Author Topic: Interesting questions (spesh) (Read 12779 times) Tweet Share

itolduso · « **Reply #105 on:** November 06, 2010, 10:03:10 pm »

ok let me make an estimation, is it pi/4 ?

brightsky · « **Reply #106 on:** November 06, 2010, 10:03:58 pm »

Brilliant estimation because that is exactly what the answer is.

Martoman · « **Reply #107 on:** November 06, 2010, 10:05:55 pm »

lol what I do is just hacky maths

itolduso · « **Reply #108 on:** November 06, 2010, 10:08:56 pm »

Quote from: brightsky on November 06, 2010, 10:03:58 pm

Brilliant estimation because that is exactly what the answer is.

I picked a number out of infinity, my lucky day (night)

Martoman · « **Reply #109 on:** November 06, 2010, 10:09:59 pm »

Are you poking fun at little old me

brightsky · « **Reply #110 on:** November 06, 2010, 10:17:55 pm »

Just to keep the train moving, I'll post up another question:

$\int^{1}_{0} \log_e(\sqrt{1 - x} + \sqrt{1 + x}) dx$

Martoman · « **Reply #111 on:** November 06, 2010, 10:42:53 pm »

again, more guesswork profits.

Let $A = ln(\sqrt{1-x} + \sqrt{1+x})$

$e^{2A} = (\sqrt{1-x} + \sqrt{1+x})^2$

$e^{2A} = 1-x + 2\sqrt{1-x^2} + 1+x$

$e^{2A} = 2 + 2\sqrt{1-x^2}$

$A = \frac{1}{2}ln(2(1+\sqrt{1-x^2}))$

$A = \frac{1}{2}ln(2) + \frac{1}{2}ln(1+\sqrt{1-x^2})$

Integrate and sub in points; im in the TT boat here, ceebs going past the point once the inspiration has been experienced.

itolduso · « **Reply #112 on:** November 06, 2010, 11:39:24 pm »

Quote from: brightsky on November 06, 2010, 10:17:55 pm

Just to keep the train moving, I'll post up another question:

$\int^{1}_{0} \log_e(\sqrt{1 - x} + \sqrt{1 + x}) dx$

let me take another guess, pi/4 - 1/2 + 1/2 ln2

Martoman · « **Reply #113 on:** November 06, 2010, 11:42:38 pm »

Quote from: itolduso on November 06, 2010, 11:39:24 pm

Quote from: brightsky on November 06, 2010, 10:17:55 pm
Just to keep the train moving, I'll post up another question:

$\int^{1}_{0} \log_e(\sqrt{1 - x} + \sqrt{1 + x}) dx$

let me take another guess, pi/4 - 1/2 + 1/2 ln2

thats just spiteful now

golden · « **Reply #114 on:** November 27, 2010, 06:08:49 pm »

Could someone please show me how to work these questions out? (The ones without ticks)..

http://img718.imageshack.us/img718/368/ccf1127201000000.jpg

Thanks!

brightsky · « **Reply #115 on:** November 27, 2010, 07:49:15 pm »

10. b) Let $\sin(\theta) = x$

Then the series is $1 + x + x^2 + x^3 + ... = \sum^{\infty}_{n = 1} x^{n - 1} = \frac{1}{1-x}$

Hence $1 + \sin(\theta) + \sin^2(\theta) + \sin^3(\theta) + ... = \frac{1}{1 - \sin(\theta)}$

We want $\frac{1}{1 - \sin(\theta)} = 2 \implies 1 = 2 - 2\sin(\theta) \implies \sin(\theta) = \frac{1}{2}$

$\therefore \theta = \sin\frac{\pi}{6}, \sin\frac{5\pi}{6}, \sin\frac{13\pi}{6}, ... = \frac{\pi(12k + 1)}{6} \text{ or } \frac{\pi(12k+5)}{6} \text{ , where } k \in \mathbb{Z}$

9. b) i) $1 + \frac{x}{2} + \frac{x^2}{4} + ... = \sum^{\infty}_{n=1} \frac{x^{n-1}}{2^{n-1}} = \frac{2}{2 -x} = S$

Hence the infinite sum exists for all $x \in \mathbb{R} \not \{2\}$

werdna · « **Reply #116 on:** November 27, 2010, 11:16:42 pm »

Are you sure you're in year 9, brightsky?

golden · « **Reply #117 on:** November 29, 2010, 12:04:21 pm »

Quote from: brightsky on November 27, 2010, 07:49:15 pm

10. b) Let $\sin(\theta) = x$

Then the series is $1 + x + x^2 + x^3 + ... = \sum^{\infty}_{n = 1} x^{n - 1} = \frac{1}{1-x}$

Hence $1 + \sin(\theta) + \sin^2(\theta) + \sin^3(\theta) + ... = \frac{1}{1 - \sin(\theta)}$

We want $\frac{1}{1 - \sin(\theta)} = 2 \implies 1 = 2 - 2\sin(\theta) \implies \sin(\theta) = \frac{1}{2}$

$\therefore \theta = \sin\frac{\pi}{6}, \sin\frac{5\pi}{6}, \sin\frac{13\pi}{6}, ... = \frac{\pi(12k + 1)}{6} \text{ or } \frac{\pi(12k+5)}{6} \text{ , where } k \in \mathbb{Z}$

9. b) i) $1 + \frac{x}{2} + \frac{x^2}{4} + ... = \sum^{\infty}_{n=1} \frac{x^{n-1}}{2^{n-1}} = \frac{2}{2 -x} = S$

Hence the infinite sum exists for all $x \in \mathbb{R} \not \{2\}$

For 9. b) i) I got the same as you, however the answers say that it is: -2 < x < 2
Are the answers wrong?

For question 10 I don't why I put it up there - in my book i've done the question.

Thanks for your time however. +1 Karma.

golden · « **Reply #118 on:** November 29, 2010, 12:20:58 pm »

I know why I put it up now, what does where k E Z mean?

iNerd · « **Reply #119 on:** November 29, 2010, 12:22:22 pm »

Quote from: golden on November 29, 2010, 12:20:58 pm

I know why I put it up now, what does where k E Z mean?

k is an element of Z where Z represents the set of integers...I think.

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