Author Topic: TT's Maths Thread (Read 157300 times) Tweet Share

Christiano · « **Reply #886 on:** February 08, 2010, 10:58:54 pm »

OH MAN sorry, I thought this was a 'general math question' thread. 'Scuse me genius'es

Thanks anyhow

TrueTears · « **Reply #887 on:** February 08, 2010, 11:00:58 pm »

na man it's cool, if you have questions ur more than welcomed to ask.

TrueTears · « **Reply #888 on:** February 09, 2010, 12:01:09 am »

yo kamil

$\lim_{n \to \infty} \frac{n^n}{n!}$

lolz

Cthulhu · « **Reply #889 on:** February 09, 2010, 12:04:26 am »

Hey TT. I have this reaaaaaaaaally hard maths problem and I cannot figure it out and I have a test tomorrow!!!!!
If Ben has two apples and Erin has 6 oranges and 12 apples and gives Ben 3 apples and 2 oranges what is the airspeed velocity of an unladen swallow?

Mao · « **Reply #890 on:** February 09, 2010, 12:11:00 am »

Quote from: Cthulhu on February 09, 2010, 12:04:26 am

Hey TT. I have this reaaaaaaaaally hard maths problem and I cannot figure it out and I have a test tomorrow!!!!!
If Ben has two apples and Erin has 6 oranges and 12 apples and gives Ben 3 apples and 2 oranges what is the airspeed velocity of an unladen swallow?

$\frac{\mbox{It's}}{9000}$

TrueTears · « **Reply #891 on:** February 09, 2010, 12:11:14 am »

HAHAHAHAHAHA

kamil9876 · « **Reply #892 on:** February 09, 2010, 12:13:50 am »

$\frac{n^n}{n!}=\frac{n}{1} \times \frac{n}{2}... \times \frac{n}{n} \geq \frac{n}{1}=n$

and obviously $lim_{n \to \infty} n=\infty$ therefore the limit in question is also infinity.

TrueTears · « **Reply #893 on:** February 09, 2010, 12:14:51 am »

/facepalm

thx kamil

kamil9876 · « **Reply #894 on:** February 09, 2010, 12:19:16 am »

Quote from: Cthulhu on February 09, 2010, 12:04:26 am

Hey TT. I have this reaaaaaaaaally hard maths problem and I cannot figure it out and I have a test tomorrow!!!!!
If Ben has two apples and Erin has 6 oranges and 12 apples and gives Ben 3 apples and 2 oranges what is the airspeed velocity of an unladen swallow?

TrueTears · « **Reply #895 on:** February 09, 2010, 03:35:33 am »

If $\sum a_n$ is a conditionally convergent series and $r \in \mathbb{R}$ then there is a rearrangement of $\sum a_n$ that has a sum equal to $r$ .

Argh, I haven't done a proof in ages, but this one really caught my attention xD

Mao · « **Reply #896 on:** February 09, 2010, 08:54:55 am »

If $\sum a_n$ is conditionally convergent, $\sum |a_n|$ is divergent, and $\lim_{n\to \infty} |a_n| = 0$ .

So basically you have two divergent series, [for simplicity sake, I'll use alternating series as an example] $\sum a_{2n-1}$ and $\sum a_{2n}$ . Since both are divergent, but made up of infinitesimals, you can make up any real number with some arrangements of $\sum a_n$ (by selecting a certain number of elements from one/both series). That's the basic gist of it.

As for how to do this mathematically, that is for pure math majors to worry about.

humph · « **Reply #897 on:** February 09, 2010, 12:11:28 pm »

Quote from: Mao on February 09, 2010, 08:54:55 am

As for how to do this mathematically, that is for pure math majors to worry about.

Oh hai thar

TrueTears · « **Reply #898 on:** February 09, 2010, 02:35:51 pm »

thx

also why doesn't this work...

Is the following series absolutely convergent?

$\sum_{n=0}^{\infty} \frac{(-10)^n}{n!}$

$\sum_{n=0}^{\infty} \frac{(-10)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n(10)^n}{n!}$

$\sum_{n=0}^{\infty} \left | \frac{(-1)^n(10)^n}{n!} \right | = \sum_{n=0}^{\infty} \frac{(10)^n}{n!}$

Since $n! > 2^n$ for $n \ge 4$

$\frac{1}{n!} < \frac{1}{2^n}$ for $n \ge 4$

So $\frac{10^n}{n!} < \frac{10^n}{2^n}$ for $n \ge 4$

Using the comparison test, $\sum_{n=0}^{\infty} \frac{10^n}{2^n}$ is not convergent... so what do I do?

humph · « **Reply #899 on:** February 09, 2010, 02:55:19 pm »

Quote from: TrueTears on February 09, 2010, 02:35:51 pm

thx

also why doesn't this work...

Is the following series absolutely convergent?

$\sum_{n=0}^{\infty} \frac{(-10)^n}{n!}$

$\sum_{n=0}^{\infty} \frac{(-10)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n(10)^n}{n!}$

$\sum_{n=0}^{\infty} \left | \frac{(-1)^n(10)^n}{n!} \right | = \sum_{n=0}^{\infty} \frac{(10)^n}{n!}$

Since $n! > 2^n$ for $n \ge 4$

$\frac{1}{n!} < \frac{1}{2^n}$ for $n \ge 4$

So $\frac{10^n}{n!} < \frac{10^n}{2^n}$ for $n \ge 4$

Using the comparison test, $\sum_{n=0}^{\infty} \frac{10^n}{2^n}$ is not convergent... so what do I do?

You want to show that $n! > 10^n$ for sufficiently large $n$ - work out when this inequality is first true, then use induction (this is when a calculator or computer comes in handy).

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