....this confuses me and I really want an answer for this. The question I want to ask is why the output seems to approach pi at the beginning and then turn out to fluctuate and eventually becoming 0 at some point rather than continuously approaching pi.......
This concept and equation seems to be sensible and correct as it approaches pi,3.1415...etc when I input big values into my calculator, by substitution.
However, I found the that when I plot the graph using wolfram alpha from n =0 to n = 1000000 I found the output of the equation started to fluctuate in stead of approaching pi.....this confuses me and I really want an answer for this. The question I want to ask is why the output seems to approach pi at the beginning and then turn out to fluctuate and eventually becoming 0 at some point rather than continuously approaching pi...............
Your reasoning is sound, but let's look at that last limit.
lim(n->inf.)(n*[2(1 - cos(pi/n))]^(1/2)) = 2^(1/2)*lim(n->inf.)(n*[1 - cos(pi/n)]^(1/2)).
Now you should be inputting C in radians, not degrees, as radians is the 'natural'
measure of angles, and you should have your calculator in radian mode. In which
case, for n = 100, I get 3.1414634; for n = 1000, I get 3.1415888 and for
n = 10000 I get 3.1413373. For n = 20000 I got 3.1419739. This is as far as I could
go with my calculator, and the numbers are wandering away from what we know to
be pi. So I agree with you; for higher n there is significant fluctuation. What might
be happening is that we are overtaxing the ability of our calculators and Wolfram
to deal with both the large and small numbers involved in the calculation. Or maybe
something else more mathematically fundamental? Interesting. I'll keep working
at it. :)
Edit: Hey, Ryan. So I plugged in 2^(1/2)*lim(n->inf.)(n*(1 - cos(pi/n))^(1/2)) into
Wolfram Alpha and got pi as the answer, so your concept works! I don't know
why we were getting the fluctuations, but I think it has to do more with machines
than mathematics. :)
Edit #2: O.k., I can actually prove it now. Multiply the top and bottom of the
limit by (1 + cos(pi/n))^(1/2); then the limit becomes
2^(1/2)*lim(n->inf.)(n*sin(pi/n) / (1 + cos(pi/n))^(1/2)) since
(1 - cos(pi/n))^(1/2) * (1 + cos(pi/n))^(1/2) = (1 - cos^2(pi/n))^(1/2) =
(sin^2(pi/n))^(1/2) = sin(pi/n) because sin(pi/n) is positive. Now the
term (1 + cos(pi/n))^(1/2) goes to (1 + 1)^(1/2) = 2^(1/2) as n -> inf., so
the limit becomes lim(n->inf.)(n*sin(pi/n)) = lim(n->inf.)(pi*sin(pi/n)/(pi/n)).
Now let x = pi/n. Then x -> 0 as n -> inf., and the limit is transformed to
pi*lim(x->0)(sin(x)/x) = pi*1 = pi, where the fact that lim(x->0)(sin(x)/x) = 1
is a well-known result. So, again, your concept is absolutely sound and the
problem is with the calculating machines. Now I can get some sleep. :D
