# Prove this limit

Posted 9 years ago
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 Can someone prove this limit for me? It would help in future calculations of the MRB constant. Limit[(3 + 2 Sqrt[2])^x - 2 Cos[x ArcCos[3]], x -> Infinity]==0 Here is my latest formula for calculating the MRB constant: (*Fastest (at MRB's end) as of 24 dEC 2014.*) prec = 20000;(*Number of required decimals.*)ClearSystemCache[]; T0 = SessionTime[]; expM[pre_] := Module[{a, d, s, k, bb, c, n, end, iprec, xvals, x, pc, cores = 12, tsize = 2^7, chunksize, start = 1, ll, ctab, pr = Floor[1.005 pre]}, chunksize = cores*tsize; n = Floor[1.32 pr]; end = Ceiling[n/chunksize]; Print["Iterations required: ", n]; Print["end ", end]; Print[end*chunksize]; d = Cos[n ArcCos[3]]; {b, c, s} = {SetPrecision[-1, 1.1*n], -d, 0}; iprec = Ceiling[pr/27]; Do[xvals = Flatten[ParallelTable[Table[ll = start + j*tsize + l; h = Log[ll]/ll; x = N[Exp[h], iprec]; pc = iprec; While[pc < pr, pc = Min[3 pc, pr]; x = SetPrecision[x, pc]; y = x^ll - ll; x = x (1 - 2 y/((ll + 1) y + 2 ll ll));];(*N[Exp[Log[ll]/ll], pr]*)x, {l, 0, tsize - 1}], {j, 0, cores - 1}, Method -> "EvaluationsPerKernel" -> 4]]; ctab = ParallelTable[Table[c = b - c; ll = start + l - 2; b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1)); c, {l, chunksize}], Method -> "EvaluationsPerKernel" -> 2]; s += ctab.(xvals - 1); start += chunksize; Print["done iter ", k*chunksize, " ", SessionTime[] - T0];, {k, 0, end - 1}]; N[-s/d, pr]]; t2 = Timing[MRBtest2 = expM[prec];]; MRBtest2 I found the Limit[Cos[n ArcCos[3]]/Cos[(n - 1) ArcCos[3]], n -> Infinity]=3 + 2 Sqrt[2]. 
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Posted 9 years ago
 I guess this is a simpler way of proving Limit[(3 + 2 Sqrt[2])^x - 2 Cos[x ArcCos[3]], x -> Infinity]==0: In[10]:= FullSimplify[(3 + 2 Sqrt[2])^x - 2 Cos[x ArcCos[3]]] Out[10]= -(3 + 2 Sqrt[2])^-x which of course goes to 0 as x-> Infinity. I would still like to see a geometric proof, though.
Posted 9 years ago
 I would still like to see a geometric proof, though. It turns out just by definition: It remained to show $\exp(-i \arccos(3))=3+2\sqrt{2}$. With the definition of InverseCosine, eq. (2), $\arccos(z)=\frac{\pi}{2}+i\ln(iz+\sqrt{1-z^2})$ one has $\exp(-i(\frac{\pi}{2}+i \ln(i 3 + \sqrt{-8})))=\exp(-i(\frac{\pi}{2}+i\ln(i (3+2\sqrt{2}))))=\exp(-i(\frac{\pi}{2}+i\frac{i \pi}{2}+i\ln(3+2\sqrt{2})))=3+2\sqrt{2}$ $\Box$
Posted 9 years ago
 Bad luck, I typed too much In[33]:= ComplexExpand[Exp[-I ArcCos[3]]] Out[33]= 3 + 2 Sqrt[2] life can be so easy ... and if one does In[35]:= ComplexExpand[(3 + 2 Sqrt[2])^x - 2 Cos[x ArcCos[3]]] Out[35]= (3 + 2 Sqrt[2])^x - 2 Cosh[x Log[3 + 2 Sqrt[2]]] one sees the result directly because of the definition of Cosh and with In[38]:= TrigToExp[ComplexExpand[(3 + 2 Sqrt[2])^x - 2 Cos[x ArcCos[3]]]] Out[38]= -(3 + 2 Sqrt[2])^-x the result of FullSimplify reappears.
Posted 9 years ago
 In principle that goes that way: In[2]:= ArcCos[3] // N Out[2]= 0. + 1.76275 I  $\cos(z)=\frac{1}{2}(\exp(iz)+\exp(-iz))$ by definition with $z=x \arccos(3)$ it follows $2 \cos(x \arccos(3)) \sim \exp(1.76275 x)$ for big $x$. Now one needs only $\exp(1.76275) \equiv 3 + \sqrt{2}$ which fits seemingly In[4]:= E^1.76275 // N Out[4]= 5.82844 In[5]:= 3 + 2 Sqrt[2] // N Out[5]= 5.82843 or written more careful In[8]:= N[E^Im[ArcCos[3]], 30] - N[3 + 2 Sqrt[2], 30] Out[8]= 0.*10^-29 this might have even a geometrical proof, because the $x$ has gone out of it.