You can use that \cos (2x) = \cos^2 x - \sin^2 x, and hence 1 - \cos \frac{\pi}{2^k} = 1 - \cos^2 \frac{\pi}{2^{k+1}} + \sin^2 \frac{\pi}{2^{k+1}} = 2\sin^2 \frac{\pi}{2^{k+1}}. Then write \sqrt{\frac{2}{1+\cos \frac{\pi}{2^k}}} = \sqrt{\frac{2(1 - \cos \frac{\pi}{2^k})}{1 - \cos^2 \frac{\pi}{2^{k}}}} = \sqrt{\frac{4\sin^2 \frac{\pi}{2^{k+1}}}{\sin^2 \frac{\pi}{2^k}}} = \frac{2\sin \frac{\pi}{2^{k+1}}}{\sin \frac{\pi}{2^k}}, ...

Notice that f'(x)=\begin{cases} 2x\sin\frac1x-\cos\frac1x &\text{ if } x\ne0\\ 0 &\text{ if } x=0 \end{cases}. Therefore |f'(x)|\le 2|x|\cdot\left|\sin\frac1x\right|+\left|\cos\frac1x\right|\le 3 \quad \forall x\ne 0. ...

I assume that 1<\alpha<2. If not, as remarked by the questioner, f(x) is Lipschitz on (0,1) and the problem is simpler. You can use the following approach. Take a small \delta>0. You wish ...

HINTS: From the Convolution Theorem , we have \int_{-\infty}^\infty f(t)g(t)\,e^{i\omega t}\,dt=\frac{1}{2\pi}\int_{-\infty}^\infty F(\omega-\omega')G(\omega')\,d\omega' Setting f(t)=g(t)=\frac{\sin(t)}{\pi t} ...

Set \mathscr{F}'(x) : = \sin^4(x) , i.e., \mathscr{F} is the antiderivative of \sin^4(x) . Then by the fundamental theorem of calculus h(x) = \mathscr{F} \left( \frac{1}{x} \right) - \mathscr{F}\left( 2 \right). ...

Let n be a positive integer and let f(x) = x^{2n} \cdot \sin\left(\frac{1}{x}\right) f(0) = 0 Then mathematical induction can be used to prove: (a) The nth derivative of f(x) exists for ...