Understanding $\arctan$, with the Leibniz series

Question

I am a newbie at complex math, and formulas that uses complex functions ($\sum_{i=a}^B$, $\arctan \theta$, $\tan \theta$ , etc.).

I mostly understand all functions that I said in the last sentance, but I don't understand the $\arctan \theta$ function. Also, I have always be amazed (since I learned what $\pi$ is) that $\pi$ has infinite digits. Some time later, I learned how people calculated more than millions digits of $\pi$.

I wanted to do that myself, so I found out for Leibniz series, that I yet, don't understand.

The Leibniz formula is : $\arctan 1 = \pi/4$. I am only able to calculate that with a calculator. I want to know how to calculate it on paper. I understand how to get $\pi$ with that formula ($4\arctan 1$), but how do I calculate $\arctan 1$?

Thanks for advance! Note: I haven't even started learning Trigonometry in school (I am $5^\mathrm{th}$ grade), so try to explain it well.

Answer 1

To make use of Leibniz formula you need to express function $\arctan x$ in the form of power series. That is $$ \arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}$$

Here I show how it is derived, but you don't know this to use it. You have, for $|x|<1$: $$ (\arctan x)' = \frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^n x^{2n}$$ where the middle eqaulity comes from calculating the sum of an infinite geometric series $\sum_{n=0}^\infty q^n = \frac{1}{1-q}$ for $|q|<1$. By taking an antiderivative of this equality we have (for $|x|<1$): $$ \arctan x= \int \sum_{n=0}^{\infty} (-1)^n x^{2n} dx = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} + C $$ For $x=0$ we have $\arctan 0 = 0$, which means that we need $C=0$ and we get $$ \arctan x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}$$

We have only proved it for $|x|<1$, but it turns out that for $|x|=1$ the series is still convergent (just not absolutely convergent) and the equality will still hold. That means that $$ \frac\pi 4 =\arctan 1 = \sum_{n=0}^{\infty} (-1)^n \frac{1}{2n+1} = 1 - \frac13 + \frac15 - \frac17 + \dots$$ $$ \pi = 4 - \frac43 + \frac45 - \frac47 + \dots$$ Note that this series converges very slowly, so you'd need a lot of terms to get a good approximation of $\pi$. To make the calculations a bit faster you can notice that $$ \frac{1}{1}- \frac{1}{3} = \frac{2}{1\cdot 3}$$ $$ \frac{1}{5}- \frac{1}{7} = \frac{2}{5\cdot 7}$$ $$ \frac{1}{9}- \frac{1}{11} = \frac{2}{9\cdot 11}\qquad \text{etc.}$$ Which gives you the formula $$\pi = 8\big(\frac{1}{1\cdot 3} + \frac{1}{5\cdot 7} + \frac{1}{9\cdot 11} + \frac{1}{13\cdot 15} +\dots\big) $$

To get a good approximation even faster, you can use for example $$ \arctan \frac{1}{\sqrt{3}} = \frac\pi 6$$ which gives you $$ \pi = 6 \arctan \frac{1}{\sqrt{3}} = 6\sum_{n=0}^{\infty} (-1)^n \frac{(1/\sqrt{3})^{2n+1}}{2n+1} = 2\sqrt{3} \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \frac{1}{3^n}$$ $$ \pi = 2\sqrt{3}\Big(1 - \frac1{3\cdot 3} + \frac{1}{5\cdot 3^2} - \frac{1}{7\cdot 3^3} + \dots\Big)$$ this series will give you a good precision much faster.

Answer 2

In practice, the first efficient formula was John Machin's, discovered in 1706:

$$\frac \pi 4=4\arctan\frac 15-\arctan\frac 1{239}.$$ When you expand the series defining $\arctan x$ with these values up to the $n^\text{th}$ term, you obtain $15$ exact decimal digits.

Other formulæ in a similar vein were found out later by various mathematicians (Euler, Gauß, Størmer,&c.) The latest was found by Hwang Chien-Lih in 2003: $$\textstyle\frac \pi 4=183\arctan\frac 1{239}+32\arctan\frac1{1023}-68\arctan\frac 1{5832}+12\arctan\frac 1{113021}-100\arctan\frac1{6826318}.$$

Another formula, converging still more rapidly, was found by Ramanujan before 1910: $$ \frac1\pi=\frac{2\sqrt 2}{9801}\sum_{n=0}^{\infty}\frac{(4n)!}{(n!)^4}\cdot\frac{1103+26390n}{(4\cdot 99)^{4n}}. $$