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Question Number 118712 by 1549442205PVT last updated on 19/Oct/20
Prove the following inequalities: 1)(((n+1)/2))^n >n! for ∀n∈N^∗ ,n>1 2)∣sinnx∣≤n∣sinx∣ for ∀n∈N^∗
$$\mathrm{Prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequalities}: \\ $$$$\left.\mathrm{1}\right)\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{n}} >\mathrm{n}!\:\mathrm{for}\:\forall\mathrm{n}\in\mathrm{N}^{\ast} ,\mathrm{n}>\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\mid\mathrm{sinnx}\mid\leqslant\mathrm{n}\mid\mathrm{sinx}\mid\:\mathrm{for}\:\forall\mathrm{n}\in\mathrm{N}^{\ast} \\ $$
Answered by Dwaipayan Shikari last updated on 19/Oct/20
((1+2+3+4+...n)/n)≥(1.2.3.4..n)^(1/n) ((n(n+1))/(2n))≥(n!)^(1/n) (((n+1)/2))^n ≥n! take n>1 ((3/2))^2 ≥2! (Which is true) So (((n+1)/2))^n >n! (n>1)
$$\frac{\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…{n}}{{n}}\geqslant\left(\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}..{n}\right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}{n}}\geqslant\left({n}!\right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)^{{n}} \geqslant{n}! \\ $$$${take}\:{n}>\mathrm{1} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \geqslant\mathrm{2}!\:\:\:\left({Which}\:{is}\:{true}\right) \\ $$$${So} \\ $$$$\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)^{{n}} >{n}!\:\:\:\:\left({n}>\mathrm{1}\right) \\ $$
Commented by 1549442205PVT last updated on 20/Oct/20
Thank Sir.
$$\mathrm{Thank}\:\mathrm{Sir}. \\ $$
Answered by TANMAY PANACEA last updated on 19/Oct/20
2)applying logic 1≥∣sinnx∣≥0 and 1≥∣sinx∣≥0 but n≥n∣sinx∣≥0 so n∣sinx∣>∣sinnx∣
$$\left.\mathrm{2}\right){applying}\:{logic} \\ $$$$\mathrm{1}\geqslant\mid{sinnx}\mid\geqslant\mathrm{0} \\ $$$$\boldsymbol{{and}}\:\mathrm{1}\geqslant\mid{sinx}\mid\geqslant\mathrm{0} \\ $$$${but}\:{n}\geqslant{n}\mid{sinx}\mid\geqslant\mathrm{0} \\ $$$${so}\:{n}\mid{sinx}\mid>\mid{sinnx}\mid \\ $$$$ \\ $$
Commented by 1549442205PVT last updated on 19/Oct/20
I don′t see logic here,havn′t final result n∣sinx∣≥∣sin(nx)∣
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{see}\:\mathrm{logic}\:\mathrm{here},\mathrm{havn}'\mathrm{t}\:\mathrm{final} \\ $$$$\mathrm{result}\:\mathrm{n}\mid\mathrm{sinx}\mid\geqslant\mid\mathrm{sin}\left(\mathrm{nx}\right)\mid \\ $$
Commented by TANMAY PANACEA last updated on 19/Oct/20
ok sir i am trying
$${ok}\:{sir}\:\:{i}\:{am}\:{trying} \\ $$
Commented by 1549442205PVT last updated on 20/Oct/20
Thank Sir.
$$\mathrm{Thank}\:\mathrm{Sir}. \\ $$
Answered by mindispower last updated on 19/Oct/20
1)⇔nln(((n+1)/2))≥ln(Π_(k=1) ^n k) Π_(k=1) ^n (k)≤(((Σ_(k=1) ^n k)/n))^n ,AM−GM ⇒ln(Π_(k=1) ^n k)≤nln(((Σ_(k=1) ^n k)/n))=nln(((n+1)/2))
$$\left.\mathrm{1}\right)\Leftrightarrow{nln}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)\geqslant{ln}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{k}\right) \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left({k}\right)\leqslant\left(\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}}{{n}}\right)^{{n}} ,{AM}−{GM} \\ $$$$\Rightarrow{ln}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{k}\right)\leqslant{nln}\left(\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}}{{n}}\right)={nln}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right) \\ $$$$ \\ $$
Commented by 1549442205PVT last updated on 20/Oct/20
Thank you sir
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by 1549442205PVT last updated on 20/Oct/20
We prove by the induction method 1−For n=2 we obtain the true inequality 2<9/4.Suppose that k!<(((k+1)/2))^k .Then by the induction hypothesis (k+1)!= k!(k+1)<(((k+1)/2))^k (k+1).If now we prove that (((k+1)/2))^k (k+1)<(((k+2)/2))^(k+1) (1) the theorem is proved because then (k+1)!<(((k+1)/2))^k (k+1)<(((k+2)/2))^(k+1) that is our inequality holds true for n=k+1 Inequality(1) can clearly be rewritten as (((k+2)^(k+1) )/((k+1)^(k+1) ))>(2^(k+1) /2^k )or (1+(1/(k+1)))^(k+1) >2 But the binomial theorem yields (1+(1/(k+1)))^(k+1) =1+(k+1)(1/(k+1))+...>2 so the inequality (1)holds and thus the oriinal inequality is proved 2)The inequality is obviously true for n=1.Assuming that ∣sinkx∣≤k∣sinx∣ ,we prove that ∣sin(k+1)x∣≤(k+1)∣sinx∣ Indeed,using the inequality∣coskx∣≤1 we have ∣sin(k+1)x∣=∣sinkx.cosx+sinxcoskx∣ ≤∣sinkx∣.∣cosx∣+∣sinx∣∣coskx∣≤ ∣sinkx∣+∣sinx∣≤k∣sinx∣+∣sinx∣ =(k+1)∣sinx∣ which shows the required is true.The proof completed.
$$\mathrm{We}\:\mathrm{prove}\:\mathrm{by}\:\mathrm{the}\:\mathrm{induction}\:\mathrm{method} \\ $$$$\mathrm{1}−\mathrm{For}\:\mathrm{n}=\mathrm{2}\:\mathrm{we}\:\mathrm{obtain}\:\mathrm{the}\:\mathrm{true}\:\mathrm{inequality} \\ $$$$\mathrm{2}<\mathrm{9}/\mathrm{4}.\mathrm{Suppose}\:\mathrm{that}\:\mathrm{k}!<\left(\frac{\mathrm{k}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{k}} .\mathrm{Then} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{induction}\:\mathrm{hypothesis}\:\left(\mathrm{k}+\mathrm{1}\right)!= \\ $$$$\mathrm{k}!\left(\mathrm{k}+\mathrm{1}\right)<\left(\frac{\mathrm{k}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{k}} \left(\mathrm{k}+\mathrm{1}\right).\mathrm{If}\:\mathrm{now}\:\mathrm{we}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\left(\frac{\mathrm{k}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{k}} \left(\mathrm{k}+\mathrm{1}\right)<\left(\frac{\mathrm{k}+\mathrm{2}}{\mathrm{2}}\right)^{\mathrm{k}+\mathrm{1}} \left(\mathrm{1}\right) \\ $$$$\mathrm{the}\:\mathrm{theorem}\:\mathrm{is}\:\mathrm{proved}\:\mathrm{because}\:\mathrm{then} \\ $$$$\left(\mathrm{k}+\mathrm{1}\right)!<\left(\frac{\mathrm{k}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{k}} \left(\mathrm{k}+\mathrm{1}\right)<\left(\frac{\mathrm{k}+\mathrm{2}}{\mathrm{2}}\right)^{\mathrm{k}+\mathrm{1}} \\ $$$$\mathrm{that}\:\mathrm{is}\:\mathrm{our}\:\mathrm{inequality}\:\mathrm{holds}\:\mathrm{true}\:\mathrm{for}\:\mathrm{n}=\mathrm{k}+\mathrm{1} \\ $$$$\mathrm{Inequality}\left(\mathrm{1}\right)\:\mathrm{can}\:\mathrm{clearly}\:\mathrm{be}\:\mathrm{rewritten} \\ $$$$\mathrm{as}\:\frac{\left(\mathrm{k}+\mathrm{2}\right)^{\mathrm{k}+\mathrm{1}} }{\left(\mathrm{k}+\mathrm{1}\right)^{\mathrm{k}+\mathrm{1}} }>\frac{\mathrm{2}^{\mathrm{k}+\mathrm{1}} }{\mathrm{2}^{\mathrm{k}} }\mathrm{or}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\right)^{\mathrm{k}+\mathrm{1}} >\mathrm{2} \\ $$$$\mathrm{But}\:\mathrm{the}\:\mathrm{binomial}\:\mathrm{theorem}\:\mathrm{yields} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}\right)^{\mathrm{k}+\mathrm{1}} =\mathrm{1}+\left(\mathrm{k}+\mathrm{1}\right)\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}+…>\mathrm{2} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{inequality}\:\left(\mathrm{1}\right)\mathrm{holds}\:\mathrm{and}\:\mathrm{thus}\:\mathrm{the}\: \\ $$$$\mathrm{oriinal}\:\mathrm{inequality}\:\mathrm{is}\:\mathrm{proved} \\ $$$$\left.\mathrm{2}\right)\mathrm{The}\:\mathrm{inequality}\:\mathrm{is}\:\mathrm{obviously}\:\mathrm{true}\:\mathrm{for} \\ $$$$\mathrm{n}=\mathrm{1}.\mathrm{Assuming}\:\mathrm{that}\:\mid\mathrm{sinkx}\mid\leqslant\mathrm{k}\mid\mathrm{sinx}\mid \\ $$$$,\mathrm{we}\:\mathrm{prove}\:\mathrm{that}\:\mid\mathrm{sin}\left(\mathrm{k}+\mathrm{1}\right)\mathrm{x}\mid\leqslant\left(\mathrm{k}+\mathrm{1}\right)\mid\mathrm{sinx}\mid \\ $$$$\mathrm{Indeed},\mathrm{using}\:\mathrm{the}\:\mathrm{inequality}\mid\mathrm{coskx}\mid\leqslant\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mid\mathrm{sin}\left(\mathrm{k}+\mathrm{1}\right)\mathrm{x}\mid=\mid\mathrm{sinkx}.\mathrm{cosx}+\mathrm{sinxcoskx}\mid \\ $$$$\leqslant\mid\mathrm{sinkx}\mid.\mid\mathrm{cosx}\mid+\mid\mathrm{sinx}\mid\mid\mathrm{coskx}\mid\leqslant \\ $$$$\mid\mathrm{sinkx}\mid+\mid\mathrm{sinx}\mid\leqslant\mathrm{k}\mid\mathrm{sinx}\mid+\mid\mathrm{sinx}\mid \\ $$$$=\left(\mathrm{k}+\mathrm{1}\right)\mid\mathrm{sinx}\mid\:\mathrm{which}\:\mathrm{shows}\:\mathrm{the}\:\mathrm{required} \\ $$$$\mathrm{is}\:\mathrm{true}.\mathrm{The}\:\mathrm{proof}\:\mathrm{completed}. \\ $$

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