# Calculate $\pi$ By Hand?

All over the internet the only hand equation i found was $$\frac\pi4 = 1 - \frac13 + \frac15 - \frac17+\cdots.$$ But this takes something like a thousand iterations to get to four digits, is there a better way to calculate pi by hand?

• There are many answers to this question; hence I added the tags soft-question and big-list, and flagged for community wiki. Please revert back to the old version if the tagging was inappropriate. – user122283 Sep 19 '14 at 23:31
• – lhf Sep 19 '14 at 23:36
• Here is a very closely related big-list. – MCT Sep 19 '14 at 23:40
• @SDevalapurkar, I don't think community wiki is appropriate for this post, any more than it would be for any other elementary math question that has at least one right answer. – Stephen Sep 20 '14 at 0:05
• Also pi314.net – Jean-Claude Arbaut Sep 20 '14 at 0:05

The fastest known formula for calculating the digits of pi is Chudnovsky formula: $$\frac{1}{\pi}=12 \sum_{k=0}^\infty \frac{(-1)^k (6k)! (163 \cdot 3344418k + 13591409)}{(3k)! (k!)^3 640320^{3k+1.5}}$$ This formula is used to create world record for the most digits of pi. This formula rapidly converges and it needs 3-4 terms to yield good approximation of pi which is possible by hand.

By hand, it's relatively easy to use the development of the arctangent, and a Machin-like formula:

Machin: $$\frac\pi4=4\arctan\frac15-\arctan\frac1{239}$$

Gauss: $$\frac\pi4=12\arctan\frac1{18}+8\arctan\frac1{57}-5\arctan\frac1{239}$$

I have done it once with Machin's formula and 24 decimals, in a few hours. It's recommended to do it by two methods, to check there is no computation error.

The arctangent is

$$\arctan x=\sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{2k+1}$$

Given a number of decimals, you find where to truncate by estimating the rest, and it's easy since it's an alternating series (so the rest is less in absolute value than the first omitted term).

Jean-Claude Arbaut has reminded us of the identity $$\frac\pi4=4\arctan\frac15-\arctan\frac1{239}.$$ Let us examine that. You learned in high school that $\tan\dfrac\pi4=1$, and that

\begin{align} \tan(\alpha+\beta) & = \dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} \tag 1 \\[10pt] & =\frac{c+d}{1-cd} \end{align}

Thus $$\arctan c+\arctan d=\alpha+\beta=\arctan\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\arctan\frac{c+d}{1-cd}$$

From $(1)$ we get $$\tan(\alpha+\beta+\gamma+\delta)=\frac{c+d+e+f-cde-cdf-cef-def}{1-cd-ce-cf-de-df-ef+cdef}$$ where $c,d,e,f$ are the respective tangents of $\alpha,\beta,\gamma,\delta$, and hence $$\tan(4\alpha) = \frac{4\tan\alpha-4\tan^3\alpha}{1 - 6\tan^2\alpha+\tan^4\alpha}.$$ Hence $$4\arctan c = \arctan\frac{4c-4c^3}{1-6c^2+c^4}.$$ So $$4\arctan\frac15 = \arctan\frac{(4/5)-(4/5^3)}{1-(6/5^2)+ (1/5^4)} = \arctan\frac{480}{476} = \arctan\frac{120}{119}.$$ Next we look at \begin{align} & 4\arctan\frac15 - \arctan\frac{1}{239} = \arctan\frac{120}{119} - \arctan\frac{1}{239} \\[15pt] = {} & \arctan\frac{(120/119)-(1/239)}{1+(120/119)(1/239)} \\[15pt] = {} & \arctan\frac{28561}{28561} = \arctan 1 = \frac\pi4. \end{align}

One easy-to-understand improvement to your method, which I I don't see used much, is:

$$\pi/6 = \arctan \left ( \frac{\sqrt{3}}{3} \right ) \\ = \int_0^{\frac{\sqrt{3}}{3}} \frac{1}{1+x^2} dx \\ = \sum_{n=0}^\infty \frac{(-1)^{n} \left ( \frac{\sqrt{3}}{3} \right )^{2n+1}}{2n+1} \\ = 3^{-1/2} \sum_{n=0}^\infty \frac{(-1)^{n} 3^{-n}}{2n+1}.$$

Consequently we have

$$\pi = 2 \sqrt{3} \sum_{n=0}^\infty \frac{(-1)^{n} 3^{-n}}{2n+1} \\ \approx 2 \sqrt{3} \sum_{n=0}^N \frac{(-1)^{n} 3^{-n}}{2n+1} \\ = 2 \sqrt{3} \left ( 1 - \frac{1}{9} + \frac{1}{45} - \frac{1}{189} + \dots \right )$$

This gives each decimal digit in slightly fewer than $\log_3(10) \approx 2.1$ steps, provided you can accurately estimate $\sqrt{3}$ to do the final multiplication.

For some fast-converging ideas, I recommend looking at this part of the wikipedia page. For example, the first option they present is $$\frac{\pi}{2} = \sum_{k=0}^\infty\frac{k!}{(2k+1)!!}=\\ \frac{1}{1} + \frac{1}{3 \cdot 1} + \frac{2\cdot1}{5\cdot3\cdot1} + \frac{3 \cdot2 \cdot 1}{7\cdot 5 \cdot 3 \cdot 1} + \cdots$$ Taking this out to the eighth step gives us $\pi \approx 3.137$

• It's not that efficient, since the summand is $2^k{2k \choose k}^{-1}\frac{1}{2k+1}$, and $$\frac{2k\choose k}{4^k}\sim\frac{1}{\sqrt{n\pi}}$$ I wonder why it's reported among efficient series like the ones by Ramanujan or Chudnovsky. – Jean-Claude Arbaut Sep 19 '14 at 23:58

You need a method to calculate exact $\pi$ by hand by an iteration? Or just some good numerical approximations, which easy to evaluate?

In the first case you can have this.

$$\pi = \left( \frac{2\sqrt{2}}{9801} \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}} \right)^{-1}.$$

This is a formula by Ramanujan and this gives 8 correct decimal digits for each $k$.

A little bit hard to evaluate Ramanujan's $\pi$ approximation by hand, but if you need just good numerical approximation, for $k=0$ you get this. $$\frac{9801\sqrt{2}}{4412} \approx \underline{3.141592}729$$

There are other formulas which are easier to evaluate by hand using just the elementary operations. $$3+\frac{4}{28}-\frac{1}{790+\frac56} \approx \underline{3.141592653}921$$

This one is good for $9$ decimal digits, and it is easy to remember because the formula contains each number from $0$ to $9$ exactly once.

This one is also good for $9$ decimal places.

$$\frac{22}{17} + \frac{37}{47} + \frac{88}{83} \approx \underline{3.141592653}467$$

This famous one is good for $6$ places, and also easy to remember:

$$\frac{355}{113} \approx \underline{3.141592}920$$

This one is also interesing, the nominator is $2^9$, and the denominator is the largest Heegner number. This is good for $3$ places:

$$\frac{512}{163} \approx \underline{3.141}104$$

The following one is used since at least Archimedes, and this is good for $2$ places:

$$\frac{22}{7} \approx \underline{3.14}285$$

Here is the proof that $22/7$ exceeds $\pi$.

You could find other approximations at Wikipedia and Wolfram MathWorld.

• I think the intent was a method which, carried sufficiently far, will compute $\pi$ to any desired degree of accuracy. – MJD Sep 21 '14 at 13:28

Very basic method based on A = πr2, A = π when r = 1. Get some really big piece of graph paper with a grid of fine lines. Draw either a circle centered on the intersection of two lines, or at least a quadrant of a circle. Then count squares at least 50% inside the circle. Normalize to the number of grid squares in a quadrant square. The finer the grid with a larger circle will give a more accurate result.

You can actually do the work in Excel - pretend it's done by hand, well in principle it could be. Use 1 million rows. Column 1 is x from 1 to R. Column 2 is the number of boxes at x. Define

Ny = INT( (R2-x2)1/2 +.5) ;

i.e., the number of boxes at x, or an integer estimate of y. Define

πest = 4*sum(Ny i)/R2, from i = 1 to R.

The accuracy should be roughly π/2*R/R2, because the quadrant arc is longer than R, and I'm estimating the number of squares in the path as a fraction of the total. In other words, the error involves the squares I have to decide are inside the arc or not, relative to the total.

But in fact, my fractional error is π/4R, because I'm getting half wrong (see 'calc fErr'). Define

fErr = (π - val)/π , of course you can simplify that.

R       val obs       fErr          calc fErr
3       2.222222222   0.292644697   0.261799388
10      2.92          0.070535132   0.078539816
30      3.066666667   0.023849682   0.026179939
100     3.1192        0.007127803   0.007853982
300     3.134577778   0.002232904   0.002617994
1000    3.13956       0.000647014   0.000785398
3000    3.140924889   0.000212556   0.000261799
10000   3.14139208    6.38446E-05   7.85398E-05
30000   3.141525667   2.13226E-05   2.61799E-05
100000  3.141572602   6.38275E-06   7.85398E-06
300000  3.141585979   2.12459E-06   2.61799E-06
1000000 3.141590652   6.37019E-07   7.85398E-07


I see Ramanujans formula here, but the one by Chudnovsky Brother's is still missing. Lets change that: $${\pi =1/12\, \left( \sum _{k=0}^{\infty }{\frac { \left( -1 \right) ^{k} \left( 6\,k \right) !\, \left( 545140134\,k+13591409 \right) \\ \mbox{}}{ \left( 3\,k \right) !\, \left( k! \right) ^{3}{640320}^{3\,k+3/2}}} \right) ^{-1}}$$ If you do it for the first iteration you get $${\frac {53360 \sqrt{640320}}{13591409}} \approx \underline{3.1415926535897}342077$$ Pretty neat, but I wouldn't want to do this "by hand".

Take your series and accelerate the convergence by applying an Euler transform to get

$$\frac\pi4=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}=\sum_{k=0}^\infty\frac1{2^{k+1}}\sum_{n=0}^k\binom kn\frac{(-1)^n}{2n+1}$$

Approximations:

p       sum(p)
1       0.666666666666666666666666666666666666666666666666666666666
2       0.733333333333333333333333333333333333333333333333333333333
3       0.761904761904761904761904761904761904761904761904761904761
4       0.774603174603174603174603174603174603174603174603174603174
5       0.780375180375180375180375180375180375180375180375180375180
10      0.785276505400344409632335328929756174338217681870932644926
20      0.785398074685084908175480624006064720023617739324001623280
30      0.785398163325289930752003674876445569039248433603744969299
40      0.785398163397386648371582346060565132165127359657157204105
50      0.785398163397448255413886763406807490779773842642064566708
100     0.785398163397448309615660845819841234165114606756625114249
150     0.785398163397448309615660845819875721049292349818655367858
200     0.785398163397448309615660845819875721049292349843776455243
...     ...........................................................
π/4     0.785398163397448309615660845819875721049292349843776455243


So that's $\pi/4$ out the first 57 digits.

(of course, other methods are recommended, but applying the Euler transform on an alternating series in general will rapidly raise convergence. and of course, I didn't do the +20 by hand)