Estimate of $\frac{\cosh^2(x)}{\sinh^2(x)}-\frac{\cosh^2(y)}{\sinh^2(y)}$ depending on $x$ and $y$

Question

Can we get any estimate of $\frac{\cosh^2(x)}{\sinh^2(x)}-\frac{\cosh^2(y)}{\sinh^2(y)}$ depending on $x$ and $y$, where $x, y$ real with $x>y.$

I have tried using increasing property of hyperbolics, get nothing.

Please help. Thanks.

Answer 1

Not sure if this helps but, if $x,y \ge a > 0$, Lagrange's theorem gives you $$ f(x)-f(y) = f'(\xi) (x-y), \quad \xi \in (x,y) $$

which would yield $$ \left|\frac{\cosh^2 x}{\sinh^2 x}-\frac{\cosh^2 y}{\sinh^2 y} \right| \leq 2 |\coth a | \textrm{csch}^2 a |x - y| $$

Answer 2

If $y$ is close to $x$, you can use Taylor series and get $$\coth ^2(x)-\coth ^2(y)=\sum_{n=1}^p a_n (y-x)^n+O\left((y-x)^{p+1}\right)$$ and the first coefficients will be $$a_1=-2 \coth ^3(x)+2 \coth (x)$$ $$a_2=3 \coth ^4(x)-4 \coth ^2(x)+1$$ $$a_3=-4 \coth ^5(x)+\frac{20}{3} \coth ^3(x)-\frac{8 }{3}\coth (x)$$ and so on.