4 Unit Revising Marathon HSC '10 (1 Viewer)

The Nomad · Jan 20, 2010

vafa said:
Can you provide a mathematical proof for this statement? We all wish, it could be that easy, but it is not. your hypothesis is cosa+cosb+cos c=sin a+sin b+sin c=0, but your point is just one of the cases, it could be the case that cosa, cosb and cos c and sin a, sin b and sin c, all add up to zero. Right? and in that case, neither of the terms could be zero but their addition would be zero.

Oh deary me...I completely misread the question. My mistake, I know what you mean now.

shaon0 · Jan 20, 2010

The Nomad said:
$\\$Sides of a triangle add to 180 degrees, so $ A + B + C = 180.\\ A = 180 - (B + C), Tan(\frac{A}{2}) = Tan(90 - \frac{B + C}{2}) = Cot(\frac{B + C}{2})\\ Tan\frac{A}{2}Tan\frac{B}{2}\cdot Tan\frac{A}{2}Tan\frac{C}{2}\cdot Tan\frac{B}{2}Tan\frac{C}{2} = \\(Tan\frac{B}{2}+Tan\frac{C}{2})Tan\frac{A}{2} + Tan\frac{B}{2}Tan\frac{C}{2}=\\ (Tan\frac{B}{2} + Tan\frac{C}{2})Cot\frac{B + C}{2} + Tan\frac{B}{2}Tan\frac{C}{2} = \\(Tan\frac{B}{2} + Tan\frac{C}{2})\frac{1-Tan\frac{B}{2}Tan\frac{C}{2}}{Tan\frac{B}{2}+Tan\frac{C}{2}} + Tan\frac{B}{2}Tan\frac{C}{2} = \\(1 - Tan\frac{B}{2}Tan\frac{C}{2}) + Tan\frac{B}{2}Tan\frac{C}{2} = 1.$

That was so annoying to type. And what do you mean Jetblack?

I got it out like this, but i'm wondering whether a polynomial proof is possible. The polynomial is degree 3 with roots tan(A/2), tan(B/2), tan(C/2) and sum of roots=product of roots.

The Nomad · Jan 20, 2010

shaon0 said:
I got it out like this, but i'm wondering whether a polynomial proof is possible. The polynomial is degree 3 with roots tan(A/2), tan(B/2), tan(C/2) and sum of roots=product of roots.

I think you mean sum of roots = product of roots for the equation with roots Tan A, Tan B and Tan C...

shaon0 · Jan 20, 2010

The Nomad said:
I think you mean sum of roots = product of roots for the equation with roots Tan A, Tan B and Tan C...

Yeah, sorry. I don't think it works though.

Trebla · Jan 20, 2010

shaon0 · Jan 20, 2010

Trebla said:
$$Prove by mathematical induction that for all positive integers of $n,\\\\\displaystyle\int_0^{\infty}x^{n-1}e^{-x}\,dx = (n-1)!$

Let n=1:
S {x=inf to x=0} e^-x dx = -[e^-(inf)-1] = -(0-1) = 1 =0!

Let n=k:
Assume, S {x=inf to x=0} x^(k-1).e^-x dx=(k-1)!

Let n=k+1:
S {x=inf to x=0} x^(k).e^-x dx
= -[e^-x.x^k] {x=inf to x=0} - k.S {x=inf to x=0} x^(k-1).e^-x dx
= -(0-0)+k(k-1)! [As e^x dominates x^k]
= k!

Trebla · Jan 20, 2010

shaon0 said:
Let n=1:
S {x=inf to x=0} e^-x dx = -[e^-(inf)-1] = -(0-1) = 1 =0!

Let n=k:
Assume, S {x=inf to x=0} x^(k-1).e^-x dx=(k-1)!

Let n=k+1:
S {x=inf to x=0} x^(k).e^-x dx
= -[e^-x.x^k] {x=inf to x=0} - k.S {x=inf to x=0} x^(k-1).e^-x dx
= -(0-0)+k(k-1)! [As e^x dominates x^k]
= k!

sign error?

shaon0 · Jan 20, 2010

Trebla said:
sign error?

Yeah, sorry. It was pretty late and my concentration levels were pretty low. Thanks

cutemouse · Jan 20, 2010

Trebla said:
$$Prove by mathematical induction that for all positive integers of $n,\\\\\displaystyle\int_0^{\infty}x^{n-1}e^{-x}\,dx = (n-1)!$

You should really write

$\int_0^{\infty}x^{n-1}e^{-x}\,dx$

as

$\lim_{b \to \infty}\int_0^{b}x^{n-1}e^{-x}\,dx$

I think...

addikaye03 · Jan 20, 2010

shaon0 said:
They're angles in a triangle.

Yeah my bad, that's what i meant lol

Alternative Answer:

Let x=A/2, y=B/2 and z=C/2

tan(x+y+z)=

[tanx+tany+tanz+(tanxtanytanz)]/[1-(tanytanx+tanxtanz+tanytanz)

But since x+y+z=90 degrees (since A+B+C=180) therefore is undefined

Hence tan(y)tan(x)+tan(x)tan(z)+tan(y)tan(z)=1

tan(A/2)tan(B/2)+tan(A/2)tan(C/2)+tan(B/2)tan(C/2)=1 #

shaon0 · Jan 20, 2010

jm01 said:
You should really write

$\int_0^{\infty}x^{n-1}e^{-x}\,dx$

as

$\lim_{b \to \infty}\int_0^{b}x^{n-1}e^{-x}\,dx$

I think...

I think both ways are correct.

@addikaye03: That's a pretty good solution

Trebla · Jan 20, 2010

jm01 said:
You should really write

$\int_0^{\infty}x^{n-1}e^{-x}\,dx$

as

$\lim_{b \to \infty}\int_0^{b}x^{n-1}e^{-x}\,dx$

I think...

$\int_0^{\infty}x^{n-1}e^{-x}\,dx$

actually means

$\lim_{b \to \infty}\int_0^{b}x^{n-1}e^{-x}\,dx$

shaon0 · Jan 20, 2010

I've just made up a question. it's quite easy though.
a) Prove by induction or otherwise:
[a1+a2+...+an]/n > (a1a2...an)^(1/n)

b) Hence or otherwise, prove that (1-e)^-n=n^n.e^(n/2)(1-n) for n large where n E Z+
HINT: Replace a1,a2,...,an by 1,e,e^2,...,e^n

addikaye03 · Jan 20, 2010

shaon0 said:
I think both ways are correct.

@addikaye03: That's a pretty good solution

Thanks i got a heap of interesting Q actually, been doing stuff from "Art and craft of problem solving"...here and there.

shaon0 · Jan 20, 2010

addikaye03 said:
Thanks i got a heap of interesting Q actually, been doing stuff from "Art and craft of problem solving"...here and there.

Yeah, nw. Fancy solving my question? It's pretty easy if you have seen the hint. Otherwise, it's pretty difficult

lolman12567 · Jan 20, 2010

(i) Find the sum 1 + w + w² + w³ +... to n terms, considering the cases n=3k, n=3k+1, n=3k+2 where k is an integer.

(ii) Show that
(1 - w+w²)(1 - w² +w⁴)(1 - w² +w⁸) ... to 2n factors = 2²ⁿ

Lukybear · Jan 21, 2010

Koroneos said:
$(a)\,\,(a+b+c)^3\\=a^3+b^3+c^3+3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b)+6abc\\=a^3+b^3+c^3+3(ab(a+b+c)+bc(a+b+c)+ac(a+b+c))-3abc\\=a^3+b^3+c^3+3(a+b+c)(ab+bc+ac)-3abc\\\\\therefore a^3+b^3+c^3-3abc\\=(a+b+c)^3-3(a+b+c)(ab+bc+ac)\\=(a+b+c)((a+b+c)^2-3(ab+bc+ac))\\=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$

$(b)\,\,a^2+b^2\geq 2ab\\b^2+c^2\geq 2bc\\a^2+c^2\geq 2ac\\\therefore 2(a^2+b^2+c^2)\geq 2(ab+bc+ac)\\a^2+b^2+c^2-ab-bc-ac\geq 0\\\\a+b+c\geq 0($assuming\,\,a,b,c>0)\\\therefore (a+b+c)(a^2+b^2+c^2-ab-bc-ac)\geq 0\\a^3+b^3+c^3-3abc\geq 0\\\frac{a^3+b^3+c^3}{3}\geq abc\\sub\,\,a\rightarrow \sqrt[3]{x},\,\,b\rightarrow \sqrt[3]{y},c\rightarrow \sqrt[3]{z}\\\frac{x+y+z}{3}\geq \sqrt[3]{xyz}$

6 Last step where it says, a+b+c>0, and it assumes it so. But wouldnt that mean solution only applies to positive integers?

The Nomad · Jan 21, 2010

Lukybear said:
6 Last step where it says, a+b+c>0, and it assumes it so. But wouldnt that mean solution only applies to positive integers?

The inequality only applies for positive integers anyway.

Cazic · Feb 12, 2010

Show that

$\int_{0}^{2\pi} \frac{2}{3-2\sqrt{2}\cos\theta} \, d\theta = 4\pi$

untouchablecuz · Feb 12, 2010

Cazic said:
Show that

$\int_{0}^{2\pi} \frac{2}{3-2\sqrt{2}\cos\theta} \, d\theta = 4\pi$

$\\$Let $\theta=2 \tan ^{-1} t\Rightarrow d\theta=\frac{2}{1+t^2}dt\\ \int \frac{2}{3-2\sqrt{2}\cos\theta} d\theta \\=\int\frac{2}{3-2\sqrt{2}\times\frac{1-t^2}{1+t^2}} \, \frac{2}{1+t^2}dt\\ = 4\int\frac{dt}{(3-2\sqrt2)+t^2(3+2\sqrt2)}\\ = \frac{4}{3+2\sqrt2}\int\frac{dt}{\frac{3-2\sqrt2}{3+2\sqrt2}+t^2}\\ = \frac{4}{3+2\sqrt2}\int\frac{dt}{(3-2\sqrt2)^2+t^2}\\ = \frac{4}{3+2\sqrt2}\left[\frac{1}{3-2\sqrt2}\tan ^{-1} \frac{t}{3+2\sqrt2}\right]+C\\ = 4\left[\tan ^{-1} \frac{\tan \frac{\theta}{2}}{3+2\sqrt2}\right]+C$

$\\\int_{0}^{2\pi} \frac{2}{3-2\sqrt{2}\cos\theta} d\theta \\=4\left[\tan ^{-1} \frac{\tan \frac{\theta}{2}}{3+2\sqrt2}\right]_0^a+4\left[\tan ^{-1} \frac{\tan \frac{\theta}{2}}{3+2\sqrt2}\right]_a^{2\pi}\\=8\tan ^{-1} \frac{\tan \frac{a}{2}}{3+2\sqrt2}\\$As $a\to \pi^{\pm}, \tan ^{-1} \left(\frac{\tan \frac{a}{2}}{3+2\sqrt2}\right)\to \frac{\pi}{2}\\\therefore \int_{0}^{2\pi} \frac{2}{3-2\sqrt{2}\cos\theta} d\theta=4\pi$

4 Unit Revising Marathon HSC '10 (1 Viewer)

Member

...

Member

...

Administrator

...

Administrator

...

Account Closed

The A-Team

...

Administrator

...

The A-Team

...

Member

Active Member

Member

Member

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)