Cfrgtkky

Note that $pi=6arcsinleft(frac12right)$. So, since$$arcsin(x)=sum_{n=0}^infty frac1{2^{2n}}binom{2n}nfrac{ x^{2n+1}}{2n+1},$$you have$$pi=sum_{n=0}^inftyfrac6{2^{4n+1}(2n+1)}binom{2n}n.$$Now, for each $Ninmathbb{Z}^+$, let$$S_N=sum_{n=0}^Nfrac6{2^{4n+1}(2n+1)}binom{2n}ntext{ and let }R_N=sum_{n=N+1}^inftyfrac6{2^{4n+1}(2n+1)}binom{2n}n.$$Then:

• 
  $(forall Ninmathbb{Z}^+):pi=S_N+R_N$;


• the sequence $(S_N)_{Ninmathbb{Z}_+}$ is strictly encreasing and $lim_{Ntoinfty}S_N=pi$. In particular, each $S_N$ is a better approximation of $pi$ than the previous one.

Since$$(forall ninmathbb N):binom{2n}n<4^n=2^{2n},$$you have$$R_N<sum_{n=N+1}^inftyfrac6{2^{2n+1}}=frac1{4^N}.$$So, taking $N=0$, you get that $pi=S_0+R_0$. But $S_0=3$ and $R_0<1$. So, the first digit of $pi$ is $3$. If you take $N=3$, then $pi=S_3+R_3$. But $S_3approx3.14116$ and $R_3<0.015625$. So, the second digit is $1$. And so on…

The simplest method to explain to a child is probably the polygon method, which states that the circumference of a circle is bounded from below by the circumference of an inscribed regular $n$-polygon and from above by the circumference of a circumscribed polygon.

Once you have a bound from below and above, you can guarantee some digits. For example, any number between $0.12345$ and $0.12346$ will begin with $0.1234$.

In his answer, José shows how to calculate pi via a specific approximation and why that works. I believe the why is rather overlooked there and I wanted to clarify and make it less specific to the computation of $pi$.

Imagine you compute $S = Sigma_0^{infty} a_n$ for some series $a_n$. And, after summing the first few terms, let's say $bar S_i = Sigma_0^i a_n$, you can also prove that the rest of the sum is below some bounds $R_i^- le Sigma_{i + 1}^infty a_n le R_i^+$. Then you know also that $bar S_i + R_i^- le S le bar S_i + R_i^+$. See how that bounds the exact sum $S$ from above and below? If now both above and below have the same leading digits, we can be sure that those are also the leading digits of $S$.

Now, have another look at what José does: he computes the sum over a series up to term $N$ - the exact series is not important here. He approximates the errors $R_N^- = 0$ - all terms are positive - and $R_N^+ = frac{1}{4^N}$. So after you summed the first $N$ terms, what I called $bar S_N$ you can definitly say $bar S_N le S le bar S_N + frac{1}{4^N}$.

The answers so far to this great question illustrate a problem that we should redress in this forum: We rush in good faith to say something smart, something that other mathematicians may enjoy for its cleverness, but something which often is difficult to digest to the OP.

*steps off the soap box

Let me try a different take that will be of use to a 15 yo. There are two parts to the question: a) Do all known methods get arbitrarily many digits correct, b) how to tell that a digit is already correct.

a) Throughout history, people have found many ingenious ways to approximate $pi$, say as $22/7$ or $sqrt{10}$. Sometimes they knew they had an approximation, sometimes they mistakenly assumed they had the actual value.
When in modern mathematics a formula is presented for $pi$, it is guaranteed to give (eventually) as many digits as desired. The keyword is to say that the formula converges.

Please note that mathematicians word things differently; we do not care that “we get arbitrarily many digits correctly”, but rather that the value computed “is arbitrarily close to the target value”. These are equivalent, but the second is not dependent on writing numbers in base 10.

b) Every formula converges at its own pace, so there is no universal way to decide when a digit given by one or another is settled. However, there are general techniques to prove convergence, and often it is possible to see at a glance (or after a brief computation) that the formula converges. Other times it is not so straightforward...

So let’s take a look at only one example; namely the formula mentioned in the question:
$$4-4/3+4/5-4/7+ldots$$

This is particularly slow, but offers a great insight into convergence. It is an example of an alternating series; i.e., you add, then subtract, then add, then subtract, in perfect alternation. Moreover, each term is smaller than the previous one, as in $4/3>4/5>4/7>ldots$. Moreover, these terms get arbitrarily small, as in
$$4/4000001 < 4/4000000 = 0.000001$$

Now given these three conditions, we know the infinite sum will converge to a final vale (which we are told is $pi$). Why? Plot the consecutive sums on the real line to see what happens. You get 4, then 2.6666, then 3.46666, etc. More, then less, then more, so that the values are nested (because each term is smaller than the previous), and overshoot the final value of $pi$. Since the terms get small, the sums are forced to get closer and closer to the final value.

Here is the kicker: when you add $4/41$ (for instance), you overshot your mark, so the current sum is closer to $pi$ than $4/41$, and similarly for any other summand.

In particular, when you add $4/4000001$, you are closer to target than 0.000001, and the first 5 digits will be guaranteed.

Disclaimer. This does not show that the final value is $pi$. That requires more math. The argument only shows that the sum converges to a final value.

The Monte Carlo method is a stochastic method, so it does not provide certain proof. All it can do is say that the probability of having a particular result, if it were wrong about the first $k$ digits of $pi$, goes to zero.

For a sequence that converges to $pi$, however, we have that there is some function $f(k)$ such that for any $k$ and $n>f(k)$, the $n$-th term is correct to $k$ digits (barring the .9999.... issue). That's just from the definition of "converges"; one formulation of what it means to converge that is equivalent to the standard definition is that given any number of digits, there is some point in the sequence such that all the terms after that point are accurate to that number of digits. So any time someone claims that a sequence converges to $pi$, they are claiming that for each digit, there is some point at which it is certain (however, some people are loose with stochastic terminology, giving such formulations as "converges with probability one", which is not a precise formulation). Generally, proofs of convergence, even if they do not explicitly construct a function $f(k)$, can be easily modified to generate such a function.

For any approximation based on a Taylor series, the is the Lagrange error bound.

No method gives $pi$ exactly, i.e. all digits of $pi$, in finite time. But many methods give arbitrarily close approximations of $pi$ if they run long enough. Such methods construct a sequence of values $x_n$ whose limit as $ntoinfty$ is $pi$. For example, the technique you've mentioned has $x_1=4,,,x_2=4-frac{4}{3}$ etc.

Now, among the sequences satisfying $lim_{ntoinfty}x_n=pi$, some are "faster" than others. For example, the aforementioned sequence has $|x_n-pi|$ roughly proportional to $frac{1}{n}$, so the number of correct decimal places in approximating $pi$ as $x_n$ is approximately $log n$, for $n$ large. For example it takes about a million ($400,000$ in fact) terms to get $6$ decimal places right.

The good news is there are much better sequences than that; for example, this method gets a number of correct decimal place approximately proportional to $9^n$. All we have to do to be sure of specific digits is use appropriate mathematical theory to know how far to run a technique for our purposes. The bad news is this theory gets a little thorny, but I'll try to keep it simple. (If you feel i've made it too simple, see here to learn more.)

If $x_n$ is a sequence of limit $L$, and some $K,,p$ exist with the large-$n$ approximation $|epsilon_{n+1}|approx K|epsilon_n|^p$ with $epsilon_n:=x_n-L$, there are three separate cases to consider:

• 
  $p=K=1$, resulting in very slow convergence such as our original example;


• 
  $p=1,,K<1$, so $|epsilon_n|$ is approximately proportional to $K^n$, and the number of correct decimal places is approximately proportional to $n$;


• 
  $p>1$, so $log|epsilon_{n+1}|approx p|epsilon_n|+log K$, and the number of correct decimal places is approximately proportional to $p^n$.

The first case is called logarithmic convergence; the second is called linear convergence; the third is called superlinear convergence. Note that among superlinearly convergent algorithms increasing $p$ only causes a fractional reduction in the value of $n$ needed to get a given number of decimal places right, and often high-$p$ algorithms have such complicated steps they aren't worth it. The real question is whether some $p>1$ is achievable.

I linked before to a $p=9$ example of superlinear convergence, but it's very complicated. Depending on your son's ambition in self-education, he may be able to understand how this $p=2$ superlinear method works. In fact I probably should have focused on $p=2$ from the start, since calculus lessons often cover a (usually) $p=2$ technique for solving equations called the Newton-Raphson method. Somewhat easier, since it only requires a few basic facts about complex numbers, is understanding certain linear methods such as this work.

We can apply Dalzell's idea to prove $pi<frac{22}{7}$ to decimal approximations as well.

The first digit of $pi$ is guaranteed by the inequality
$$3<pi<4,$$

which can be proven from integrals
$$pi=3+2int_0^1frac{x(1-x)^2}{1+x^2}dx$$

and

$$pi=4-4int_0^1 frac{x^2}{1+x^2}dx$$

Similarly, the second digit being $1$ is equivalent to

$$3.1<pi<3.2$$

or

$$frac{31}{10}<pi<frac{16}{5},$$

which is proven by

$$pi=frac{31}{10}+2int_0^1 frac{x^2(1-x)^2(1-x+x^2)}{1+x^2}dx$$

and

$$pi=frac{16}{5}-int_0^1 frac{x^2(1-x)^2(1+2x+x^2)}{1+x^2}dx$$

Similar double inequalities can be written for every digit. For example, the answer https://math.stackexchange.com/a/2485646/134791 shows an integral for $pi>3.14$.

I wish to remind you of this formula: pi/4 = arctan(1) = 4 * arctan(1/5) - 1 * arctan(1/239) . This is easily proven with highschool math.
Then with the Taylor formula for the arctan() function you can see that this converges quickly (much quicker than arctan(1) itself), and you can even calculate how many digits you gain (on average) for each iteration. It all depends on starting with a good formula !

Assuming you can explain to you kid that:

$$
a_n rightarrowpi iff exists n_0 , text{such that} , n>n_0 rightarrow |pi -a_n|<varepsilon
$$

Then it is possible to state that $varepsilon$ is the precision of the "approximation" $a_n$.

Thus, you can compare the digits of $a_n+varepsilon$ and of $a_n-varepsilon$. All unchanged digits are certain.