Author Topic: Specialist Problem (Read 11081 times) Tweet Share

Tea.bag · « **Reply #15 on:** March 22, 2008, 05:38:07 pm »

A super hard question..

1) Let $a_1,a_2,...$ be positive real numbers in geometric progression. For each $n$ , let $A_n, G_n, H_n$ be respectively, the arithmetic mean, geometric means and harmonic mean of $a_1, a_2, ... , a_n$ . Find an expression for the geometric mean of $G_1, G_2, ... ,G_n$ in terms of $A_1, A_2, ... ,A_n, H_1, H_2, ... ,H_n.$

Ahmad · « **Reply #16 on:** March 22, 2008, 06:35:01 pm »

Let the geometric progression be $a,\ ar,\ ar^2, \ldots ,\ ar^{n-1}$

$A_n = \frac{1}{n} \displaystyle \sum_{i=0}^{n-1} ar^i = \frac{a(r^n - 1)}{n(r-1)}$

$H_n = n \cdot \left( \displaystyle \sum_{i=0}^{n-1} \frac{1}{ar^i} \right) ^{-1} = \frac{an(r^{-1} - 1)}{r^{-n} - 1}$

$G_n = \displaystyle \sqrt[n]{\prod_{i=0}^{n-1} (ar^i)} = ar^{\frac{n-1}{2}}$

$G(G_1,\ G_2,\ \ldots,\ G_n) = \displaystyle \prod_{i=1}^{n} (ar^\frac{i-1}{2}) = ar^\frac{n-1}{4}$

$= \sqrt[4]{A_1^2\cdot A_n\cdot H_n}$

Tea.bag · « **Reply #17 on:** March 22, 2008, 08:20:18 pm »

is this the same to ur solution?

Let $G_m$ be the geometric mean of $G_1, G_2, ... , G_3$

$\implies G_m = (G_1×G_2 .... G_n)^{\frac{1}{n}}$

$=[(a_1)×(a_1×a_1r)^{\frac{1}{2}}×(a_1×a_1r×a_1r^2)^{\frac{1}{3}} ... (a_1×a_1r×a_1r^2 ... a_1r^{n-1})^{\frac{1}{n}}]^{\frac{1}{n}}$

where $r$ is the common ratio of $G.P. a_1, a_2, ... , a_n$

$=[(a_1×a_1 ... n times)(r^{\frac{1}{2}}×r^{\frac{3}{3}}×r^{\frac{6}{4}} ... r^{\frac{n(n-1)}{2n}})]^{\frac{1}{n}}$

$=[a_1^n×r^{\frac{1}{2}+1+\frac{3}{2}+ ... \frac{n-1}{2}}]^{\frac{1}{n}} = a_1×[r^{\frac{1}{2}×[\frac{n(n-1)}{2}]}]^{\frac{1}{n}} = a_1×[r^{\frac{n-1}{4}}]$

Now, $A_n = \frac{a_1+a_2+ ... +a_n}{n} = \frac{a_1(1-r^n)}{n(1-r)}$

And, $H_n = \frac{n}{(\frac{1}{a_1}+\frac{1}{a_2}+ ... +\frac{1}{a_n})} = \frac{n}{\frac{1}{a_1}(1+\frac{1}{r}+ ... + \frac{1}{r^{n-1}})} = \frac{a_1n(1-r)r^{n-1}}{1-r^n}$

Again, $A_n×H_n = \frac{a_1(1-r^n)}{n(1-r)}×\frac{a_1n(1-r)r^{n-1}}{1-r^n} = a_1^2r^{n-1}$

$\implies \prod_{k=1}^{n}A_kH_k = \prod_{k=1}^{n}(a_1^2r^{k-1}) = (a_1^2×a_1^2×a_1^2 ... n times)× r^0×r^1×r^2 ... r^{n-1}$

$=a_1^{2n}×r^{1+2+ ... +(n-1)} = a_1^{2n}r^{\frac{n(n-1)}{2}} = [a_1r^{\frac{n-1}{4}}]^{2n} = [G_m]^{2m}$

$\implies G_m = [\prod_{k=1}^{n}A_kH_k]^{\frac{1}{2n}}$

$\implies G_k = (A_1A_2 ... A_n × H_1H_2 ... H_n)^{\frac{1}{2n}}$

AppleXY · « **Reply #18 on:** March 22, 2008, 08:39:01 pm »

Are you sure this is in the study design of specialist? I have never seen such a question in an any exam. (I mean, you only have 3 hours in total exam time)

cara.mel · « **Reply #19 on:** March 22, 2008, 08:45:20 pm »

Quote from: AppleXY on March 22, 2008, 08:39:01 pm

Are you sure this is in the study design of specialist? I have never seen such a question in an any exam. (I mean, you only have 3 hours in total exam time)

The last two questions aren't. The rest on this thread use spec knowledge but their only place is on like a problem solving SAC, they wouldn't be on the exams.
I do not like my familiarity with VCAA >.<

gfb · « **Reply #20 on:** March 22, 2008, 09:00:25 pm »

Good rittens! .

No way this is in the spesh study design because we are going to do them in semester 2? at Uni lol.

Ahmad · « **Reply #21 on:** March 22, 2008, 09:03:26 pm »

Yup it's the same solution expressed in a different way.

Tea.bag · « **Reply #22 on:** March 22, 2008, 11:35:29 pm »

Im pretty sure its not in the spesh course.
i was just letting people have a crack at it.

Ahmad · « **Reply #23 on:** March 22, 2008, 11:43:29 pm »

Possibly better place to put them in the future is,

Recreational Problems:
http://vcenotes.com/forum/index.php/topic,11.0.html

Tea.bag · « **Reply #24 on:** March 23, 2008, 10:05:37 pm »

how do u simplify?

Simplify:

1) $tan\left(\displaystyle\frac{\pi}{2}-A\right)$

2) $cot\left(\displaystyle\frac{3\pi}{2}+A\right)$

3) $cot\left(\displaystyle\frac{\pi}{2}+y\right)$

i seem to get undefined answers..

Mao · « **Reply #25 on:** March 23, 2008, 10:52:00 pm »

try looking at the transformations:

1) $tan\left( \frac{\pi}{2} -A\right)=tan\left(-\left(A-\frac{\pi}{2}\right) \right)$

so essentially it was flipped along y after the translation,
and because we have learnt about the relationship between tan(x) and cot(x), we know that this transformation describes cot(A)

2)remember the period for cot is $\pi$ , also looking at the transformations involved here, cot is shifted to the left by one period and $\frac{\pi}{2}$

3)notice the resemblence to the above?

Ahmad · « **Reply #26 on:** March 24, 2008, 12:48:35 am »

Another way,

$\tan (\frac{\pi}{2} - A) = \frac{\sin (\frac{\pi}{2} - A)}{\cos (\frac{\pi}{2} - A)} = \frac{\cos A}{\sin A} = \cot A$

Tea.bag · « **Reply #27 on:** March 24, 2008, 12:05:59 pm »

thnx.

More questions

Prove:

1) $secx-tanx=\frac{cosx}{1+sinx}$

2) $\sqrt\frac{1+sinx}{1-sinx}=secx+tanx$

3) $(secx-tanx)^2=\frac{1-sinx}{1+sinx}$

4) $sin^6x+cos^6x=1-3sin^2x cos^2x$

5) $(cosec\theta+cot\theta)^2=\frac{sec\theta+1}{sec\theta-1}$

Mao · « **Reply #28 on:** March 24, 2008, 01:57:23 pm »

1) $secx-tanx=\frac{cosx}{1+sinx}$

$sec(x)-tan(x)$

$\implies \frac{1}{cos(x)}-\frac{sin(x)}{cos(x)}$

$\implies \frac{1-sin(x)}{cos(x)}$

if we multiplied by $1+sin(x)$ on top and bottom:

$\implies \frac{\left( 1-sin(x) \right) \left( 1+sin(x) \right)}{cos(x)\left( 1+sin(x) \right)}$

notice that resembles difference between two squares, which simplifies to $1-sin^2 (x) \implies cos^2(x)$

$\implies \frac{cos^2(x)}{cos(x)\left( 1+sin(x) \right)}$

$\implies \frac{cos(x)}{1+sin(x)}$

$secx-tanx=\frac{cosx}{1+sinx}$

2) $\sqrt\frac{1+sinx}{1-sinx}=secx+tanx$

it will be very problematic expanding that square root, so we'll show that RHS is equal to LHS.

$sec(x)+tan(x)$

$\implies \frac{1+sin(x)}{cos(x)}$

using the pythagorean identity, $sin^2(x)+cos^2(x)=1 \implies cos(x)=\sqrt{1-sin^2(x)}$ , so we use this for the denominator:

$\implies \frac{1+sin(x)}{\sqrt{1-sin^2(x)}}$

now we include the numerator inside the square root:

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

notice how the denominator is difference between two squares

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

$\implies \sqrt\frac{1+sinx}{1-sinx}$

$\sqrt\frac{1+sinx}{1-sinx}=secx+tanx$

3) $(secx-tanx)^2=\frac{1-sinx}{1+sinx}$

from 1):

$secx-tanx=\frac{cosx}{1+sinx}$

$(sec(x)-tan(x))^2=\frac{cos^2(x)}{(1+sin(x))^2}$

$(sec(x)-tan(x))^2=\frac{1-sin^2(x)}{(1+sin(x))^2}$

$(sec(x)-tan(x))^2=\frac{(1+sin(x))(1-sin(x))}{(1+sin(x))^2}$

$(secx-tanx)^2=\frac{1-sinx}{1+sinx}$

4) $sin^6x+cos^6x=1-3sin^2x cos^2x$

given that: $sin^2(x)=1-cos^2(x)$

we can rewrite $sin^6(x)$ as:

$sin^6(x)=\left( sin^2(x) \right) ^3$

$sin^6(x)=\left( 1-cos^2(x) \right) ^3$

$sin^6(x)=1-3cos^2(x)+3cos^4(x)-cos^6(x)$

$sin^6(x)=1-3cos^2(x)+3\cdot cos^2(x) \cdot cos^2(x) -cos^6(x)$

$sin^6(x)=1-3cos^2(x)+3\cdot \left( 1-sin^2(x) \right) \cdot cos^2(x) -cos^6(x)$

$sin^6(x)=1-3cos^2(x)+3cos^2(x)-3sin^2(x)cos^2(x)-cos^6(x)$

$sin^6(x)+cos^6(x)=1-3sin^2(x)cos^2(x)$

5) $(cosec\theta+cot\theta)^2=\frac{sec\theta+1}{sec\theta-1}$

$cosec(\theta)+cot(\theta)=\frac{1+cos(\theta)}{sin(\theta)}$

$(cosec\theta+cot\theta)^2=\frac{(1+cos(\theta)^2}{sin^2(\theta)}$

$(cosec\theta+cot\theta)^2=\frac{(1+cos(\theta)^2}{1-cos^2(\theta)}$

$(cosec\theta+cot\theta)^2=\frac{(1+cos(\theta)^2}{(1+cos(\theta))(1-cos(\theta))}$

$(cosec\theta+cot\theta)^2=\frac{1+cos(\theta}{1-cos(\theta)}$

$(cosec\theta+cot\theta)^2=\frac{1+sec^{-1}(\theta}{1-sec^{-1}(\theta)}$

$(cosec\theta+cot\theta)^2=\frac{sec(\theta) \cdot sec^{-1}(\theta)+sec^{-1}(\theta}{sec(\theta) \cdot sec^{-1}(\theta)-sec^{-1}(\theta)}$

$(cosec\theta+cot\theta)^2=\frac{sec^{-1}(\theta) \cdot (sec(\theta)+1)}{sec^{-1}(\theta)\cdot (sec(\theta)-1)}$

$(cosec\theta+cot\theta)^2=\frac{sec\theta+1}{sec\theta-1}$

Ahmad · « **Reply #29 on:** March 24, 2008, 02:13:37 pm »

2. Multiply inside the square root of the LHS by $\frac{1 + \sin x}{1 + \sin x}$

4. $(\sin^2 x + \cos^2 x)^3 = 1$

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