Question Number 116437 by ZiYangLee last updated on 04/Oct/20
$$\mathrm{Prove}\:\mathrm{that}\:\pi<\frac{\mathrm{22}}{\mathrm{7}} \\ $$
Answered by floor(10²Eta[1]) last updated on 04/Oct/20
![let′s look at the integral and solve it: I(x)=∫_0 ^1 ((x^4 (1−x)^4 )/(1+x^2 ))dx =∫_0 ^1 ((x^8 −4x^7 +6x^6 −4x^5 +x^4 )/(1+x^2 ))dx =∫_0 ^1 (x^6 −4x^5 +5x^4 −4x^2 +4−(4/(1+x^2 )))dx =[(x^7 /7)−((2x^6 )/3)+x^5 −((4x^3 )/3)+4x−4tan^(−1) (x)]_0 ^1 =((22)/7)−π and since I(x)>0 so ((22)/7)−π>0⇒π<((22)/7)](https://www.tinkutara.com/question/Q116438.png)
$$\mathrm{let}'\mathrm{s}\:\mathrm{look}\:\mathrm{at}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{and}\:\mathrm{solve}\:\mathrm{it}: \\ $$$$\mathrm{I}\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{4}} \left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{4}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{8}} −\mathrm{4x}^{\mathrm{7}} +\mathrm{6x}^{\mathrm{6}} −\mathrm{4x}^{\mathrm{5}} +\mathrm{x}^{\mathrm{4}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{x}^{\mathrm{6}} −\mathrm{4x}^{\mathrm{5}} +\mathrm{5x}^{\mathrm{4}} −\mathrm{4x}^{\mathrm{2}} +\mathrm{4}−\frac{\mathrm{4}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)\mathrm{dx} \\ $$$$=\left[\frac{\mathrm{x}^{\mathrm{7}} }{\mathrm{7}}−\frac{\mathrm{2x}^{\mathrm{6}} }{\mathrm{3}}+\mathrm{x}^{\mathrm{5}} −\frac{\mathrm{4x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{4x}−\mathrm{4tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{22}}{\mathrm{7}}−\pi \\ $$$$\mathrm{and}\:\mathrm{since}\:\mathrm{I}\left(\mathrm{x}\right)>\mathrm{0}\:\mathrm{so} \\ $$$$\frac{\mathrm{22}}{\mathrm{7}}−\pi>\mathrm{0}\Rightarrow\pi<\frac{\mathrm{22}}{\mathrm{7}} \\ $$
Commented by ZiYangLee last updated on 04/Oct/20
$$\mathrm{wow} \\ $$
Answered by MJS_new last updated on 04/Oct/20
![the area of a regular polygon with n sides and an incircle of radius 1 is A_n =n(√((1−cos ((2π)/n))/(1+cos ((2π)/n)))) lim_(n→∞) A_n =π [ _(approximates the circle)^(obvious because the polygon) ] if ((22)/7)<π ⇒ A_n =((22)/7) has no solution ∈R ⇔ if A_n =((22)/7) ⇒ ((22)/7)>π A_n =((22)/7) ⇒ n≈90.4298 ⇒ ((22)/7)>π](https://www.tinkutara.com/question/Q116477.png)
$$\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{polygon}\:\mathrm{with}\:{n}\:\mathrm{sides} \\ $$$$\mathrm{and}\:\mathrm{an}\:\mathrm{incircle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{1}\:\mathrm{is} \\ $$$${A}_{{n}} ={n}\sqrt{\frac{\mathrm{1}−\mathrm{cos}\:\frac{\mathrm{2}\pi}{{n}}}{\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{2}\pi}{{n}}}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{A}_{{n}} =\pi\:\left[\:_{\mathrm{approximates}\:\mathrm{the}\:\mathrm{circle}} ^{\mathrm{obvious}\:\mathrm{because}\:\mathrm{the}\:\mathrm{polygon}} \right] \\ $$$$\mathrm{if}\:\frac{\mathrm{22}}{\mathrm{7}}<\pi\:\Rightarrow\:{A}_{{n}} =\frac{\mathrm{22}}{\mathrm{7}}\:\mathrm{has}\:\mathrm{no}\:\mathrm{solution}\:\in\mathbb{R} \\ $$$$\Leftrightarrow\:\mathrm{if}\:{A}_{{n}} =\frac{\mathrm{22}}{\mathrm{7}}\:\Rightarrow\:\frac{\mathrm{22}}{\mathrm{7}}>\pi \\ $$$${A}_{{n}} =\frac{\mathrm{22}}{\mathrm{7}}\:\Rightarrow\:{n}\approx\mathrm{90}.\mathrm{4298} \\ $$$$\Rightarrow\:\frac{\mathrm{22}}{\mathrm{7}}>\pi \\ $$