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 Post subject: Calculus (12)
PostPosted: Wed, 9 Apr 2008 03:51:26 UTC 
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For any real number t > 0, define f(t) to be the number of pairs of integers (x,y) which satisfy x^2+y^2 < t.

Evaluate $ \lim_{t\to\infty} \frac{f(t)}{t}.


Hint:
Spoiler:
You'll eventually need to evaluate a some sort of Riemann sum!


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 Post subject: Re: Calculus (12)
PostPosted: Sun, 4 Jan 2009 15:51:51 UTC 
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commutative wrote:
For any real number t > 0, define f(t) to be the number of pairs of integers (x,y) which satisfy x^2+y^2 < t.

Evaluate $ \lim_{t\to\infty} \frac{f(t)}{t}.


Hint:
Spoiler:
You'll eventually need to evaluate a some sort of Riemann sum!


No, you don't really need any sort of Riemann sums...
Spoiler:
f(t) is the number of integral points in the open disc centred at origin with radius sqrt(t), so considering the unit squares centred at integral points, we can bound f(t) from above by Area(disc of radius sqrt(t)+1/sqrt(2)) and below by Area(disc of radius sqrt(t)-1/sqrt(2)), i.e.
\pi(\sqrt{t}-\frac{1}{\sqrt{2}})^2\leq f(t)\leq \pi(\sqrt{t}+\frac{1}{\sqrt{2}})^2
and hence
\displaystyle
\pi=\lim_{t\to+\infty}\dfrac{\pi(\sqrt{t}-\frac{1}{\sqrt{2}})^2}{t}
\leq\lim_{t\to+\infty}\dfrac{f(t)}{t}
\leq\lim_{t\to+\infty}\dfrac{\pi(\sqrt{t}+\frac{1}{\sqrt{2}})^2}{t}=\pi
so
\displaystyle\lim_{t\to+\infty}\dfrac{f(t)}{t}=\pi


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