1
$\begingroup$

I want to find the smallest constant C such that for all x and y we have$$\frac{\cosh(x)e^y + e^{-y}}{e^{Cx^2}+1} \leq e^{Cy^2}$$

Playing around with a graphing utility online, it looks like $C \approx .6$

Is there a good analytic way to approach these types of questions?

I don't know if it helps, but when $x = 0$ we get $\cosh(y) \leq e^{Cy^2}$ which is true when $C \geq .5$

$\endgroup$
0
$\begingroup$

First, note the following: We have to optimize the globally minimal difference between two multivariable functions, subject to a constraint. Such problems often require a sufficient amount of brute force computation to render an analytical approach impractical.

Generalizing the problem a bit, we have a global inequality between two functions $f:\mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ and $g:\mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$: $$f_{C}(x,y) \leq g_{C}(x,y),\ \forall (x,y) \in \mathbb{R}^{2},$$ where the subscript denotes the particular constant $C$ used. We can simplify by defining a function $h_{C}(x,y) = g_{C}(x,y)-f_{C}(x,y)$, and require that

$$h_{C}(x,y) \geq 0, \ \forall (x,y) \in \mathbb{R}^{2}.$$

Let's not forget about the end goal of this model: We want the minimal constant $C_{min}$ such that the above is satisfied. Define a function $H:\mathbb{R} \rightarrow \mathbb{R}$ which maps $C$ to the global min of $h_{C}$: $$H(C) = \min_{(x,y)\in \mathbb{R}^{2}}\{h_{C}(x,y)\}.$$ Then the entire problem can be formulated as $$C_{min} = \min_{C \in \mathbb{R}}\{C\},$$ subject to $$H(C_{min}) \geq 0.$$ Why is this approach often impractical? First of all, finding $H(C)$ and then minimizing $H(C)$ are tasks which contain several messy steps. Let's just start with the first: Deriving $H(C)$.

To find $H(C)$, we have to find the global minimum $(x_{min},y_{min})$ for $h_{C}$ in terms of $C$, where $h_{C}(x,y)$ for your problem is

$$h_{C}(x,y) = e^{Cy^2}-\frac{\cosh(x)e^y + e^{-y}}{e^{Cx^2}+1}.$$

On the first calculation (solving $\frac{\partial h_{C}}{\partial x} = 0, \frac{\partial h_{C}}{\partial y}=0$) even Wolfram Alpha runs out of allowed time, which is usually a good indicator that the problem's brute force requires computational methods. enter image description here

$\endgroup$

Your Answer

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

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