Gemm94 0 Posted April 25 (edited) Hi everyone, I have a problem when doing an approximation. The problem comes in the final results that I have to demonstrate two functions below equal each other frac{1}{8\sqrt{2}\cos{\frac{\phi-\phi_0}{2}}(1-\sin{\frac{\phi-\phi_0}{2}})\sqrt{1-\sin{\frac{\phi-\phi_0}{2}}}} = \frac{1}{[1+\cos{(\phi-\phi_0)}]^2}. \begin{equation} \frac{1}{8\sqrt{2}\cos{\frac{\phi-\phi_0}{2}}(1-\sin{\frac{\phi-\phi_0}{2}})\sqrt{1-\sin{\frac{\phi-\phi_0}{2}}}} = \frac{1}{[1+\cos{(\phi-\phi_0)}]^2}. \end{equation} I have checked the two functions by numerical calculation to a graph and see that two functions give exactly the same shape with the $\phi\leq \pi$ as shown in the figure. \begin{figure}[!t] \centering \includegraphics[scale=1]{Nonuniform-ka20-E.eps} \caption{Comparison between two functions} \end{figure} Edited April 25 by Gemm94 0 Share this post Link to post Share on other sites

taeto 92 Posted April 25 (edited) When I copy/paste the first equation and enclose it between \ [ and \ ] (without the blanks) I get \[\frac{1}{8\sqrt{2}\cos{\frac{\phi-\phi_0}{2}}(1-\sin{\frac{\phi-\phi_0}{2}})\sqrt{1-\sin{\frac{\phi-\phi_0}{2}}}} = \frac{1}{[1+\cos{(\phi-\phi_0)}]^2}.\] Neither expression is defined for \(\phi = \phi_0 +\pi.\) For all other values of \(\phi\) the equation is equivalent to \[8\sqrt{2}\cos\theta (1-\sin \theta)\sqrt{1-\sin \theta} = [1+\cos (2\theta)]^2,\] where \(\theta = \frac{\phi-\phi_0}{2}.\) For \(\theta = 0\) this is not right, since \(8\sqrt{2} \neq 4.\) Did you copy your equation correctly? You say that your graphs coincide, so they must show something different from this. Or just that they "have the same shape", but one might be a multiple of the other? We can try to use \(\cos (2\theta) = 2\cos^2 \theta - 1,\) which makes \([1+\cos (2\theta)]^2 = 4\cos^4 \theta.\) Edited April 25 by taeto 0 Share this post Link to post Share on other sites

taeto 92 Posted April 25 (edited) Then \[4\cos\theta (1-\sin \theta)\sqrt{1-\sin \theta} = [1+\cos (2\theta)]^2 = 4\cos^4 \theta\] becomes equivalent to \(\cos \theta = \sqrt{1-\sin \theta},\) and by Pythagoras's \(\cos^2 + \sin^2 = 1,\) this is equivalent to \(-\pi/2 \leq \theta \leq \pi/2\) if we assume \(|\theta|\leq \pi.\) That is since if \(\theta\) lies outside this interval, \(\cos \theta\) is negative, whereas \(\sqrt{1-\sin \theta}\) is non-negative. Edited April 25 by taeto 0 Share this post Link to post Share on other sites

taeto 92 Posted April 25 (edited) In conclusion, if we replace your \(8\sqrt{2}\) by \(4,\) then the two functions agree whenever \(|\phi-\phi_0| < \pi,\) where difference is taken modulo \(\pi,\) and otherwise they do not. If you replace \(8\sqrt{2}\) by \(4,\) and replace the LHS by its absolute value, then the functions agree everywhere they are defined. Edited April 25 by taeto 0 Share this post Link to post Share on other sites