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

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    $\begingroup$ 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. $\endgroup$ – Semiclassical Nov 11 '14 at 16:49
  • $\begingroup$ 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. $\endgroup$ – sxpmaths May 29 '18 at 9:00
  • $\begingroup$ 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. $\endgroup$ – 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.

I. Algorithm 1. Start with seed values:

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

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

and two iterative rules,

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

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


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

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.

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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 $T_k[\tan]$

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  • $\begingroup$ 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. $\endgroup$ – It'sRainingMen Nov 12 '14 at 10:50
  • $\begingroup$ @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. $\endgroup$ – AlexR Nov 12 '14 at 11:06
  • $\begingroup$ 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 $\endgroup$ – It'sRainingMen Nov 12 '14 at 13:39
  • $\begingroup$ 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 $\sin(x)$. $\endgroup$ – 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).

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