# Is this a valid way to calculate pi ?Watch

#1
Ok so i was thinking for a bit and i came up with this

The circle, centre origin, radius one. Has the equation y^2=1-x^2. So if we consider the top right quadrant, we can say that the equation is y=sqrt(1-x^2)

The area here is given as pi/4. So if i integrate this between 0 and 1 this is what i should get; pi/4. I haven't dealt with how to integrate these kinds of expressions so i had to resort to more rudimentary tactics. I did the binomial expansion of (1-x^2)^.5 to get something more easily integrated. Then i integrated it and ended up with something that seems to converge on pi, if only very slowly with each term.

Reason i'm asking whether it's valid is because when you integrate, one of the limits is 1, which strictly speaking isn't good maths because the binomial expansion is only valid for a |x|<1. A reasonable approximation seems to come out with enough iterations but i just want to know whether the method i used was mathematically sound.
0
10 years ago
#2
The binomial expansion converges for |x| < 1

But you're dealing with the integral of said expansion, which is a whole different expansion, no?

Why not use the ratio test and see what range of values this new expansion converges for?
0
10 years ago
#3
A good question. I'm no expert, but I think all you should really implicitly be doing is taking a left-hand limit; set x = 1 - e, and let e tend to 0 from above. This seems to give you the exact same answer.

I'm having a bit of a mathematical blank here, so better wait till someone more experienced comes along.
0
10 years ago
#4
(Original post by sussexy)
The binomial expansion converges for |x| < 1

But you're dealing with the integral of said expansion, which is a whole different expansion, no?

Why not use the ratio test and see what range of values this new expansion converges for?
Any values that the antiderivative converges for, the original series will also converge for.
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10 years ago
#5
That last statement isn't strictly true, GE.

e.g. x - x^2/2+x^3/3 - ... (=log(1+x)) converges when x = 1.

But it's derivative 1-x+x^2-... (=1/(1+x)) diverges when x = 1.

To prove the result is OK here, you'd want to put a bound on the size of the binomial coefficients. I'm fairly sure you can show the series is abs convergent when x = 1, in which case integrating term by term is provably OK.
0
10 years ago
#6
Meh. I think if you use Stirling, you find that sqrt(1+x^2) is only conditionally convergent when x = 1. I think that's still OK; it's an alternating series and you integrate the first N terms and the remainder term -> 0 so it's integral -> 0 as well.
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10 years ago
#7
(Original post by DFranklin)
That last statement isn't strictly true, GE.

e.g. x - x^2/2+x^3/3 - ... (=log(1+x)) converges when x = 1.

But it's derivative 1-x+x^2-... (=1/(1+x)) diverges when x = 1.

To prove the result is OK here, you'd want to put a bound on the size of the binomial coefficients. I'm fairly sure you can show the series is abs convergent when x = 1, in which case integrating term by term is provably OK.
Oops, you're right. Amendment: the radius of convergence of the two will be the same, so there'll be a disc in the complex plane, inside which both will converge and outside which both will diverge. It's just that boundary that's a bit fuzzy round the edges.
0
#8
Sorry to drag this old thing back from the dead* but i still didn't get an answer. Is my maths kosher ?

*not really
0
9 years ago
#9
Well, we're pretty sure it's possible to prove it's OK. (And the mere fact that it 'seems' to converge to pi is actually a pretty good indicator here).

But seeing as you didn't prove it's OK, your maths is not kosher.
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#10
(Original post by DFranklin)
Well, we're pretty sure it's possible to prove it's OK. (And the mere fact that it 'seems' to converge to pi is actually a pretty good indicator here).

But seeing as you didn't prove it's OK, your maths is not kosher.
I still expect my OBE in the post for "services to mathematics and that".
0
9 years ago
#11
There was a similar question solved by Leibniz where he encountered a similar problem. Why don't you look it up, it may be worth reading.

However, why do all that when a substitution, such as would work?
0
9 years ago
#12
(Original post by technicolour)
There was a similar question solved by Leibniz where he encountered a similar problem. Why don't you look it up, it may be worth reading.
The 2nd Wiki solution is basically what I suggested in post #6 (although the series under discussion here is a little more complex than the Leibniz one).

However, why do all that when a substitution, such as would work?
How would that "work"? I don't see how you'd get a formula for pi that way.
0
9 years ago
#13
(Original post by DFranklin)
How would that "work"? I don't see how you'd get a formula for pi that way.
Woops, disregard that part. A momentary brain fart.
0
9 years ago
#14
(Original post by technicolour)
However, why do all that when a substitution, such as would work?
One problem with using trig to calculate pi is that your calculator has got pi built in, and uses it to calculate things like sin x.

Which is fine, it gets the right answer, but it was easier just to press the pi button on the calculator.
0
9 years ago
#15
(Original post by stevencarrwork)
One problem with using trig to calculate pi is that your calculator has got pi built in, and uses it to calculate things like sin x.

Which is fine, it gets the right answer, but it was easier just to press the pi button on the calculator.
Yeah, I realised that just after I made that post.
0
9 years ago
#16
I actually did this for my numerical methods coursework. Although at the time we only used midpoint, trapezium and then Simpson's rule.

What I did with the coursework though, was intergate the quarter circle as usual, then act all confused and then come up with an engenius (although already figured out awhile ago for the sake of improvement during coursework) solution of making a square of sides root(0.5) inside the quarter circle, obviously of area 0.5, then only intergrate of 0 to root(0.5). Obvously this answer needs to be multiplied by four and then added on to 2. Perhaps you could try that?

Although (and I'm probably missing something here) why can't you intergrate from 0 to 1? At point 1 the y co-ordinate is 0 so there's nothing to intergrate there surely?
0
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