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 December 15th, 2017, 11:23 AM #1 Senior Member   Joined: Jan 2015 From: usa Posts: 103 Thanks: 1 Taylor's series 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.
 December 15th, 2017, 01:59 PM #2 Global Moderator   Joined: May 2007 Posts: 6,660 Thanks: 647 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.
 December 15th, 2017, 02:09 PM #3 Senior Member   Joined: Jan 2015 From: usa Posts: 103 Thanks: 1 @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.
 December 16th, 2017, 02:33 PM #4 Global Moderator   Joined: May 2007 Posts: 6,660 Thanks: 647 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.

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