPARENTLY the index in your loop, then a1*i and a2*i can be quite large, depending on those numbers. In that case, you might have to worry about overflow or underflow.

[Shaibal Ahmed]

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]

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.

Answer 1

[Torsten]

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.

Answer 2

[John D'Errico]

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.

[Shaibal Ahmed]

I'm sorry, it got a bit complicated, This is how the calculation should be in a simpler way,

1. h1 = 1 - (((beta*exp(-a1*1)*(a1*1+1)) + ((1-beta)*exp(-a2*1)*(a2*1+1)))
2. h2= 1 - h1 - (((beta*exp(-a1*2)*(a1*2+1)) + ((1-beta)*exp(-a2*2)*(a2*2+1)))
3. h3 = 1 - h1 - h2 - (((beta*exp(-a1*3)*(a1*3+1)) + ((1-beta)*exp(-a2*3)*(a2*3+1)))
4. h4 = 1 - h1 - h2 - h3 - (so on...)

does it makes sense now?

[John D'Errico]

Those exponential terms are simple to write, but not really that relevant. You have 36 equations that relate h1 to h2 to h3, etc. But then you tell us that the sum of the h elements must be 1.

That effectively makes this a linear system of 37 equations in 36 unknowns. There is no exact solution. Just because you want a solution to exist will not make it so.

So you need to decide what you can give up. Either drop that last constraint, or accept that your solution will not satisfy all 37 equations exactly. You cannot have both.

[Shaibal Ahmed]

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]

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.