An elementary proof for the Basel Problem

An elementary proof for the Basel Problem

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The Basel problem is a famous problem that was posed in 1644 and solved by the genius mathematician Leonard Euler in 1734. The problem requires the sum of the infinite series:

1+122+132+142+=?

Euler later proved that this sum is equal to π26.

If you search for the string “Basel problem”, you will find many different ways to prove this problem. While wandering around the internet, I accidentally found a quite interesting and “beautiful” elementary proof for this problem by R J Ransford in Eureka magazine No. 42, published in the summer of 1982. Below is the note of the solution.

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Link to the paper here

According to Euler: eix=cosx+isinx

Then: (eix)n=(cosx+isinx)n=ei(nx)=cosnx+isinnx

Implies: cosnx+isinnx=(cosx+isinx)n

Expanding the right side of the formula we have:

cosnx+isinnx=(n0)cosnx+(n1)cosn1.i.sinx+=((n0)cosnx+(n2)cosn2x.i2.sin2x++)+i((n1)cosn1x.sinx+(n3)cosn3x.i2.sin3x+(n5)cosn5x.i4.sin5x+)=((n0)cosnx(n2)cosn2x.sin2x+)+i((n1)cosn1x.sinx(n3)cosn3x.sin3x+(n5)cosn5x.sin5x)

Comparing the imaginary parts of the right and left sides, we deduce:

sinnx=(n1)cosn1xsinx(n3)cosn3xsin3x+(n5)cosn5xsin5x

Given a polynomial of degree n with coefficients aiC P(x)=anxn+an1xn1++a0 and those complex roots rn,rn1,,r1, then for all 0kn, we have:

1i1i2nri1ri2rik=(1)kankan

Specific case of the above formula:

i=1nri=an1an

First, let's prove the formula:

cot2(π2m+1)+cot2(2π2m+1)++cot2(mπ2m+1)=13m(2m1),(m1)

Indeed, reapplying the above sinnx formula:

sinnx=(n1)cosn1xsinx(n3)cosn3xsin3x+(n5)cosn5xsin5x

Let n=2m+1 and x=rπ2m+1,(r=1,2,,m), when sinnx=0,sinx>0, devide both sides to sinnx:

0=(2m+11)cot2mx(2m+13)cot2m2x+

Let t=cot2x, the formula becomes a polynomial with parameter t

P(t)=(2m+11)tm(2m+13)tm1+

P(t)=0 then t=cot2(rπ2m+1),(r=1,2,,m). Applying Vieta's formula to the sum of the above solutions, we can calculate the value of the sum of the solutions as (2m+13)(2m+11)=13m(2m1). We have proven the formula:

cot2(π2m+1)+cot2(2π2m+1)++cot2(mπ2m+1)=13m(2m1),(m1)

Add m to both sides:

m+cot2(π2m+1)+cot2(2π2m+1)++cot2(mπ2m+1)=m+13m(2m1)(1+cot2(π2m+1))+(1+cot2(2π2m+1))++(1+cot2(mπ2m+1))=13m(2m+2)

Apply csc2y=1+cot2y:

csc2(π2m+1)+csc2(2π2m+1)++csc2(mπ2m+1)=13m(2m+2)

Seeing that when 0<y<π/2 then 0<siny<y<tany, therefore cscy>1/y>coty:

13m(2m1)<(2m+1π)2+(2m+12π)2++(2m+1mπ)2<13m(2m+2)13m(2m1)<(2m+1)2π2+(2m+1)222π2++(2m+1)2m2π2<13m(2m+2)

Multiply all terms to π2(2m+1)2 to get:

π23[12(112m+1)(122m+1)]<1+122+132++1m2<π23[12(112m+1)(1+12m+1)]π26(112m+1)(122m+1)<1+122+132++1m2<π26(112m+1)(1+12m+1)π26(122m+1)<1+122+132++1m2<π26(1+12m+1)

When m, then 12m+10, 22m+10. We conclude:

1+122+132+=π26

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