Author Topic: TT's Maths Thread (Read 157748 times) Tweet Share

enwiabe · « **Reply #840 on:** February 03, 2010, 03:39:43 pm »

You need to have x -> N where N is a very large number, I believe, and then tend N to infinity.

TrueTears · « **Reply #841 on:** February 03, 2010, 03:41:30 pm »

Quote from: enwiabe on February 03, 2010, 03:39:43 pm

You need to have x -> N where N is a very large number, I belive, and then tend N to infinity.

Ah yeah, that makes more sense, thanks

TrueTears · « **Reply #842 on:** February 04, 2010, 12:57:23 am »

1. Find the limit of the sequence $\{\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}} \cdots \}$

2. Show that $\sqrt{2} = 1+\frac{1}{2+\frac{1}{2+ \cdots}}$

Many thanks!!!

Mao · « **Reply #843 on:** February 04, 2010, 01:01:56 am »

1. It can be shown that the sequence is $a_n = 2^{(2^n-1)/(2^n)}$ , hence as $n\to \infty$ , $a_n \to 2$

2. Show that $\sqrt{2} = 1+\frac{1}{2+\frac{1}{2+ \cdots}}$

Let $x=1+\frac{1}{2+\frac{1}{2+ \cdots}}$

$\implies \frac{1}{x-1} = 2 + \frac{1}{2+ \frac{1}{2+ \cdots}}$

$\implies \left(\frac{1}{x-1} - 2\right)^{-1} = 2 + \frac{1}{2+ \frac{1}{2+ \cdots}}$

$\implies \frac{1}{x-1} = \left(\frac{1}{x-1} - 2\right)^{-1}$

$\implies x-1 +2 = \frac{1}{x-1}$

$\implies x^2 - 1 = 1$

$\implies x = \sqrt{2}$ (discarding negative solution)

kamil9876 · « **Reply #844 on:** February 04, 2010, 01:34:00 am »

some nice paradoxes:

1.)

$x=2\times 2\times 2\times 2...$

$\implies x=2x$

$\implies x=0$

2.)

$x=1+2+4+8....$

$\implies x=1+2x$

$\implies x=\frac{-1}{2}$

Moral: check convergence.

TrueTears · « **Reply #845 on:** February 04, 2010, 10:27:57 pm »

Thanks Mao very clever!

How do I prove the sequence $\frac{n!}{2^n}$ where $n \in \mathbb{Z}^+$ is diverging?

Clearly by listing a few terms it is diverging, but how to prove it rigorously?

Nvm thanks Ahmad

$n! > 3^n$ for $n \ge 7$

Clearly this holds for $n = 7$

Assume it's true for $n = k$

$k! > 3^k$

Now $k!(k+1) > 3^{k}(k+1) > 3^k \cdot 3$

Thus proved by induction.

Now $\frac{n!}{2^n} > \left(\frac{3}{2}\right)^n$

Since $\left(\frac{3}{2}\right)^n$ is diverging for $n \in \mathbb{Z^+}$

$\frac{n!}{2^n}$ is also diverging.

kamil9876 · « **Reply #846 on:** February 04, 2010, 11:03:53 pm »

even easier:

let

$f(n)=\frac{n!}{2^n}$

$\frac{f(n+1)}{f(n)}=\frac{n+1}{2}\geq 2 \forall n\geq 4$

any sequence $f(4), f(5)...$ with this property must diverge. (as long as $f(4)\neq 0$ ) why?

TrueTears · « **Reply #847 on:** February 05, 2010, 11:52:07 pm »

Paradox:

$0 = 0 + 0 + 0 + 0 \cdots = (1-1)+(1-1)+(1-1)+(1-1)+ \cdots = 1-1+1-1+1-1+1-1+\cdots = 1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+\cdots = 1+0+0+0+0+ \cdots = 1$

hahaha why is this wrong?

kamil9876 · « **Reply #848 on:** February 06, 2010, 12:00:25 am »

Similair reason as to why the ones in my previous post are. Convergence is the key word.

TrueTears · « **Reply #849 on:** February 06, 2010, 12:01:07 am »

Quote from: kamil9876 on February 06, 2010, 12:00:25 am

Similair reason as to why the ones in my previous post are. Convergence is the key word.

Don't you mean divergent?

kamil9876 · « **Reply #850 on:** February 06, 2010, 12:03:18 am »

I used the noun "convergence" (hinting at the topic that relates to this) rather than used the adjective "convergent" to describe this particular case.

TrueTears · « **Reply #851 on:** February 06, 2010, 12:21:00 am »

Quote from: kamil9876 on February 06, 2010, 12:03:18 am

I used the noun "convergence" (hinting at the topic that relates to this) rather than used the adjective "convergent" to describe this particular case.

But the series is divergent

TrueTears · « **Reply #852 on:** February 06, 2010, 01:57:36 am »

Find the values of $x$ such that $\sum_{n=0}^{\infty} \frac{\cos^n(x)}{2^n}$ is convergent.

kamil9876 · « **Reply #853 on:** February 06, 2010, 02:07:00 am »

$n^{th} summand=(\frac{cosx}{2})^n$

for convergence we require $-1<\frac{cos(x)}{2}<1$

TrueTears · « **Reply #854 on:** February 06, 2010, 09:23:06 pm »

Ah yeah, nice thanks!

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