Math challenge

lagrangian1125 SG

By using appropriate series, prove that 1+(1/3)^2+(1/5)^2+(1/7)^2.....=(pi^2)/8

sorry, cant paste my pi symbol.

popmynutsy SG

i btc 1. Laughing

eagle SG

First, take a look at the Basel Problem

http://en.wikipedia.org/wiki/Basel_problem

From here, we understand that
summation of 1/n^2 from n=1 to infinity, i.e.
1 + (1/2)^2 + (1/3)^2 + (1/4)^2 + ... = (pi^2)/6

we need to take away (1/2)^2 + (1/4)^2 + (1/6)^2 + ...

This can be seen as
summation of (1/2n)^2 from n=1 to n = infinity
which is also equal to
summation of (1/n)^2 * 1/4 from n=1 to n=infinity
This means that
(1/2)^2 + (1/4)^2 + (1/6)^2 + ... = 1/4 * (pi^2)/6 = (pi^2)/24

Hence,

1 + (1/3)^2 + (1/5)^2 + (1/7)^2..... = (pi^2)/6 - (pi^2)/24 = (pi^2)/8
(solved)

stellazio SG

Shocked Shocked Shocked

tanjun SG

I am not sure whether this method is the righ way of doing it. However, based on my guts feeling, this is the way. However, I can't get the answer.

The above series suit this simple term.
Upper limit n, r = 0 summation (1/[2r+1]^2)

In A levels, we learn that integration is a form of summation and vice versa.
Hence, I shall draw a curve of y = 1/(1+2x)^2

After drawing a curve, I will draw n rectangles from 0 to 1 to find the estimated area of rectangles. If n stretches on to infinity, it will be similar to integration.

Estimated area of curve = 1/n[n^2](1/(n+2)^2 + 1/(n+4)^2 + 1/(n+6)^2 + ...) for integral values of n = n(1/[n+2]^2 + 1/[n+4]^2 + 1/[n+6]^2 + ...)

In the end, I will integrate 1/(2x+1)^2 from 0 to 1 but the answer ends up to be 4/3. Lazy to type the middle step. Can help me check anything wrong?

tanjun SG

lol. Didn’t know there was a person who have discovered a method to do this and the theory is named after him. Waste one night doing this problem.

eagle SG

integrating is different from summation

If we see from the area,
integrating gives you the whole area, whereas summation gives you the sum of the individual values at that particular x value (value of x = 1, 2, 3... delta x =1), not those in between.

Concept wise,
integrating gives you the summation of sum of all individual values at that particular x value (value of x = 1, 1.0000001,... delta x = infinitesimally small)

eagle SG

it was a slightly different method.... Euler's really good Wink

tanjun SG

lol. Yup. That's why ans is different. Saw this similar qn in my A lvl notes that give me some sort of idea to tackle this qn under Integration Notes. However, I realised that in my steps, I just got an extra n as my common factor.

lagrangian1125 SG

Correct.
In addition, if we were ask to derive the series 1 + (1/2)^2 + (1/3)^2 + (1/4)^2 + ... = (pi^2)/6, there is an alternative other than Euler method.

Use fourier cosine series for y=x 0
and fourier sine series for y=x too

we get 1 + (1/3)^2 + (1/5)^2 + (1/7)^2..... =(pi^2)/8 and

(1/2)^2 + (1/4)^2 + (1/6)^2 + ... = (pi^2)/24 respectively.

Then, adding them up gave the solution.

The beauty of Fourier series lie in any repeating piecewise continuous function can be written as linear combi of some other repeating funct say trig function.

lpx88 SG

Shocked Shocked Shocked Shocked Shocked Shocked Shocked Shocked Shocked Shocked

eagle SG

I forgot all my year 1 maths Crying or Very sad

I will keep this nice statement in mind Very Happy

wishboy SG

this thread is too chim for mi to understand Embarassed Surprised

eagle SG

It's something that is not taught <= A levels.

Only year 1 uni engineering/maths/science(physics only I think) will learn...