Let $$H(x)=\dfrac{\sin{\frac{\pi}{6}x}}{x}$$
Find the smallest postive ineteger $n$ such $$H(n)<H(n+1)$$
My approach is the following: I use wolframalpha found $n=9?$
Now I don't know how to prove it..
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1What is $f$? $f=H$ ? – chubakueno Mar 25 '15 at 16:57
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Write out the first few values of $\sin\frac\pi6x$. – Akiva Weinberger Mar 25 '15 at 16:57
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Now that you have edited it, columbus8's comment finishes it. $n=9$ doesn't need that many trial verifications. – chubakueno Mar 25 '15 at 17:02
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$$\begin{align} 0&\to\frac\pi6=&\sqrt{{\pi^2705600}/{25401600}}\\ 1&\to\frac12=&\sqrt{{4762800}/{25401600}}\\ 2&\to\frac{\sqrt3}4=&\sqrt{{6350400}/{25401600}}\\ 3&\to\frac13=&\sqrt{{2822400}/{25401600}}\\ 4&\to\frac{\sqrt3}8=&\sqrt{{1190700}/{25401600}}\\ 5&\to\frac1{10}=&\sqrt{{254016}/{25401600}}\\ 6&\to0=&\sqrt{0/{25401600}}\\ 7&\to-\frac1{14}=&-\sqrt{{129600}/{25401600}}\\ 8&\to-\frac{\sqrt3}{16}=&-\sqrt{{297675}/{25401600}}\\ 9&\to-\frac19=&-\sqrt{\color{green}{313600}/{25401600}}\\ 10&\to-\frac{\sqrt3}{20}=&-\sqrt{{190512}/{25401600}}\\ \end{align}$$