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November 20

Counting system

Resolved

StuRat (talk) 18:27, 20 November 2009 (UTC)[reply]

What's the name of the common counting system where you draw 1-4 lines and then put a slash through the four lines when you get to 5 ? It looks something like this:

 ||||/  ||||   |||   ||   | 
 |||/   ||||   |||   ||   |
 ||/|=5 ||||=4 |||=3 ||=2 |=1
 |/||   ||||   |||   ||   |
 /|||   ||||   |||   ||   |
/||||   ||||   |||   ||   |

StuRat (talk) 10:15, 20 November 2009 (UTC)[reply]

Tally. --PS T 10:39, 20 November 2009 (UTC)[reply]

Thanks. StuRat (talk) 18:25, 20 November 2009 (UTC)[reply]

Also five-barred gate, rather more specific than tally.→86.148.187.62 (talk) 21:03, 20 November 2009 (UTC)[reply]

Investigation of an integral

I'm working my way through some assorted questions, and stumbled across the following:
${\mbox{Investigate the integral: }}\int _{0}^{3}{\frac {1}{(1-x)^{2}}}\,dx$
I'm afraid I can't really get anywhere with it, and when I finally succombed to Mathematica it told me it didn't converge between 0 and 3. Any suggestions for what I should try? --80.229.152.246 (talk) 22:24, 20 November 2009 (UTC)[reply]

Without knowing what level you're at, I can't guess what investigations you're supposed to carry out, but the value of the integral is fairly obviously +∞. Algebraist 22:31, 20 November 2009 (UTC)[reply]

Yeah, I got that much, but didn't really get anything else apart from that. Also, how would you go about setting out a proof of that fact? As for my current level, just before undergraduate. --80.229.152.246 (talk) 23:07, 20 November 2009 (UTC)[reply]

I would prove it by noting that the function is clearly measurable, that the Lebesgue monotone convergence theorem implies that the integral you want is the limit of

\int _{0}^{1-1/n}{\frac {1}{(1-x)^{2}}}\,dx+\int _{1+1/n}^{3}{\frac {1}{(1-x)^{2}}}\,dx

, and that the Fundamental theorem of calculus allows you to calculate those integrals and see that the limit is indeed plus infinity. That's not what I'd've written before I was an undergrad, though. Algebraist 23:14, 20 November 2009 (UTC)[reply]

Thank you for that. While I agree that it's not exactly what a pre-undergrad would write, I'll still look into it. It's sure to be interesting, regardless. --80.229.152.246 (talk) 23:35, 20 November 2009 (UTC)[reply]

Since as a pre-undergrad. you are likely to be more familiar with Riemann integral, rather than Lebesgue integral, here is a proof you can follow based on those ideas:

Show that for any $0<h<1$ , $\int _{0}^{3}{\frac {1}{(1-x)^{2}}}\,dx>\int _{1-h}^{1+h}{\frac {1}{(1-x)^{2}}}\,dx$ (Hint: the integrand is positive)
Observe that, $\int _{1-h}^{1+h}{\frac {1}{(1-x)^{2}}}\,dx\geq \int _{1-h}^{1+h}{\frac {1}{h^{2}}}\,dx={\frac {2}{h}}$ .

Now by choosing h small enough, you can make the lower bound for the integral as large as you want. Hence the original integral evaluates to +∞. QED. Abecedare (talk)

Also, if you can do simple linear changes of variable, you may observe that

\int _{a/2}^{a}{\frac {1}{x^{2}}}\,dx=2\int _{a}^{2a}{\frac {1}{x^{2}}}\,dx>0,

that also makes it clear that the integral diverges.--pma (talk) 05:07, 21 November 2009 (UTC)[reply]

Or, even simplier, if you haven't any problem with a linear change of variable for an "improper Riemann integral", as it is the integral of 1/x² over (0,1]:

\int _{0}^{1}{\frac {1}{x^{2}}}\,dx=2\int _{0}^{2}{\frac {1}{x^{2}}}\,dx>0,

thus the only possible value of the LHS is

+\infty

. --pma (talk) 10:29, 21 November 2009 (UTC)[reply]

Thanks for all the help. --80.229.152.246 (talk) 17:26, 21 November 2009 (UTC)[reply]