0
$\begingroup$

I have the following problem: I have this equation: $$f(x)=(1-x) \left (\frac {ac+bdx}{b+c}\right) \left (\frac{(1-gx)^{11} -1+gx} {-gx}\right) \left (\frac{(1+\frac{a-dx}{b+c})^{10} (\frac{a-dx}{b+c})}{(1+\frac{a-dx}{b+c})^{10}-1}\right) -\left (\frac {\frac {ac+bdx}{b+c}}{1-\mu}\right )\left (1+ \left (\frac {i(\frac{a-dx}{b+c})^{10}}{(1+(\frac{a-dx}{b+c})^{10})^{11} -1} \right)-\left (\frac{(1+\frac{a-dx}{b+c})^{10}(1-gx)^{10}-1}{-(\frac{a-dx}{b+c})^{10}g}\right)-\frac {1}{(\frac{a-dx}{b+c})^{10}}\left (\frac{ (1+\frac{a-dx}{b+c})^{10} ( 1-gx)^{10}-1}{-(1+\frac{a-dx}{b+c})^{10}(1-gx)^{10}} \right )\right) $$

$a, b, c, d, g, \mu, i $ are parameters which are all bigger than zero.

I have two questions: 1. Is there a useful rule or approximation which I can use to find the optimal value for x which maximizes the function f(x)? I already saw this post: How to solve an nth degree polynomial equation But I don't really know which method is the best for my formula. Of course I need to take the first derivation first but maybe there is a way to simplify this function.

  1. Is there an easy way to say: f(x) has only one maximum and this is unique for a given value of x. (maybe this can help? https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs)

The things I already know: The functions has an inverted U-shape and I can solve it practically for different parameter values using the generalized reduced gradient method. Apart from that $ \frac {ac+bdx}{b+c}$ is monotonically increasing in $x$ and $(\frac{(1+\frac{a-dx}{b+c})^{10} (\frac{a-dx}{b+c})}{(1+\frac{a-dx}{b+c})^{10}-1})$ is monotonically decreasing with $x$.

Thank your for your help and sry if something is unclear

ok, this function above is just horrible. Maybe I can rewrite it a little bit better with things I already know:

$$f(x)=(1-x)L(x)\left(\frac {(1-g_x)^{11}-1+gx}{-gx}\right)A(x)-\frac {L(x)}{1-\mu}\left(1+\frac {i r(x)}{(1+r(x))^{10}-1)}\left(\frac {(1+r(x))^{10} (1-gx)^{10}}{-gx}-\frac {1}{r(x)} \left (\frac {(1+r(x))^{10} (1-gx)^{10}}{(1+r(x))^{10} (1-gx)^{10}-1}\right)\right)\right)$$

and $L(x)$ is monotonically increasing with x and $r(x)$ is monotonically decreasing with x and $A(x)$ is monotonically decreasing with x

There is another way to look at this equation from above and maybe I can bring in some intuition. The above function can be rewritten as $$f(x)= (1-x)y_1(x)-y_2(x)$$ and $y_1(x)$ is continues and has an inverted U-shape and the same holds for $y_2(x)$.

$\endgroup$
4
  • 1
    $\begingroup$ If I may ask : where did you catch such a monster ? $\endgroup$ Feb 18, 2017 at 13:59
  • $\begingroup$ The function is derived from another function with some sigma-signs. $\endgroup$
    – PAS
    Feb 18, 2017 at 14:34
  • $\begingroup$ I imagine the equation is $f(x)=0$? Also, please use \left( and \right) to give the correct size to your brackets. it improves readability. $\endgroup$ Feb 18, 2017 at 14:38
  • $\begingroup$ Ok, thank you for your answer. I'm very sorry for the confusion. I had a mistake in my original post. I edited the main post and I hope everything is clear now. $\endgroup$
    – PAS
    Feb 18, 2017 at 14:59

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.