# Could we calculate pi using an iterative series

I know that, as a hobbyist mathematician, this is generally a term we can use to express pi

\begin{equation*} \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \frac{1}{13} - \frac{1}{15} + \frac{1}{17} - \cdots \end{equation*}

This is great representation , and it works just fine

However, I've been introduced to some iterative series recently (they're great for finding the roots to an equation by the way) and I was wondering, if there was a iterative series for pi, could we get a more justified value?

I've looked online, and there only seems to be geometric expressions using sine and cosine.

I was wondering if a numerical formula could be derived.

Edit: An iterative series is a series much like an algorithm, for example (N+1) = root(N+ 2/N) the idea being this series will converge on a value

• Can you clarify what you mean by 'iterative series'? If I had to guess, I'd assume you meant using Newton's method to find the roots of a polynomial via iteration. To make your idea work, all you need do is find an equation with $\pi$ (or some rational multiple thereof) as one of its roots. – Semiclassical Nov 11 '14 at 16:49
• The fact that π is transcendental means that there is no polynomial (with rational coefficients) that has π as a root, so I'm not sure you'd find an iterative formula that's just a simple polynomial or rational function. – sxpmaths May 29 '18 at 9:00
• Iterative or recurrence? Leibniz formula for $\pi$ you stated, can be written as a recurrence $x_{n+1}=x_n+\frac{(-1)^n}{2n+1}$, $x_0=1$ by the way. – rtybase Jul 22 '18 at 9:54

Well, there are iterative algorithms. There are two beautiful ones by the Borwein brothers, based on work by Ramanujan. Algorithm 1 involves the silver ratio, and Algorithm 2 involves the cube of the golden ratio.

$$y_0 = -1+\sqrt{2}$$

$$a_0 = 2(-1+\sqrt{2})^2$$

and two iterative rules,

$$y_{n+1} = \frac{1-(1-{y_n}^4)^{1/4}}{1+(1-{y_n}^4)^{1/4}}\tag1$$

$$a_{n+1} = a_n(y_{n+1}+1)^4-2^{2n+3}\,y_{n+1}\big(y_{n+1}^2+y_{n+1}+1\big)\tag2$$

Then,

$$\quad\quad\quad\lim_{n\to\infty} \frac{1}{a_n} = \pi\quad\text{(very fast)}$$

The difference grows quartically,

$$\quad\quad\quad\quad\frac{1}{a_n} - \pi \approx 4^{n+2} q^{4^n},\quad \text{where}\;q = e^{-2\pi}$$

Thus for $n=1,2,3$, the difference is about $10^{-10},\,10^{-42},\,10^{-172},$ or more than the fourth power of the previous. It's that fast.

Not sure what exactly it is that you want, but if you take newton's method and the power series of $\sin$ and $\cos$, you'll get $\pi$ as the limit of the newton iterations for $x_0 = 3$ and $x_{k+1} = x_k - \tan(x_k)$ where $\cot$ has a power series wich you can chose to evaluate up to an increasing degree, say $k$ to get $$x_{k+1} = x_k - T_k[\tan](x_k)$$ Where $T_k[f]$ is the $k$-th sum of the Taylor series of $f$ at a predefined point near $\pi$ ($3$ for example). This will give you a new polynomial term each iteration but it forms a sequence with $\lim_{k\to\infty} x_k = \pi$.

See here for $T_k[\tan]$

• That taylor series looks scary.. I just saw a method online which was something like x + sin(x). That did the job just fine on my calculator. – It'sRainingMen Nov 12 '14 at 10:50
• @It'sRainingMen If that is allowed (using trigs), you could just use $x_{k+1} = x_k - \tan(x_k)$. I interpreted your question as to specifically ask to avoid trigonometric functions in the iteration. – AlexR Nov 12 '14 at 11:06
• Yes, the trigometric soloutions are less preferable since something like x + sin(x) is worthless when the calculator treats x as pi hence sin(x) as 0. Although that tan(x) sequence also works, i guess – It'sRainingMen Nov 12 '14 at 13:39
• You could also take $\tan(x) \approx \frac{T_k[\sin](x)}{T_k[\cos](x)}$ wich is a rational function and the series expansions are well known (Take a higher order to prevent moving out of the neighborhood of $\pi$ so you don't get a different root of $\sin(x)$. – AlexR Nov 12 '14 at 14:14

The Basel infinite convergent serie (1/k squared) has the result of Pi squared over 6.By iterating actually you refine Pi value.There may be similar series that have a greater CONVERGING SPEED (you get PI decimals quicker).