Finding a root with interval constraint

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Chien-Chia Huang
Chien-Chia Huang 2014-1-24
评论： Walter Roberson 2014-1-25
Hello there!
I am trying to find a point x within the time interval [t-1,t] (for some t, say t = 3) so that the function attains value zero. That is, I want to solve "Q_0 + integral(a+b*sin(c*t+d)-mu,t-1,x) = 0" for x in [t-1,t]. My code is the following
y = fsolve(@(x) Q_0+(a-mu)*(x-t+1)-(b/c)*cos(c*x+d)+(b/c)*cos(c*(t-1)+d),0,optimset('Display','off'))
wherein (a,b,c,d) satisfy a + b*sin(c*t+d), and Q_0 and mu are constants. This code has no problem. However, the solution may sometimes be outside the time interval [t-1,t], which is not what I want.
So, my question is if there is a way to restrict the routine to find a solution that lies within [t-1,t] exactly?
Thanks!

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Walter Roberson
Walter Roberson 2014-1-24
As your x0 is a scalar (0), your x are scalar, and that implies you can use fzero() instead of fsolve(). With fzero() you can pass the interval [t-1 t] as your x0.
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Chien-Chia Huang
Chien-Chia Huang 2014-1-25
编辑：Chien-Chia Huang 2014-1-25
Thanks, Walter. My code now goes like this (value of mu changes)
[FirstVanish,~,exitflag] = fzero(@(x) queuelength(counter_qln)+(a(j)-mu)*(x-t+1)-(b(j)/c(j))*cos(c(j)*x+d(j))+(b(j)/c(j))*cos(c(j)*(t-1)+d(j)),[t-1,t])
However, it showed the error msg
Error using fzero (line 274) The function values at the interval endpoints must differ in sign.
This is why I will need to know how to get things going without seeing the above.

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