How to prove: $2^\frac{3}{2}<\pi$ without writing the explicit values of $\sqrt{2}$ and $\pi$.
I am trying by calculus but don't know how to use here in this problem. Any idea?
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pjs36
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5Consider a regular $n$-gon inscribed in a circle of radius $1$. If $n = 8$ you can show that this has area $\sqrt{8}$ which gives $\pi > \sqrt{8} = 2^{3/2}$ – Winther Jan 24 '17 at 04:08
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ohhhhh!! Iet me work over this. This geometrical idea seems nice! – Subhash Chand Bhoria Jan 24 '17 at 04:10
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Let $ABCD$ be a square inscribed in the unit circle. Then the length of arc $AB$ is $\cfrac{\pi}{2}$, and the length of side $AB$ is $\sqrt{2}$. Since the shortest path between two points is the straight line segment between them, it follows that $\sqrt{2} \lt \cfrac{\pi}{2} \;\iff\; \sqrt{8} \lt \pi\,$.
dxiv
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It's pretty easy to see that $\pi > 3$, by inscribing a circle inside a regular hexagon. Then squaring both sides gives $\pi^2>9>8$. Taking square roots again gives $\pi > \sqrt{8}=2^\frac{3}{2}$
Stella Biderman
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how do we know $\pi>\sqrt{8}$? This I also thought but the condition is we don't have to use value of $\pi$. I think some idea of calculus or analysis involved here. – Subhash Chand Bhoria Jan 24 '17 at 04:07
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If you know $\pi^2>8$ then you can just take square roots of both sides. – Stella Biderman Jan 24 '17 at 04:10
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Say $\sin x\leq x$ with $x=\dfrac{\pi}{6}$ we get $\dfrac12\leq\dfrac{\pi}{6}$ or $3\leq\pi$ and $8<9\leq\pi^2$.
Nosrati
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Well yes, but it takes some explanation why. Even if you just say because of the degree 3 taylor expansion. – Stella Biderman Jan 24 '17 at 04:23
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It holds because the distance in the Cartesian plane from $(\cos x, 0)$ to $(\cos x, \sin x)$ is $\sin x,$ which is less then the distance from $(1,0)$ to $(\cos x, \sin x)$ which is less than the length of the circular arc from $(1,0)$ to $(\cos x, \sin x),$ which is $x.$ – DanielWainfleet Jan 24 '17 at 05:10
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If we take the definition of $\pi$ to be based on the circumference of a circle (namely that $\frac{C}{d} = \pi$), then we can see a nice geometric proof of this fact.
Consider a square with side length $1$. Now, draw a circle around this square such that all four corners of the square fall on the circle. This circle has a diameter of $\sqrt{2}$. The perimeter of the square involved is $4$.
$\frac{4}{\sqrt{2}} = 2^{\frac{3}{2}}$
Now, by looking at this picture it is quite obvious that the circumference of the circle is larger than the perimeter of the square. If you were constructing these shapes this could be easily verified. But now we can say that: \begin{align} C &> P_{square} \\ C &> 4 \\ \frac{C}{d} &> \frac{4}{d} \ \ \ \ \textbf{Since $d > 0$} \\ \pi &> 2^{\frac{3}{2}} \end{align}
Dylan
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Notice that $\sin x<x$ on $(0,\pi/2)$ by applying the Mean Value Theorem on $\sin x - \sin 0$ (and noting that $\cos t<1$ whenever $0<t<x<\pi/2$). Hence $$ \int\limits_{0}^{\pi/2}{\sin x\,dx} < \int\limits_{0}^{\pi/2}{x\,dx}. $$ Notice that the left-hand side is $1$, while the right-hand side is $\pi^2/8$.
Joey Zou
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Without invoking integration, the argument can also be made as follows: we have that $x^2/2 + \cos x -1 > 0$ for $x\in(0,\pi)$, since it is zero at $x=0$ and the derivative is positive in $(0,\pi)$ for the same reason mentioned above. Plugging in $x = \pi/2$ gives the inequality. – Joey Zou Jan 24 '17 at 05:05