Monday, August 28, 2006

∑ 1/n2 = π2/6

In today's blog, I show the proof that ∑ 1/n2 = π2/6. For background on Basel's problem and Leonhard Euler's technique for arriving at the solution, see here. In today's blog, I show a proof that Euler's solution is correct.

Most of the content in today's blog is taken from the Wikipedia article on the Basel Problem. See references below for more information.

Lemma 1: cot2(x) is one-to-one on the interval (0,π/2)

Proof:

(1) Suppose there exists x,y such that cot2(x) = cot2(y) in the interval (0,π/2)

(2) By definition, cot(x) = (cos x)/(sin x) so that we have:

(cos2 x)/(sin2 x) = (cos2 y)/(sin2 y)

which means that:

(cos2 x)/(cos2 y) = (sin2 x)/(sin2 y)

(3) We also know (see Corollary 2, here), cos2(x) + sin2(x) = 1 which is the same as:

1 - sin2(x) = cos2(x)

(4) Now, we can make the same argument for y and then we can divide by y to get:

[1 - sin2(x)]/[1 - sin2(y)] = [cos2(x)]/[cos2(y)]

(5) Now, we can substitute in step #2 to give us:

[1 - sin2(x)]/[1 - sin2(y)] = (sin2 x)/(sin2 y)

(6) This amounts to:

(sin2 y)(1 - sin2 x) = (sin2 x)(1 - sin2 y)

(7) Adding (sin2 x)(sin2 y) to both sides gives us:

(sin2 y)(1 - sin2 x) + (sin2 x)(sin2 y) =

= (sin2 y)(1 - sin2 x + sin2 x) = sin2 y

(sin2x)(1 - sin2 y) + (sin2 x)(sin2 y) =

= (sin2x)(1 - sin2y + sin2 y) = sin2 x

So that we have:

sin2 y = sin2 x

(8) Squaring both sides gives us:

±(sin y) = ±(sin x)

(9) Since we know that the sin function is nonnegative on the interval (0, π/2), we know that the square has to be nonnegative. This gives us:

sin y = sin x

(10) Finally, since sin x is one-to-one on the interval (0,π/2) (see here for properties of sin), we can see that x = y.

The basic idea here is that the sin function between 0 and π/2 does not repeat any of its values. This proves that if sin x = sin y, then x = y.

QED

Lemma 2: Sum of coefficients for a polynomial

if :

f(t) = amtm + am-1tm-1 + ... + a1t + a0 where am ≠ 0

then:

the sum of the roots of f(t) [counting multiplicities] is -am-1/am

Proof:

(1) Using the Fundamental Theorem of Algebra (see here), we know that we can divide up f(t) into a sequence of m roots such that:

f(t) = am(t - r1)(t - r2)*...*(t - rm)

(2) Now when we multiply all these values together it is clear that the sum of coefficients for tm-1 = the sum of all the roots multiplied by am.

We can see that if we carry out the multiplication of (t - r1)(t - r2)*...*(t - rm), we get:

tm + tm-1(-r1 + -r2 + ... + -rm) + ... + (-r1)(-r2)*...*(-rm)

(3) We can see that the coefficient for tm-1, am-1 = -am(r1 + r2 + ... + rm)

(4) But then r1 + r2 + ... + rm = -am-1/am

QED

Lemma 3: csc2(x) = 1 + cot2(x)

Proof:

(1) sin2x + cos2x = 1 (see Corollary 2, here)

(2) Divide both sides by sin2

Then we have:

1/(sin2x) = 1 + (cos2x)/(sin2x)

(3) Since by definition, cot2x = (cos2 x)/(sin2 x) and csc2x = 1/(sin2x), step #2 gives us:

csc2 x = 1 + cot2 x

QED

Lemma 4: cot2x is less than 1/x2 which is less than csc2 x on the interval (0,π/2)

Proof:

(1) If x is in (0, π/2), then (see Theorem, here):

sin x is less than x which is less than tan x.

(2) Taking the inverse of each (since x ≠ 0 → sin x ≠ 0 and tan x ≠ 0) gives us:

csc x is greater than 1/x which is greater than cot x.

(3) Now, squaring each value gives us:

cot2 x is less than 1/x2 which is less than csc2 x.

QED

Lemma 5: as m approaches infinity, both (2m)(2m+2)/(2m+1)2 and (2m)(2m-1)/(2m+1)2 approaches 1

Proof:

(1) Putting all values into polynomial form, we have:

(2m)(2m+2) = 4m2 + 4m

(2m)(2m - 1) = 4m2 - 2m

(2m+1)2 = 4m2 + 4m + 1

(2) Now, we can calculate the limits using L'Hopital's rule (see here)

lim (m → ∞) [(4m2 - 2m)/(4m2 + 4m + 1)] = lim (m → ∞) [(8m - 2)/(8m + 4)] = lim (m → ∞) [8/8] = 1.

lim (m → ∞) [(4m2 + 4m)/(4m2 + 4m + 1)] = lim (m → ∞) [(8m + 4)/(8m + 4)] = 1.

QED

Theorem: ∑ 1/n2 = π2/6

Proof:

(1) Using De Moivre's Formula (see Corollary, here for proof) we know that:

(cos x + i sin x)n = cos(nx) + i sin(nx)

(2) Dividing both sides by (sin x)n gives us:

[cos(nx) + i sin(nx)]/(sin x)n = (cos x + i sin x)n/(sin x)n = [(cos x + i sin x)/sin x]n

(3) Since cot x = (cos x)/(sin x) [by definition, cot(x) = 1/tan(x)], we have:

[(cos x + i sin x)/sin x] = cot x + i

(4) Putting this all together gives us:

[ cos(nx) + i sin(nx) ]/(sin x)n = (cot x + i)n

(5) Using the Binomial Theorem [See here], we know that:

(cot x + i)n = cotn x + (n,1)(cotn-1x)i + ... + (n,n-1)(cot x)in-1 + in

(6) Since i2 = -1, we can now divide up all terms in step #5 so that we have:

cotn x + (n,1)(cotn-1x)i + ... + (n,n-1)(cot x)in-1 + in = [(cotnx) - (n,2)(cotn-2 x) ± ... ] + i[(n,1)(cotn-1x) - (n,3)(cotn-3x) ± ... ]

(7) Since for any identity a + bi = c + di, we know that a=c and b=d, we can build the following equation from step #2 and step #6:

[isin(nx)]/(sin x)n = i[(n,1)(cotn-1x) - (n,3)(cotn-3x) ± ... ]

(8) Dividing both sides by i gives us:

sin(nx)/(sin x)n = [(n,1)(cotn-1x) - (n,3)(cotn-3x) ± ... ]

(9) Now, since step #8 is an identity, it is true for all values n,x.

Let n = 2m + 1 where m is a positive integer.
Let x = rπ/(2m + 1) where r can = 1, 2, ..., m

From this, we can see that:

nx = (2m + 1)*(rπ)/(2m+1) = rπ

(10) From a property of sin (see Lemma 1, here), we can see that:

sin(nx) =0

(11) Applying these values to step #8 gives us:

sin(nx)/(sin x)n = 0/(sin x)n = 0 =

= [(n,1)(cotn-1x) - (n,3)(cotn-3x) ± ... ] =

= [(2m+1,1)(cot2mx) - (2m+1,3)(cot2m-2x) ± ... + (-1)m ]

(12) Using Lemma 1 above, we can conclude that the values for x in step #11 are distinct. By the above equation we know that each of these m distinct solutions for x are roots to the following equation:

(2m + 1, 1)tm - (2m + 1,3)tm-1 ± ... + (-1)m = 0

(13) Using Lemma 2 to calculate the sum of roots using the coefficients, we have:

cot2(π/[2m+1]) + cot2(2π/[2m + 1]) + ... + cot2(mπ/[2m + 1]) = (2m+1,3)/(2m+1,1) = [(2m+1)!/(2m-2)!(3!)]/[(2m+1)!/(2m)!] = [(2m+1)(2m)(2m-1)/6]/(2m+1)] = (2m)(2m-1)/6.

(14) Using Lemma 3 above, we know that:

csc2 x = cot2x + 1

which means that:

cot2x = csc2 x - 1

(15) Applying (cot2x = csc2x - 1) to step #13 gives us:

cot2(π/[2m+1]) + cot2(2π/[2m + 1]) + ... + cot2(mπ/[2m + 1]) =

csc2(π/[2m+1]) + csc2(2π/[2m + 1]) + ... + csc2(mπ/[2m + 1]) - m =

= (2m)(2m-1)/6.

(16) If we add m to both sides we get:

csc2(π/[2m+1]) + csc2(2π/[2m + 1]) + ... + csc2(mπ/[2m + 1]) = (2m)(2m-1)/6 + m = (2m)(2m-1)/6 + 6m/6 = (4m2 - 2m + 6m)/6 = (4m2 + 4m)/6 = (2m)(2m+2)/6

(17) From Lemma 4 above, we know that for the interval (0,π/2) we have the following inequality:

cot2x is less than 1/x2 which is less than csc2 x

(18) But this implies that:

cot2(π/[2m+1]) + cot2(2π/[2m + 1]) + ... + cot2(mπ/[2m + 1]) is less than ([2m+1]/π)2 + ([2m+1]/2π)2 + ... + ([2m+1]/mπ)2 which is less than csc2(π/[2m+1]) + csc2(2π/[2m + 1]) + ... + csc2(mπ/[2m + 1])

(19) Now since we know that

(a) cot2(π/[2m+1]) + cot2(2π/[2m + 1]) + ... + cot2(mπ/[2m + 1]) = (2m)(2m-1)/6

(b) csc2(π/[2m+1]) + csc2(2π/[2m + 1]) + ... + csc2(mπ/[2m + 1]) = (2m)(2m+2)/6

We have:

(2m)(2m-1)/6 is less than ([2m+1]/π)2 + ([2m+1]/2π)2 + ... + ([2m+1]/mπ)2 which is less than (2m)(2m+2)/6

(20) If we multiply on all sides by (π/(2m+1))2, then we have:

π2/(2m+1)2(2m)(2m-1)/6 = (π2)(2m)(2m-1)/(6)(2m+1)2

2)/(2m+1)2([2m+1]/π)2 + ([2m+1]/2π)2 + ... + ([2m+1]/mπ)2 = 1/1 + 1/4 + .. + 1/m2

2)/(2m+1)2(2m)(2m+2)/6 = π2(2m)(2m-1)/(6)(2m+1)2

(21) By Lemma 5 above, as m approaches infinity, both the left and right hand expressions approach 2)/6 so by the Squeeze Theorem (see Lemma 3, here for proof), we have:

∑ (k=1,∞) (1/k2) = lim (m → ∞) (1/12 + 1/22 + ... + 1/m2) = π2/6

QED

References

2 comments:

Ali Riza ARICAN said...

Hi,
Thank you for this nice blog. I really enjoyed sigma(1/n^2)=pi^2/6 and its proof...
Best wishes
Ali

Babak Sairafi said...

hello
please tell me sigma(1/i) from i=2 to n
thanx
Babak Sairafi
babaksairafi@yahoo.com