16
$\begingroup$

I was given this problem on series by a friend.

If

$$y=\sqrt{4 + \sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\ldots}}}}$$

then how to solve such problem.

I don’t want the full answer, rather, insights, mathematical facts, theorems, and relationships that would help me solve it on my own.

My efforts: I thought that the whole thing inside the square bracket must be a perfect square so we have [$4~+$ something] should be a positive perfect square but that would be like finding a trivial solution by trial and error method so I don't know how to solve it.

I also tried by squaring and checking like this $$y^2 - 4=\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\ldots}}}$$

so we get two factors $y-2$ and $y+2$, but still it was like same trial and error method of finding factors . So can any one help. Thanks in advance.

Edit: The only reasonable interpretation is the recurrence $y_n=\sqrt[n]{4+y_{n+1}}$ (Thanks @par)

| cite | improve this question | |
$\endgroup$
  • 1
    $\begingroup$ Can you write a recurrence relation for $y$? $\endgroup$ – dbanet Jun 29 '16 at 17:49
  • 2
    $\begingroup$ The only reasonable interpretation is the recurrence $y_{n}=\sqrt[{n}]{4+y_{n+1}}$ so that $y_{n+1}=y_{n}^{n}-4$. The problem now is that we have no initial condition. $\endgroup$ – parsiad Jun 29 '16 at 17:50
  • 1
    $\begingroup$ Finding a closed form for expressions like this is notoriously difficult if possible at all. Take a look at some of the other similar problems (and links within) as you will see that not many of them have a known closed form solution: Q1, Q2, Q3 $\endgroup$ – Winther Jun 29 '16 at 18:04
  • $\begingroup$ Some more related question: Q4, Q5, Q6. Q6 is basically your question but with the number $2$ instead of $4$. $\endgroup$ – Winther Jun 29 '16 at 18:10
  • 1
    $\begingroup$ Searching in Approach0 returns a few posts about the same nested radical: math.stackexchange.com/questions/1782032/… and math.stackexchange.com/questions/875203/… $\endgroup$ – Martin Sleziak Feb 3 '17 at 13:38
5
$\begingroup$

Too long for a comment, so

$\begin{array}\\ y &=\sqrt{4 + \sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+...............}}}}\\ &=2\sqrt{1 + \frac14\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+...............}}}}\\ &=2\sqrt{1 + \frac14\sqrt[3]{4}\sqrt[3]{1+\frac1{\sqrt[3]{4}}\sqrt[4]{4+\sqrt[5]{4+...............}}}}\\ &=2\sqrt{1 + \frac1{4^{2/3}}\sqrt[3]{1+\frac1{\sqrt[3]{4}}\sqrt[4]{4}\sqrt[4]{1+\frac1{\sqrt[4]{4}}\sqrt[5]{4+...............}}}}\\ &=2\sqrt{1 + \frac1{4^{2/3}}\sqrt[3]{1+\frac1{4^{1/12}}\sqrt[4]{1+\frac1{\sqrt[4]{4}}\sqrt[5]{4+...............}}}}\\ &=2\sqrt{1 + \frac1{4^{2/3}}\sqrt[3]{1+\frac1{4^{1/12}}\sqrt[4]{1+\frac1{\sqrt[4]{4}}\sqrt[5]{4}\sqrt[5]{1+...............}}}}\\ &=2\sqrt{1 + \frac1{4^{2/3}}\sqrt[3]{1+\frac1{4^{1/12}}\sqrt[4]{1+\frac1{4^{1/20}}\sqrt[5]{1+...............}}}}\\ \end{array} $

It looks like there is a pattern of $\dfrac1{4^{1/n-1/(n+1)}} =\dfrac1{4^{1/(n(n+1))}} $ which might make it easier to get a more solvable recurrence.

And, of course, $4$ can be replaced by any value, probably preferably a square.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ I don't understand what to get out of your last line with the recurrence, as it is merely a true statement. $\endgroup$ – Simply Beautiful Art Jul 1 '16 at 13:32
4
+50
$\begingroup$

This is only a hint, because the following argumentation is a rough calculation.

$y^n_n=4+y_{n+1}$

If we assume a value $>1$ for $y^n_n$ we can set $y_n\approx 1+\frac{a}{n}$ for large $n$ so that we get

$(1+\frac{a}{n})^n\approx 4+(1+\frac{a}{n+1})$ which means for $n\to\infty $ the equation $e^a=5$ or simply

$a=\ln 5\approx 1.6094379...$ .

It follows (for small $n$, here $n:=2$) $\enspace y:=y_2\approx \sqrt{4+(1+\frac{\ln 5}{2})}\approx 2,409298...$ .

I think this value is a good upper bound for $y$ .

A lower bound should be $\sqrt{4+(1+\frac{\ln 5}{3})}\approx 2,35297...$ .

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

I got with high precision $$y \approx 2.401615526026297355671587971656109034017...$$ when started with any value $x_0$ between $10$ and $1e20$ and the root-index of 1000.
The simple continued fraction of the approximation having 793 terms was (reading from left to right and then top-down):

    2   2    2   24   2     1    3    1   1     5
    1   1    7    1   2     1    1    4   3     1
    1   1   15  274   2    49    1    1   1    16
    2   3    1    2   1    48   10    4   1    14
    7   1    2   19  27     3    4    1   8    11
   20   1    5    3   4     1   52    1   2     1
    2   8    3    2   2     1   60    1   1     4
    3   8    2    1  69     1    5    1   3    10
   10   2    1   21  12     1    4    1  15     2
    3   1    1    9   3     8    1    1   1     2
    6   1  100    4  11     3    1    3   6     1
    1   4    3    2   2    16    2    5   2     6
    2   1    2   11   6     2   40    1   1     8
    1   5    1    1   3     2    7    1   1    15
    3   1    2    2   2     1    1    1   4     7
    1   1    6    3   2    12   13  890   1     3
    2   1    3    7   2    14  563    3  36     2
    3   9    7    1   1     1    3    1   2    13
    9   2    1    3   2     2    7   12  11     2
    1   4    1    1   1     7    5    1   4     5
    1   5    1   60   1     2    3    1  14     1
    1   6    1    1   1     4    2   59   1     2
    1  26   41    1   6     6    1    1   1     1
    2   1    2    2   1     4    1   12   1     2
    2   2   20    5   4     8    6    1   5     1
    2   1    4    2   3     2    2    2   3     3
    1   4    2    1   1    15   27    1   2     1
    1   1    3    1   4     1    4    1   1     6
   18   9    1   23   3     1    1    2  11     1
    1  24    1    5   1     1    2   13   1    22
    6   1    8    1   3     1    3    1   4     2
    1  33    4    1   2     1    2    1   2     8
  170   2    2    6   6    17    1    2   1     2
    2   6    1    1   2     1    2    1   4    13
    1   2    1    2   3     1    1    3   2     1
    2   6    5    1   2    15    2    2  13     1
   19   1    2    3   1     1    1    5   3     1
   28   1    3    1   2     1    1    2   5     1
    3   2    1    2   2     1   22    2   2     4
    1  21    2    1   9     1    7    1   6     7
    5   1   18    4   2     2    1    2   1     1
    1   1    3    1   1     1    6   27  18     1
    6   1    9    1   4     3    1   28   2     8
    6   1   35    1   1     2   56    1   1     1
    2   1    1    7   1     1    4    2   1     2
    1   1    1    3   1     1   37    1   8  3154
    1   1    3    1   3     1   11    1   9     1
    3   2    7    2   1    56    2   11   8     1
    2   1    1    2   1     4   16    1   2     3
    1   1    2    1   1    15    1    1   1     1
  254   1    1    7   1     4    1    2   4     1
    2   1    1    1   1     1    2    2   1     2
    1  11    5    2   1    89    1   10   1     1
    3   3    1   11   1     1    1    1  13     1
    5  37    1    6   1     7    1   11   1     2
    6   2    2    4   2     1    2    1   4     3
    3   4    7    1   7     1    3   14  17     2
    1   4    1    4   3     1   80    6   1     4
    1   2    1    9   5     1    1    1   7     3
    2   3    1    1  87     1    1   12   1   177
    1   2    1    1   2     6    1   47   7     7
    2   4    1    6   1     2    2    1   4     2
    1   4    1    3   4     4   69    1  10     5
    1   4    1    3   4     1   14    3   1     1
    2  11    1    1   3     1    2    1   2     1
   10   2    1    1  11    25    2    6   3    13
    1  12   12    4   1     4    4    1   3     1
    1   6    1   11   3     5    1    1   1     4
    2   3    1   11   4     2    1    1   1     1
    4   4    2    1   4     4  190    2  20    14
    1   5    1   14   1     1    1    2   2     1
    1   5    2    1   1  2289    2   14   6    37
    1   5    1   21   2    16    1    9   2     2
   11  13    6    1   2     4    3    6   3    55
    1   1   10    1   2    30    4   14   6     1
    3   2    1    1   2     1    1    2   1     1
   13   1    1    5   1     2    1    1   2    17
    3   2    2    1   1     5    2    5   2     3
    1   1    1    1   7     2    1    1   1     2
   ...

A rough procedure to test is the following in Pari/GP. Assume $\small x_1=1$ as lower bound, $\small x_2=1e200 $ as upperbound, and some initial root-index, say 30. Then compute the nested radical down to index 2:

index=30;x1=1;x2=1e200
forstep(k=index,2,-1, x1=(4+x1)^(1/k); x2=(4+x2)^(1/k); print([x1,x2,x2-x1]) );

Output: 

   x1(incr.)       x2(decr.)        difference
 ------------------------------------------------
  1.05511306354  4641588.83361           4641587.77850
  1.05746641188  1.69779429681          0.640327884932
  1.05959637667  1.06411735764        0.00452098097191
  1.06188715066  1.06192227795      0.0000351272888663
  1.06436097020  1.06436125429    0.000000284083700916
  1.06704070281  1.06704070521  0.00000000239421214644
  1.06995316160  1.06995316162       2.10650144506E-11
  1.07313002353  1.07313002353       1.93857373231E-13
  1.07660898434  1.07660898434       1.87000018397E-15
  1.08043525836  1.08043525836       1.89516657781E-17
  1.08466355021  1.08466355021       2.02307283136E-19
  1.08936068075  1.08936068075       2.28122116120E-21
  1.09460913536  1.09460913536       2.72577952262E-23
  1.10051193561  1.10051193561       3.46358372476E-25
  1.10719944779  1.10719944779       4.69913367993E-27
  1.11483908939  1.11483908939       6.83842220013E-29
  1.12364948065  1.12364948065       1.07306679076E-30
  1.13392160888  1.13392160888       1.82678325796E-32
  1.14605141776  1.14605141776       3.39829163458E-34
  1.16059171388  1.16059171388       6.96744109268E-36
  1.17833818093  1.17833818093       1.59090319833E-37
  1.20047878604  1.20047878604       4.09793745817E-39
  1.22886851186  1.22886851186       1.21042372084E-40
  1.26657139203  1.26657139203       4.18852719443E-42
  1.31902996773  1.31902996773       1.74838380889E-43
  1.39690376355  1.39690376355       9.18334334489E-45
  1.52417968370  1.52417968370       6.48384053520E-46
  1.76775713485  1.76775713485       6.91617339503E-47
  2.40161552603  2.40161552603       1.43990020886E-47
| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I tried finding a minimal polynomial for such an approximation, but the best I found is the following which should be good up to 122 digits but won't match after that: $$8258028589600 y^{8} - 10398489573954 y^{7} - 16164983819372 y^{6} - 8219788311584 y^{5} - 12767863340898 y^{4} - 11126864974160 y^{3} + 3533772609633 y^{2} - 17611448551760 y + 11820477102949$$ $\endgroup$ – MvG Feb 9 '17 at 17:27
1
$\begingroup$

$$\lim_{n\to\infty}\sqrt[n+1]{const}=1,$$ so $$y=\sqrt{4+\sqrt[3]{4+\sqrt[4]{\dots+\sqrt[n]{5+o(1)}}}} = \lim_{n\to\infty} y_n,$$ where $$y_n=\sqrt{4+\sqrt[3]{4+\sqrt[4]{\dots+\sqrt[n]{5}}}}.$$ Using calculator, one can get: $$y_2 = \sqrt 5 \approx 2,2360679774997896964091736687313,$$ $$y_3=\sqrt{4+\sqrt[3]5} \approx 2,3895555960631460037703850634224,$$ $$y_4=\sqrt{4+\sqrt[3]{4+\sqrt[4]5}}\approx 2,4009740604119511706503467195743,$$ $$y_5=\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{5}}}}\approx 2,4015885636119483509721951488115,$$ $$y_6=\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\sqrt[6]{5}}}}}\approx 2,4016145888373557072804406968762\dots,$$ indicating a good practical convergence of the sequence $\{y_n\}$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

The question is pretty close to this one, just replace $\,2\,$ by $\,4\,$ in:

There is a little (Delphi Pascal) computer program in one of the answers. At the end of that piece of code we find:

begin
  anatoly(2,13); { at double precision }
end.
Replace this by:

begin
  anatoly(4,13); { at double precision }
end.
To get:

 4.00000000000000E+0000  2.40161552602630E+0000 +/- 1.80950091432944E-0018
Which is - no big surprise - in concordance with the numerical value as given by Gottfried Helms in his answer .

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.