Math challenge
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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. -
i btc 1.
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First, take a look at the Basel ProblemOriginally posted by lagrangian1125: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.
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) -
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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? -
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.
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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) -
it was a slightly different method.... Euler's really goodOriginally posted by tanjun: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.
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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.Originally posted by eagle: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) -
Correct.Originally posted by eagle: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)
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 0and 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. -
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I forgot all my year 1 mathsOriginally posted by lagrangian1125: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.
I will keep this nice statement in mind
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this thread is too chim for mi to understand
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It's something that is not taught <= A levels.Originally posted by wishboy:this thread is too chim for mi to understand
Only year 1 uni engineering/maths/science(physics only I think) will learn...