find-all-n-N-such-that-0-1-sinx-2n-2-cosx-2n-dx-1-4-1011-

Question Number 219563 by hardmath last updated on 28/Apr/25

$$\mathrm{find}\:\mathrm{all}\:\:\:\mathrm{n}\:\in\:\mathbb{N}^{\ast} \\ $$$$\mathrm{such}\:\mathrm{that}\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\left(\mathrm{sinx}\right)^{\mathrm{2n}−\mathrm{2}} \:\centerdot\:\left(\mathrm{cosx}\right)^{\mathrm{2}\boldsymbol{\mathrm{n}}} \:\mathrm{dx}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{1011}} } \\ $$

Answered by Nicholas666 last updated on 28/Apr/25

((Γ(n−(1/2))Γ(n+(1/2)))/(2Γ(2n))) ≥ (1/4^(1011) ) (((2n−1))/4) (([Γ(n−(1/2))]^2 )/(Γ(2n))) ≥(1/4^(1011) ) (π/4)≥(1/4^(1011) ) (with n=1) (π/(32)) ≥ (1/4^(1011) ) (with n=2) N^∗

$$\:\:\frac{\Gamma\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\mathrm{2}{n}\right)}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{1011}} } \\ $$$$\:\:\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{4}}\:\frac{\left[\Gamma\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)\right]^{\mathrm{2}} }{\Gamma\left(\mathrm{2}{n}\right)}\:\geqslant\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{1011}} } \\ $$$$\:\:\:\:\frac{\pi}{\mathrm{4}}\geqslant\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{1011}} }\:\:\:\:\:\left({with}\:{n}=\mathrm{1}\right) \\ $$$$\:\:\:\:\:\frac{\pi}{\mathrm{32}}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{1011}} }\:\:\:\:\:\:\left({with}\:{n}=\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathbb{N}^{\ast} \\ $$$$ \\ $$

Answered by maths2 last updated on 29/Apr/25

∫_0 ^1 sin^(2n−2) (x)cos^(2n) (x)dx≤∫_0 ^1 sin^(2n−2) (x)cos^(2n−2) (x)dx sin(x)cos(x)≤(1/2)(sin^2 (x)+cos^2 (x)=(1/2) ⇒0≤U_n =∫_0 ^1 sin^(2n−2) (x)cos^(2n−2) (x)dx≤∫_0 ^1 (1/2^(2n−2) )dx=(1/2^(2n−2) ) ⇒lim_(n→∞) ∫_0 ^1 sin^(2n−2) cos^(2n−2) (x)dx=0 ⇒∃N∈N ∀n≥N ;U_n <(1/4^(1011) ) Absurd ⇒∀n∈N ∄m∈N ∀k≥m U_k ≥(1/4^(1011) )

$$\int_{\mathrm{0}} ^{\mathrm{1}} {sin}^{\mathrm{2}{n}−\mathrm{2}} \left({x}\right){cos}^{\mathrm{2}{n}} \left({x}\right){dx}\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} {sin}^{\mathrm{2}{n}−\mathrm{2}} \left(\boldsymbol{{x}}\right)\boldsymbol{{cos}}^{\mathrm{2}\boldsymbol{{n}}−\mathrm{2}} \left({x}\right){dx} \\ $$$${sin}\left({x}\right){cos}\left({x}\right)\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left({sin}^{\mathrm{2}} \left({x}\right)+{cos}^{\mathrm{2}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\right. \\ $$$$\Rightarrow\mathrm{0}\leqslant{U}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {sin}^{\mathrm{2}{n}−\mathrm{2}} \left({x}\right){cos}^{\mathrm{2}{n}−\mathrm{2}} \left(\boldsymbol{{x}}\right)\boldsymbol{{dx}}\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{2}} } \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} {sin}^{\mathrm{2}{n}−\mathrm{2}} {cos}^{\mathrm{2}{n}−\mathrm{2}} \left({x}\right){dx}=\mathrm{0} \\ $$$$\Rightarrow\exists{N}\in\mathbb{N}\:\forall{n}\geqslant{N}\:;{U}_{{n}} <\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{1011}} }\:{Absurd} \\ $$$$\Rightarrow\forall{n}\in\mathbb{N}\:\nexists{m}\in\mathbb{N}\:\forall{k}\geqslant{m}\:{U}_{{k}} \geqslant\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{1011}} } \\ $$

Commented by hardmath last updated on 02/May/25

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$

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