0
$\begingroup$

Hi everyone! I have a question about how to find the closed form of a function defined by

$$\phi(\theta)=\inf_{x\geq 2}f(x;\theta)\equiv\inf_{x\geq 2}\frac{(x+2)^2}{\frac{1}{\theta}\left(\frac{x-1}{2}-\frac{1}{x}\right)+\frac{x^2-1}{16}},\ \ \theta>0$$

Since finding the minimum of $f(x;\theta)$ leads to solving a cubic polynomial. Could someone help me to characterize $\phi(\theta)$ please? Many thanks!

$\endgroup$

1 Answer 1

1
$\begingroup$

If $\theta>2$ then $\phi(\theta)=16$. For $\theta<2$, let $$t(x)=\frac{4(x^3-4x^2-6x-4)}{x^2(1+2x)},$$ $$g(x)=f(x,t(x)).$$ The function t is invertible from $[x_m,\infty)$ to $[0,2)$, where $x_m$ is the root of t(x)=0, approximately 5.28. We have $\phi(\theta)=g(t^{-1}(\theta))$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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