Vtyjkyuk

There's an old-school pocket calculator trick to calculate $x^n$ on a pocket calculator, where both, $x$ and $n$ are real numbers. So, things like $,0.751^3.2131$ can be calculated, which is awesome.

This provides endless possibilities, including calculating nth roots on a simple pocket calculator.

The trick goes like this:

1. Type $x$ in the calculator


2. Take the square root twelve times


3. Subtract one


4. Multiply by $n$


5. Add one


6. Raise the number to the 2nd power twelve times (press * and = key eleven times)

Example:

I want to calculate $sqrt[3]20$ which is the same as $20^1/3$. So $x=20$ and $n=0.3333333$. After each of the six steps, the display on the calculator will look like this:

1. $;;;20$


2. $;;;1.0007315$


3. $;;;0.0007315$


4. $;;;0.0002438$


5. $;;;1.0002438$


6. $;;;2.7136203$

The actual answer is $20^1/3approx2.7144176$. So, our trick worked to three significant figures. It's not perfect, because of the errors accumulated from the calculator's 8 digit limit, but it's good enough for most situations.

Question:

So the question is now, why does this trick work? More specifically, how do we prove that:
$$x^napprox Big(n(x^1/4096-1)+1Big)^4096$$

Note: $4096=2^12$.

I sat in front of a piece of paper trying to manipulate the expression in different ways, but got nowhere.

I also noticed that if we take the square root in step 1 more than twelve times, but on a better-precision calculator, and respectively square the number more than twelve times in the sixth step, then the result tends to the actual value we are trying to get, i.e.:

$$lim_atoinftyBig(n(x^1/2^a-1)+1Big)^(2^a)=x^n$$

This, of course doesn't mean that doing this more times is encouraged on a pocket calculator, because the error from the limited precision propagates with every operation. $a=12$ is found to be the optimal value for most calculations of this type i.e. the best possible answer taking all errors into consideration. Even though 12 is the optimal value on a pocket calculator, taking the limit with $atoinfty$ can be useful in proving why this trick works, however I still can't of think a formal proof for this.

Thank you for your time :)

A standard trick is to calculate the natural logarithm first to get the exponent under control:

$$log(lim_atoinfty(n(x^1/a-1)+1)^a)=lim_atoinftyalog(nx^1/a-n+1)$$
Set $u=1/a$.
We get
$$lim_uto 0fraclog (nx^u-n+1)u$$
Use L'Hopital:
$$lim_uto 0fracnx^ulog xnx^u-n+1=nlog x=log x^n$$
Here we just plugged in $u=0$ to calculate the limit!

So the original limit goes to $x^n$ as desired.

For fixed $x > 0$ and $n$, let $t = 1/2^a to 0$. Then we need to prove that
$$
lim_t to 0 left( n (x^t - 1) + 1 right)^1/t = x^n.
$$
In fact, we have
$$
ln left[lim_t to 0 left( n (x^t - 1) + 1 right)^1/tright] =
lim_t to 0 fracln (1 + n(x^t - 1))t = lim_t to 0 fracn(x^t - 1)t = nln x,
$$
where the first equality follows from the continuity of $ln(x)$, and the second equality has used the fact that $ln(1 + x) sim x$ when $x to 0$.

If $x$ (actually $ln x$) is relatively small, then $x^1/4096=e^(ln x)/4096approx1+(ln x)/4096$, in which case

$$n(x^1/4096-1)+1)approx1+nln xover4096$$

If $n$ is also relatively small, then

$$(n(x^1/4096-1)+1)^4096approxleft(1+nln xover4096right)^4096approx e^nln x=x^n$$

Remark: When I carried out the OP's procedure on a pocket calculator, I got the same approximation as the the OP, $2.7136203$, which is less than the exact value, $20^1/3=2.7144176ldots$. Curiously, the exact value (according to Wolfram Alpha) for the approximating formula,

$$left(1over3(20^1/4096-1)+1right)^4096=2.7150785662ldots$$

is more than the exact value. On the other hand, if you take the square root of $x=20$ eleven times instead of twelve -- i.e., if you use $2048$ instead of $4096$ -- the calculator gives $2.7152613$ while WA gives $2.715739784ldots$, both of which are too large. Very curiously, if you average the two calculator results, you get

$$2.7136203+2.7152613over2=2.7144408$$

which is quite close to the true value!