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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].
POSTED BY: Marvin Ray Burns
4 Replies

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 BY: Marvin Ray Burns

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 BY: Udo Krause

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 BY: Udo Krause

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.

POSTED BY: Udo Krause
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