mona123

Let $\nu\ge 1$ be a parameter.

For all $t>0,$ we consider
\begin{align}
A(t) & =\frac{1-\cos(t\sqrt{4\nu-1})}{4\nu-1}-(\cosh(t)-1) \\[10pt]
g(t) & =\frac{\frac{\sin(t\sqrt{4\nu-1})}{\sqrt{4\nu-1}}+\sinh(t)}{A(t)}
\end{align}


By using Taylor's series, I want to prove that there exists a constant $c>0$ which doesn't depend on $\nu$ such that $$\frac{\nu t^3}{\ln\left(1-\frac{2}{g(t)+1}\right)}\le c$$ for all $t\le \frac{1}{\sqrt{4\nu-1}}$


Can you please help me to do so.

Thanks.

mathman

First step: Let $\displaystyle T=\frac{1}{\sqrt{4\nu -1}}\ so\ that\ \nu=\frac{1}{4}(1+\frac{1}{T^2})$. The algebra should be easier.

mona123

@mathman,

This is what i wrote:
for all $t\le \frac{1}{\sqrt{c}}$

$$sin(t\sqrt{c})=t\sqrt{c}-\frac{(t\sqrt{c})^3}{3!}+..+(-1)^n\frac{(t\sqrt{c})^{2n+1}}{(2n+1)!}+O((t\sqrt{c })^{2n+3})$$ for all $t\le \frac{1}{\sqrt{c}}$
for all $t\le \frac{1}{\sqrt{c}}$

$$cos(t\sqrt{c})=1-\frac{(t\sqrt{c})^2}{2!}+..+(-1)^n\frac{(t\sqrt{c})^{2n}}{(2n)!}+O((t\sqrt{c})^{ 2n+1})$$
$$\cosh(t)=1+\frac{t^2}{2!}+..+\frac{t^{2n}}{(2n)! }+O(t^{2n+1})$$



$$\sinh(t)=1+\frac{t^3}{3!}+..+\frac{t^{2n+1}}{(2n +1)!}+O(t^{2n+3})$$

But i don't know how to find the result.

mathman

Study the behavior of $\displaystyle 1-\frac{2}{g(t)+1}$. For ln to be > 0 for all t, the expressions must be > 1 for all t.