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Question Number 6247 by Yozzii last updated on 20/Jun/16
Show that, ∀x∈R, the sequence {f(n)}_0 ^∞   defined by f(0)=cosx, f(1)=sin(cosx) and  f(n)= { ((sin(cos(f(n−2)))  if n≥3 is odd)),((cos(sin(f(n−2)))  if n≥2 is even,)) :}  converges to a limit l∈(0.6,0.7).  If you can, determine the exact value  of l.
$${Show}\:{that},\:\forall{x}\in\mathbb{R},\:{the}\:{sequence}\:\left\{{f}\left({n}\right)\right\}_{\mathrm{0}} ^{\infty} \\ $$$${defined}\:{by}\:{f}\left(\mathrm{0}\right)={cosx},\:{f}\left(\mathrm{1}\right)={sin}\left({cosx}\right)\:{and} \\ $$$${f}\left({n}\right)=\begin{cases}{{sin}\left({cos}\left({f}\left({n}−\mathrm{2}\right)\right)\right)\:\:{if}\:{n}\geqslant\mathrm{3}\:{is}\:{odd}}\\{{cos}\left({sin}\left({f}\left({n}−\mathrm{2}\right)\right)\right)\:\:{if}\:{n}\geqslant\mathrm{2}\:{is}\:{even},}\end{cases} \\ $$$${converges}\:{to}\:{a}\:{limit}\:{l}\in\left(\mathrm{0}.\mathrm{6},\mathrm{0}.\mathrm{7}\right). \\ $$$${If}\:{you}\:{can},\:{determine}\:{the}\:{exact}\:{value} \\ $$$${of}\:{l}. \\ $$
Commented by Yozzii last updated on 20/Jun/16
f(0)=cosx  f(1)=sin(cosx)  f(2)=cos(sin(cosx))  f(3)=sin(cos(sin(cosx)))  f(4)=cos(sin(cos(sin(cosx))))  f(5)=sin(cos(sin(cos(sin(cosx)))))  …  lim_(n→∞) f(n)=?  Graphically, the curves f_n (x) flatten  to the line y≈0.695 for sufficiently  large n. I am yet to figure out how  to show this behaviour formally.
$${f}\left(\mathrm{0}\right)={cosx} \\ $$$${f}\left(\mathrm{1}\right)={sin}\left({cosx}\right) \\ $$$${f}\left(\mathrm{2}\right)={cos}\left({sin}\left({cosx}\right)\right) \\ $$$${f}\left(\mathrm{3}\right)={sin}\left({cos}\left({sin}\left({cosx}\right)\right)\right) \\ $$$${f}\left(\mathrm{4}\right)={cos}\left({sin}\left({cos}\left({sin}\left({cosx}\right)\right)\right)\right) \\ $$$${f}\left(\mathrm{5}\right)={sin}\left({cos}\left({sin}\left({cos}\left({sin}\left({cosx}\right)\right)\right)\right)\right) \\ $$$$\ldots \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}\left({n}\right)=? \\ $$$${Graphically},\:{the}\:{curves}\:{f}_{{n}} \left({x}\right)\:{flatten} \\ $$$${to}\:{the}\:{line}\:{y}\approx\mathrm{0}.\mathrm{695}\:{for}\:{sufficiently} \\ $$$${large}\:{n}.\:{I}\:{am}\:{yet}\:{to}\:{figure}\:{out}\:{how} \\ $$$${to}\:{show}\:{this}\:{behaviour}\:{formally}. \\ $$
Commented by prakash jain last updated on 20/Jun/16
The value will become constant and take  value u if for some u  sin (cos u)=u  cos (sin u)=u  Then the value will be a constant.  However  sin (cos x)=x⇒x=.695  cos (sin x)=x⇒x=.768  Also for cos (.695)=.768  So the graph should show 2 lines  .695 and .768
$$\mathrm{The}\:\mathrm{value}\:\mathrm{will}\:\mathrm{become}\:\mathrm{constant}\:\mathrm{and}\:\mathrm{take} \\ $$$$\mathrm{value}\:{u}\:\mathrm{if}\:\mathrm{for}\:\mathrm{some}\:{u} \\ $$$$\mathrm{sin}\:\left(\mathrm{cos}\:{u}\right)={u} \\ $$$$\mathrm{cos}\:\left(\mathrm{sin}\:{u}\right)={u} \\ $$$$\mathrm{Then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{will}\:\mathrm{be}\:\mathrm{a}\:\mathrm{constant}. \\ $$$$\mathrm{However} \\ $$$$\mathrm{sin}\:\left(\mathrm{cos}\:{x}\right)={x}\Rightarrow{x}=.\mathrm{695} \\ $$$$\mathrm{cos}\:\left(\mathrm{sin}\:{x}\right)={x}\Rightarrow{x}=.\mathrm{768} \\ $$$$\mathrm{Also}\:\mathrm{for}\:\mathrm{cos}\:\left(.\mathrm{695}\right)=.\mathrm{768} \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{should}\:\mathrm{show}\:\mathrm{2}\:\mathrm{lines} \\ $$$$.\mathrm{695}\:\mathrm{and}\:.\mathrm{768} \\ $$
Commented by Yozzii last updated on 20/Jun/16
The result of having two values of the limit   u from the two recurrence formulae   indicates that no limit exists?
$${The}\:{result}\:{of}\:{having}\:{two}\:{values}\:{of}\:{the}\:{limit}\: \\ $$$${u}\:{from}\:{the}\:{two}\:{recurrence}\:{formulae}\: \\ $$$${indicates}\:{that}\:{no}\:{limit}\:{exists}? \\ $$
Commented by Yozzii last updated on 21/Jun/16
u=sin(cosu)  u=cos(sinu)  ⇒2u=sin(cosu)+cos(sinu)   u turns out to be approximately 0.731  according to this equation.  −−−−−−−−−−−−−−−−−−−−−−−−  sin^(−1) u=cosu  cos^(−1) u=sinu  ⇒(sin^(−1) u)^2 +(cos^(−1) u)^2 =1   (∗)  Wolfram gives complex solutions of u  for (∗).  sin^(−1) u=0.5π−cos^(−1) u  j=cos^(−1) u⇒ (0.5π−j)^2 +j^2 =1  2j^2 −πj+0.25π^2 −1=0  j=((π±(√(π^2 −4(2)((π^2 /4)−1))))/4)  j=((π±(√(−π^2 +8)))/4)=((π±i(√(π^2 −8)))/4)  ∴ u=cos(((π±i(√(π^2 −8)))/4))
$${u}={sin}\left({cosu}\right) \\ $$$${u}={cos}\left({sinu}\right) \\ $$$$\Rightarrow\mathrm{2}{u}={sin}\left({cosu}\right)+{cos}\left({sinu}\right)\: \\ $$$${u}\:{turns}\:{out}\:{to}\:{be}\:{approximately}\:\mathrm{0}.\mathrm{731} \\ $$$${according}\:{to}\:{this}\:{equation}. \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${sin}^{−\mathrm{1}} {u}={cosu} \\ $$$${cos}^{−\mathrm{1}} {u}={sinu} \\ $$$$\Rightarrow\left({sin}^{−\mathrm{1}} {u}\right)^{\mathrm{2}} +\left({cos}^{−\mathrm{1}} {u}\right)^{\mathrm{2}} =\mathrm{1}\:\:\:\left(\ast\right) \\ $$$${Wolfram}\:{gives}\:{complex}\:{solutions}\:{of}\:{u} \\ $$$${for}\:\left(\ast\right). \\ $$$${sin}^{−\mathrm{1}} {u}=\mathrm{0}.\mathrm{5}\pi−{cos}^{−\mathrm{1}} {u} \\ $$$${j}={cos}^{−\mathrm{1}} {u}\Rightarrow\:\left(\mathrm{0}.\mathrm{5}\pi−{j}\right)^{\mathrm{2}} +{j}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{2}{j}^{\mathrm{2}} −\pi{j}+\mathrm{0}.\mathrm{25}\pi^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${j}=\frac{\pi\pm\sqrt{\pi^{\mathrm{2}} −\mathrm{4}\left(\mathrm{2}\right)\left(\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\mathrm{1}\right)}}{\mathrm{4}} \\ $$$${j}=\frac{\pi\pm\sqrt{−\pi^{\mathrm{2}} +\mathrm{8}}}{\mathrm{4}}=\frac{\pi\pm{i}\sqrt{\pi^{\mathrm{2}} −\mathrm{8}}}{\mathrm{4}} \\ $$$$\therefore\:{u}={cos}\left(\frac{\pi\pm{i}\sqrt{\pi^{\mathrm{2}} −\mathrm{8}}}{\mathrm{4}}\right) \\ $$$$ \\ $$
Commented by prakash jain last updated on 21/Jun/16
u=sin (cos u)  u=cos (sin u)  Both equation are not simulatenoeus.  u=.695 will statify one equation.
$${u}=\mathrm{sin}\:\left(\mathrm{cos}\:{u}\right) \\ $$$${u}=\mathrm{cos}\:\left(\mathrm{sin}\:{u}\right) \\ $$$$\mathrm{Both}\:\mathrm{equation}\:\mathrm{are}\:\mathrm{not}\:\mathrm{simulatenoeus}. \\ $$$${u}=.\mathrm{695}\:{will}\:{statify}\:{one}\:{equation}. \\ $$
Commented by Yozzii last updated on 21/Jun/16
No limit exists then?
$${No}\:{limit}\:{exists}\:{then}? \\ $$

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