Hiya
How would I evaluate p to as many decimal places as possible using an appropriate integral? I've got| ò02 | ____ Ö4-x2 | dx=p |
| ____ Ö4-x2 |
Well, (4-x2 )1/2 =
2 (1-x2 /4)1/2
=2 ( 1 + (1/2) (-x2 /4) + (1/2)(-1/2)/2
(-x2 /4)2 + (1/2)(-1/2)(-3/2)/3!
(-x2 /4)3 ... )
=2( 1 - x2 /8 - 1/128 x4 - 1/1024
x6 - 5/32768 x8 ...)
and so on. Now you can integrate each term of this as a
polynomial, carefully forgetting to mention to your teacher that
what you are doing is highly dodgy mathematically (it does work,
but you need to prove carefully in higher maths why it works).
This will give you a reasonable approximation to pi.
David
Well, if you are going to do this
properly, there are two things you need to prove, and both of
them are university-level maths.
First, you need to prove that the series expansion of
(4-x2 )1/2 is valid. This alone is well
beyond A-level standard (your teachers will probably pretend they
can prove it, but they won't do it properly). It is more or less
identical to question 11 from the 2000 Mathematics 1A exam paper
at Cambridge (the exam I will be doing in 2 weeks time). The way
to do it is to apply Taylor's theorem, and find an exact
expression in terms of an integral for the difference between the
first N terms of the series and the value of the function. You
can then show that this remainder term tends to zero.
Secondly, you need to show that it is valid to integrate the
series termwise. It seems sort of obvious that it should be, but
the standard proof that you can differentiate polynomials only
works for ones of finite degree, so there is some work to be
done. Anyway, this is justified by something called uniform
convergence, which is slightly technical but again is in the
Cambridge 1st year maths course, and probably in equivalent
courses at other universities.
David
Er... I'm pretty sure you will be
sitting the 2002 paper
unfortunately, since 2000 seems to have been an easy year (compared to 1998 for example)
And also the fact you know what's on the 2000 paper...