Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the requirements to minimize the following:

$$ (f(x)_1 + f(x)_2 + f(x)_3) $$

where:

$$ f(x)_1 = y_1 - (\exp(b+m_1) \times x) $$ $$ f(x)_2 = y_2 - (\exp(b+m_2) \times x) $$ $$ f(x)_3 = y_3 - (\exp(b+m_3) \times x) $$

given the range of $x$:

$$ a = 1.191206112 $$ $$ b = 1.321909214 $$ $$ x \in R \space|\space a \le x \le b $$

Is there a way to estimate the value of $x$ that returns the minimized sum of the three functions? As you can see, my $y$ and $m$ values are specific to the function but the $b$ is constant across all three.

Currently, I am testing random values between $[a,b]$ and recording the smallest sum. This takes about 50,000 iterations before I start approaching the asymptote. There has got to be a better way!

share|improve this question
    
Shouldn't the minimum always be $f(b)_1 + f(b)_2 + f(b)_3$ ? Looks like there is a missing $x$... –  Amine Aug 23 '12 at 21:39
    
You have so many variables here...For which values are you seeking the solution (e.g. x or yi or b or mi)? –  Emmad Kareem Aug 23 '12 at 23:07
    
@EmmadKareem I am looking to find $x$ in the three functions (single value), that returns the smallest possible sum of $f(x)_1$+$f(x)_2$+$f(x)_3$ –  Michael Markieta Aug 24 '12 at 2:23
    
Looks like the optimal value of x: x1=x2=x3=b. this way, each f is minimized, and hence the sum is minimized. –  Emmad Kareem Aug 24 '12 at 9:02
    
But $x$ must fall between the range stated. At the moment I test all values between the range min/max until I find the best value. –  Michael Markieta Aug 25 '12 at 0:10
add comment

1 Answer 1

What about finding zeroes of the derivative of the sum of your three functions to find a candidate for the minimum? Are you sure the $x$ does not appear in the argument of the exponential function? If this is really not the case and if the third function is really of the same form as the first two, then, since $\exp(...)$ is always positive, the minimum is at the right end of the interval for $x$. Am I missing something?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.