4 Unit Revising Marathon HSC '10 (2 Viewers)

addikaye03 · Jan 16, 2010

Well done ^ ! Nice first post lol Heres a new question:

(Harder 3U)
i) Prove: 1 - r² + r⁴ - r⁶ + r⁸ - ... = 1 / (1 + r²), for | r | < 1
ii) Hence find a series for tan inverse
iii) Hence show that pi = 4 - 4/3 + 4/5 - 4/7 + 4/9 ...

adomad · Jan 16, 2010

Trebla said:
Here's an interesting one (in my opinion anyway):

$$Suppose that $:\\\\S_{n}=\sum_{j=0}^{n}\dfrac{\sin[(2j+1)x]}{2j+1}\\\\(a)$ Show that $S_{n} = \displaystyle\int_{0}^{x} \left (\sum_{j=0}^{n} \cos [(2j+1)y] \right )\,dy\\\\(b)$ Use the fact that $e^{i\theta}=\cos \theta + i\sin \theta$ to show that $\\\\ (i) \cos [(2j+1)y] = \dfrac{1}{2}\left ( e^{iy(2j+1)}+e^{-iy(2j+1)} \right )\\\\$and$\\\\(ii) \sin [(2j+1)y] = \dfrac{1}{2i}\left ( e^{iy(2j+1)}-e^{-iy(2j+1)} \right )\\\\(c) $ Use the result from (b)(i) in (a) to show that$\\\\S_{n}=\dfrac{1}{2}\displaystyle\int_{0}^{x}\left [e^{iy}\left (\dfrac{e^{2iy(n+1)}-1}{e^{2iy}-1} \right )+e^{-iy}\left (\dfrac{e^{-2iy(n+1)}-1}{e^{-2iy}-1} \right ) \right ]\,dy\\\\(d) $ Hence, use the result in (b)(ii) to deduce that $\\\\ \sum_{j=0}^{n}\dfrac{\sin[(2j+1)x]}{2j+1} = \dfrac{1}{2} \int_{0}^{x}\dfrac{\sin[2y(n+1)]}{\sin y}\,dy$

hmm,,, what topic does this fall under???

untouchablecuz · Jan 16, 2010

adomad said:
hmm,,, what topic does this fall under???

harder 3u

tho, its not that hard, more interesting

gurmies · Jan 16, 2010

Koroneos said:
$(a)\,\,S_n=\sum_{j=0}^{n}\int_{0}^{x}\cos[(2j+1)y]dy\\=\sum_{j=0}^{n}\frac{\sin[(2j+1)x]}{2j+1}$

$(b)(i)\,\, 2\cos\theta=cis\theta+cis(-\theta)\\2\cos\theta=e^{i\theta}+e^{-i\theta}\\\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})\\\cos[(2j+1)y]=\frac{1}{2}(e^{iy(2j+1)}+e^{-iy(2j+1)})$by letting$\,\,\theta=[(2j+1)y]$

$(ii)\,\,2i\sin\theta=cis\theta-cis(-\theta)\\2i\sin\theta=e^{i\theta}-e^{-i\theta}\\\sin\theta=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})\\\sin[(2j+1)y]=\frac{1}{2i}(e^{iy(2j+1)}-e^{-iy(2j+1)})$by letting$\,\,\theta=[(2j+1)y]$

$(c)\,\,S_n=\int_{0}^{x}\sum_{j=0}^{n}\cos[(2j+1)y]dy\\=\frac{1}{2}\int_{0}^{x}(\sum_{j=0}^{n}e^{iy(2j+1)}-\sum_{j=0}^{n}e^{-iy(2j+1)})dy\\=\frac{1}{2}\int_{0}^{x}\left [ e^{iy}\left ( \frac{e^{2iy(n+1)}-1}{e^{2iy}-1} \right )+e^{-iy}\left ( \frac{e^{-2iy(n+1)}-1}{e^{-2iy}-1} \right ) \right ]dy\,\,$using sum of GP$$

$(d)\,\,\frac{1}{2}\int_{0}^{x}\frac{\sin[2y(n+1)]}{\sin y}dy\\=\frac{1}{2}\int_{0}^{x}\frac{e^{2iy(n+1)}-e^{-2iy(n+1)}}{e^{iy}-e^{-iy}}dy\\=\frac{1}{2}\int_{0}^{x}\frac{e^{2iny+3iy}-e^{-2iny-iy}}{e^{2iy}-1}dy\\=\frac{1}{2}\int_{0}^{x}(\frac{e^{2iny+3iy}-e^{iy}}{e^{2iy}-1}-\frac{e^{-2iny-iy}-e^{iy}}{e^{2iy}-1})dy\\=\frac{1}{2}\int_{0}^{x}\left [ e^{iy}\left ( \frac{e^{2iy(n+1)}-1}{e^{2iy}-1} \right )+e^{-iy}\left ( \frac{e^{-2iy(n+1)}-1}{e^{-2iy}-1} \right ) \right ]dy\\=S_n\,\,$from part c$\\=\sum_{j=0}^{n}\frac{\sin[(2j+1)x]}{2j+1}$

I applaud your patience with LaTeX =] Well done.

@ adomad - this is stuff not really learnt in MEII, but it guides you through it quite nicely, so there's no reason why a competent MEII student can't do this.

$\left ( i \right ) \\\\ 1-r^{2}+r^{4}-r^{6}+r^{8}-...=\frac{1}{1-\left ( -r^{2} \right )}=\frac{1}{1+r^{2}}; \,\, \left ( \text{Sum of a Geometric Progression with infinite terms} \right ) \\\\ \left ( ii \right ) \\\\ \int \left ( 1-r^{2}+r^{4}-r^{6}+r^{8}-...\right )dr=\int \frac{dr}{1+r^{2}} \\\\ = \tan^{-1}r+C \\\\ \therefore \tan^{-1}r+C=r-\frac{r^{3}}{3}+\frac{r^{5}}{5}-\frac{r^{7}}{7}+\frac{r^{9}}{9}-... \\\\ \text{Put} \,\, r=0\Rightarrow C=0 \\\\ \text{So} \tan^{-1}r=r-\frac{r^{3}}{3}+\frac{r^{5}}{5}-\frac{r^{7}}{7}+\frac{r^{9}}{9}-...\\\\ \left ( iii \right ) \\\\ \text{Put} \,\, r=1\Rightarrow \frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-... \\\\ \therefore \pi = 4-\frac{4}{3}+\frac{4}{5}-\frac{4}{7}+\frac{4}{9}-...$

Sorry, I don't currently have an interesting question - cover for me Trebla or others?

untouchablecuz · Jan 16, 2010

addikaye03 said:
Well done ^ ! Nice first post lol Heres a new question:

(Harder 3U)
i) Prove: 1 - r² + r⁴ - r⁶ + r⁸ - ... = 1 / (1 + r²), for | r | < 1
ii) Hence find a series for tan inverse
iii) Hence show that pi = 4 - 4/3 + 4/5 - 4/7 + 4/9 ...

$\\S=1-r^2+r^4-...\\$For $|r|<1, S $ has a limiting sum with $a=1 $ and $ R=-r^2\\\therefore S=1-r^2+r^4-...=\frac{1}{1+r^2}\\$Integrating $S $ with respect to $r $ from $0 $ to $x, $ noting that $|x|<1\\\int_{0}^{x}\frac{1}{1+r^2}dr=\int_{0}^{x}(1-r^2+r^4-...)dr\\\ \tan^{-1}x =x-\frac{x^3}{3}+\frac{x^5}{5}-...\\$Let $x=1,\\\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-...\\ \pi =4-\frac{4}{3}+\frac{4}{5}-...$

edit:beaten

gurmies · Jan 16, 2010

untouchablecuz said:
$\\s=1-r^2+r^4-...\\$for $|r|<1, s $ has a limiting sum with $a=1 $ and $ r=-r^2\\\therefore s=1-r^2+r^4-...=\frac{1}{1+r^2}\\$integrating $s $ with respect to $r $ from $0 $ to $x, $ noting that $|x|<1\\\int_{0}^{x}\frac{1}{1+r^2}dr=\int_{0}^{x}(1-r^2+r^4-...)dr\\\ \tan^{-1}x =x-\frac{x^3}{3}+\frac{x^5}{5}-...\\$let $x=1,\\\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-...\\ \pi =4-\frac{4}{3}+\frac{4}{5}-...$

edit:beaten

<3

untouchablecuz · Jan 16, 2010

$$Prove that the Fibonacci sequence, defined by: $u_0=0, u_{1}=1 $ and $u_n=u_{n-1}+u_{n-2} $ has as the formula for its $n$th term:$\\\\u_n=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n-\frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^n$

note: requires strong induction

untouchablecuz · Jan 16, 2010

gurmies said:
<3

s2

jet · Jan 16, 2010

The golden ratio and it's conjugate should be to the power of "n", not 2

untouchablecuz · Jan 16, 2010

jetblack2007 said:
The golden ratio and it's conjugate should be to the power of "n", not 2

oops

shaon0 · Jan 16, 2010

untouchablecuz said:
oops

Maybe setting a question on it's derivation would be good as the induction is quite easy.

untouchablecuz · Jan 16, 2010

shaon0 said:
Maybe setting a question on it's derivation would be good as the induction is quite easy.

ill try

gurmies · Jan 16, 2010

shaon0 · Jan 16, 2010

untouchablecuz said:
ill try

Let F{n}=x^n

The Nomad · Jan 16, 2010

jet · Jan 16, 2010

jet · Jan 16, 2010

shaon0 said:
Maybe setting a question on it's derivation would be good as the induction is quite easy.

But see, you're approaching it after having covered the complete 4 unit course with hours of practice under your belt. The 10ers don't have this amount of experience

lychnobity · Jan 16, 2010

How many different sums of money can be formed from the following:

18 $5 notes
25 $10 notes
66 $20 notes
32 $50 notes
41 $100 notes

96 $1 coins
58 $2 coins ?

Go forth and multiply

Trebla · Jan 16, 2010

Don't think I can match the interesting results of my previous questions, yet. So I'll just settle with this...enjoy!

$$Suppose that $T_n(x)= \cos (n\cos^{-1}x)\\\\$Show that $\\\int_{-1}^{1}\dfrac{T_m(x)T_n(x)}{\sqrt{1-x^2}}\,dx = \begin{cases} 0& $ if $ m \neq n \\\\\dfrac{\pi}{2} &$ if $ m = n \neq 0\\\\\pi & $ if $ m = n = 0 \end{cases}$

gurmies · Jan 16, 2010

The Nomad said:
$\\ x^5 + x^4 + 1 = (x^5 + x^4 + x^3 + x^2 + x + 1) - (x^3+x^2+x)\\ = \frac{x^6-1}{x-1} - x(x^2+x+1)\\ =\frac{(x^3-1)(x^3+1)}{x-1} - x\frac{x^3-1}{x-1}\\ = (x^3-x+1)\frac{x^3-1}{x-1}\\ = (x^3-x+1)(x^2+x+1)$

Nice!

I did this:

x^5 + x^4 + x^3 + 1 - x^3 = x^3(1 + x + x^2) + (1 - x)(1 + x + x^2)

===> (1 + x + x^2)(x^3 + 1 - x)

4 Unit Revising Marathon HSC '10 (2 Viewers)

The A-Team

HSC!!

Active Member

Drover

Active Member

Drover

Active Member

Active Member

Banned

Active Member

...

Active Member

Drover

...

Member

Banned

Banned

Active Member

Administrator

Drover

Users Who Are Viewing This Thread (Users: 0, Guests: 2)