Madhava–Leibniz Series

Poetry of Logical Ideas

Author

Abhirup Moitra

Published

July 30, 2023

"It has long been an axiom of mine that the little things are infinitely the most important." ~Sherlock Holmes (A Case of Identity)

Madhava–Leibniz Formula for π

In mathematics, the Leibniz formula for π , named after Gottfried Wilhelm Leibniz, states thatπ4=113+1517+ . an alternating series. It is sometimes called the Madhava–Leibniz series as it was first discovered by the Indian mathematician Madhava of Sangamagrama or his followers in the 14th–15th century (see Madhava series), and was later independently rediscovered by James Gregory in 1671 and Leibniz in 1673.

The Fascinating π4

π4=113+1517+

But how do we get here?

Let’s take a look at the series, n=1sin(nx)n. We’re looking at this sum because if we evaluate at π2 we get our sum in the question. i.e.,

π4=113+1517+...

=n=1sin(n.π2)n    ...(1)Then all we have to do is figure out what this n=1sin(nx)n series will be all set. Getting there might be tricky though we’re going to use a different representation for sine. In fact, we’re going to use one of the famous Euler’s formula eix=cosx+isinx .

instead of this x replace with nx i.e., einx=cos(nx)+isin(nx). Now we’ll replace sin(nx) with einx. However, in Euler’s formula, only the sine part is imaginary. So the way we get around this, we replace sine with e. However, we restrict it to only the imaginary part.

n=1sin(nx)n=Im(n=1einxn)

=Im(n=1(eix)nn) This time with the n in the exponent rearranging things slightly looks like a fairly well-known power series. In fact, this is the Maclaurin series representation of ln(1x)=n=1xnn just with eix substituted in. So, the imaginary part of this series will be written as,

ln(1eix)=n=1(eix)nn        ...(2)

Now, our job comes down to evaluating a complex logarithm. We know any complex number can be represented in its polar form i.e.,

z=x+iy

z=reiθ

here, r=x2+y2  ,  θ=tan1(yx)

Hence, ln(z)=ln(reiθ)=ln(r)+iθ     ...(3)

So, we need to figure out what’s r and what’s θ for our complex value.

In this case, we have ln(1eix) . We need to figure out what’s the real part and what’s the imaginary part. Fortunately, again Euler’s formula comes to the rescue.

Therefore, ln(1eix)=ln(1cosxisinx)

=ln((1cosx)2+sin2x)i tan1(sinx1cosx)

So, we have just replaced the value of eix. Now, you can see that inside the natural logarithm, 1cosx is the real part and sinx is the imaginary part. In this case, r=(1cosx)2+sin2x and θ=tan1(sinx1cosx) . As we are only concerned with the imaginary part of the above equation i.e.,itan1(sinx1cosx) , so we don’t need to look for the real part.

Im(n=1(eix)nn)=tan1(sinx1cosx)        ...(4)

If you string this all the way back to the beginning in eqation (1) that was connected to the original series by substituting π2 . If we plug in π2 in both side of equation (4) we’ll get,

Im(n=1(eix)nn)=tan1(sinx1cosx)

=tan1(sin(π2)1cos(π2))

=tan1(110)=π4

I think this is a nice way to elaborate but it would be better if you do it yourself in pen and paper.

Prove that (1132)(1152)(1172)=π4

One can solve this problem by resolving cosθ into factors i.e., expressing cosθ as an infinite product form. You can see Resolve sinθ Into Factors. The rigorous solution of expressingcosθ as infinite products I’m leaving for the readers to provide some hints.

Hint

  • Encounter the problem as it is shown in Resolve sinθ Into Factors. Till equation 4 there will be no change.

  • As you are willing to resolve cosθ instead of sinθ. Replace θ by π2+θ in equation 4. and you will see something like the equation below.

    cosθ=2p1sinπ+2θ2psin3π+2θ2psin(2p1)π+2θ2p()

  • In equation () the last factor sin(2p1)π+2θ2p=sinπ2θ2p and the last but one-factor sin(2p3)π+2θ2p=sin3π2θ2p and so on.

  • Hence by combining the first and the last factors, the second and the last but one factors and so on, the equation () becomes,

    cosθ=2p1[sinπ+2θ2p sinπ2θ2p][sin3π+2θ2p sin3π2θ2p]

    =2p1[sin2π2psin22θ2p][sin23π2psin22θ2p]

  • Finally, use the limit laws as it’s shown in resolution into factors of sinθ.

Now, we will solve the problem which is mentioned here. If you resolve cosθ into factors you will get,

cosθ=(14θ2π2)(14θ232π2)(14θ252π2)(14θ272π2)

cosθ(14θ2π2)=(14θ232π2)(14θ252π2)(14θ272π2)

Puttingθ=π2,i.e.,2θπ=1,we get,

limθ  π/2cosθ(14θ2π2)=limθ  π/2(14θ232π2)(14θ252π2)(14θ272π2)

Using L’Hospital’s Rule in the L.H.S., we get

limθ  π/2sinθ8θπ2=(1132)(1152)(1172)

π4=(1132)(1152)(1172)

(1132)(1152)(1172)=π4(Proved)

Let’s play with some integral !

EvaluateRx2dx dywhereRis the circlex2+y2=1.

x=rcosθ,y=rsinθ Again,xr=cosθ,xθ=rsinθ,yr=sinθ,yθ=rcosθ

(x,y)(r,θ)=|xrxθyryθ|=|cosθrsinθsinθrcosθ|=r

Now, we know that

Rf(x,y) dx dy=Rf(ϕ(u,v),ψ(u,v))|(x,y)(u,v)|du dv

Rx2dx dy=r=0r=1θ=0θ=2π(rcosθ)2|(x,y)(r,θ)|dr dθ

Since for a circular regionx2+y2=1,rvaries from 0 to 1 andθvaries from 0 to2π.

=r=0r=1θ=0θ=2πr3cos2θ dr dθ=01r3dr02πcos2θdθ

=[r44]011202π(1+cos2θ)dθ=18[θ+sin2θ2]02π=π4

Rx2dx dy=π4whereRis a circlex2+y2=1

Evaluate00e(x2+y2)dx dy

Letx=rcosθ,y=rsinθ,so thatx2+y2=r2Again, r2=trdr=12dt

Also, dx dy=(x,y)(r,θ) dr dθ=|cosθrsinθsinθrcosθ|dr dθ=r dr dθ=r dr dθ

Here the limits ofxare from 0 toand those of y are also from 0 to ,so the region of integration covers the positive quadrants in which r varies from 0 to . Since θ varies from 0 to π2,sor2=t varies from 0 to .

the given integral=θ=0θ=π2r=0r=er2r.dr dθ

=12θ=0θ=π2t=0t=etdt dθ

=12θ=0θ = π/2[et]0dθ

=12θ=0θ = π/2dθ=12[θ]0π2=π4

00e(x2+y2)dx dy=π4

Conclusion

There are lots of fascinating fact related to this π4, may be in another problem solving episode I will discuss about that or I will provide and update.

See Also

  1. Leibniz formula for π.

  2. List of formulae involving π.

  3. Machin-like formula.

  4. Euler product.

References

  1. Roy, Ranjan (1990). “The Discovery of the Series Formula for π by Leibniz, Gregory and Nilakantha” (PDF). Mathematics Magazine.63(5): 291–306. doi:10.1080/0025570X.1990.11977541.

  2. Ian G. Pearce (2002). Madhava of SangamagrammaMacTutor History of Mathematics archiveUniversity of St Andrews.

  3. Horvath, Miklos (1983). “On the Leibnizian quadrature of the circle” (PDF). Annales Universitatis Scientiarum Budapestiensis (Sectio Computatorica)4: 75–83.

  4. Debnath, Lokenath (2010), The Legacy of Leonhard Euler: A Tricentennial Tribute, World Scientific, p. 214, ISBN 9781848165267.

  5. Plofker, Kim (November 2012), “Tantrasaṅgraha of Nīlakaṇṭha Somayājī by K. Ramasubramanian and M. S. Sriram”, The Mathematical Intelligencer, 35 (1): 86–88, doi:10.1007/s00283-012-9344-6, S2CID 124507583.