Calculating infinite series of roots
08-26-2019, 04:58 PM
Post: #1
 KeithB Member Posts: 198 Joined: Jan 2017
Calculating infinite series of roots
https://skullsinthestars.com/2019/08/18/...-identity/

Is there a way to calculate this easily on the Prime?

i.e. showing that (2^.5)^(2^.5)^(2^.5)... = 2?
08-29-2019, 04:56 PM
Post: #2
 Helge Gabert Senior Member Posts: 460 Joined: Dec 2013
RE: Calculating infinite series of roots
There is no symbol for repeated exponentiation on the calc, as opposed to the repeated summation and product operator, so I'm not sure if you can show that symbolically with the limit function.

But, of course, you can always write a program and show that numerically, you get very close to 2. For example, with a REPEAT ... UNTIL loop, using approx(sqrt(2)), and setting your tolerance at 1E-12, you get to 2 after 71 iterations, after which the result doesn't change any more.
08-30-2019, 11:57 PM (This post was last modified: 08-31-2019 12:07 AM by ijabbott.)
Post: #3
 ijabbott Senior Member Posts: 703 Joined: Jul 2015
RE: Calculating infinite series of roots
A "power tower" of $$n$$ $$x$$'s $$\underbrace{x^{x^{x^{\dots^x}}}}_n$$ is also known as the $$n$$th tetration of $$x$$. The operation is called "tetration" because it is the fourth in a series of operations: addition, multiplication, exponentiation, tetration. There are other terms for the same thing. There are also several notations, but one such notation is $$^{n}x$$.

Now we come on to the "infinite tetration" of $$x$$, or $$^{\infty}x$$, which is $$\lim\limits_{n\to\infty}{^{n}x}$$. For certain real values of $$x$$ such as $$x=\sqrt{2}$$, this infinite tetration converges (to $$2$$ in this case). In fact it only converges for real $$x$$ in the interval $$\big[e^{-e}, e^\frac{1}{e}\big]$$ (approx. $$[0.066, 1.445]$$).

The limit of convergence can evaluated using the Lambert W function:

$^{\infty}x = \lim_{n\to\infty}{^{n}x} = \frac{W(-\ln(x))}{-\ln(x)} \big|_{e^{-e} \le x \le e^\frac{1}{e}}$

where the Lambert W function is defined by $$W(z{e^z})=z$$, or by $$z_0 = W(z_0)e^{W(z_0)}$$.

Unfortunately, the Lambert W function is not yet built-in on the HP Prime, although it is available in Xcas/Giac, so maybe later....

— Ian Abbott
08-31-2019, 12:59 AM
Post: #4
 Helge Gabert Senior Member Posts: 460 Joined: Dec 2013
RE: Calculating infinite series of roots
(08-30-2019 11:57 PM)ijabbott Wrote:  In fact it only converges for real $$x$$ in the interval $$\big[e^{-e}, e^\frac{1}{e}\big]$$ (approx. $$[0.066, 1.445]$$).

Yes, very true, and that was shown by Euler (back in the days . . . ).

Support for the Lambert W function would be nice, if there is ever an update again!
08-31-2019, 04:07 PM
Post: #5
 Stevetuc Member Posts: 133 Joined: Jan 2014
RE: Calculating infinite series of roots
(08-31-2019 12:59 AM)Helge Gabert Wrote:
(08-30-2019 11:57 PM)ijabbott Wrote:  In fact it only converges for real $$x$$ in the interval $$\big[e^{-e}, e^\frac{1}{e}\big]$$ (approx. $$[0.066, 1.445]$$).

Yes, very true, and that was shown by Euler (back in the days . . . ).

Support for the Lambert W function would be nice, if there is ever an update again!

This cas program uses fsolve to calc Lambert fn

Code:
 #cas lmb(z):= fsolve(equal(w*e^w-z,0),w) #end

Graph from wikipedia

If z <0 lmb(z) returns both principal and negative branch solutions :

lmb(-0.5/e) returns [−2.67834699002,−0.231960952987]

lmb(-1/e) returns −0.999999842597 (negative and principal branch converge at -1 when z=-1/e)

lmb(1) returns 0.56714329041

lmb(2) returns 0.852605502014
08-31-2019, 07:13 PM (This post was last modified: 08-31-2019 07:14 PM by ijabbott.)
Post: #6
 ijabbott Senior Member Posts: 703 Joined: Jul 2015
RE: Calculating infinite series of roots
(08-30-2019 11:57 PM)ijabbott Wrote:  The limit of convergence can evaluated using the Lambert W function:

$^{\infty}x = \lim_{n\to\infty}{^{n}x} = \frac{W(-\ln(x))}{-\ln(x)} \big|_{e^{-e} \le x \le e^\frac{1}{e}}$

where the Lambert W function is defined by $$W(z{e^z})=z$$, or by $$z_0 = W(z_0)e^{W(z_0)}$$.

I've just noticed that this evaluates to $$\frac{0}{0}$$ for $$x=1$$. Hmm... more work needed? Also when $$1 \lt x \le e^\frac{1}{e}$$, then letting $$z=-\ln(x)$$, $$-\frac{1}{e} \le z \lt 0$$, and there are two branches of the $$W$$ function in this interval. I guess it uses the upper branch?

— Ian Abbott
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