HSC Mathematics Marathon (4 Viewers)

hup · Sep 1, 2011

sum

$\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+....$

thoth1 · Sep 1, 2011

can u post sln plz.

hup · Sep 1, 2011

thoth1 said:
can u post sln plz.

sum $x+x^2 +x^3+....$
then sum $1+2x+3x^2+....$
result should come in a few more steps

thoth1 · Sep 1, 2011

try this question:
$I_{n}=\int_{0}^{\pi }\frac{1-\cos nx}{1-\cos x}dx$
and $I_{3}=3\pi$
show that:
$\int_{0}^{\frac{\pi }{2}}\frac{\sin ^{2}3\theta }{\sin ^{2}\theta }d\theta = \frac{3\pi }{2}$

ohexploitable · Sep 1, 2011

hup said:
sum

$\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+....$

$x=\frac{1}{5}+\frac{2}{5^2}+...+\frac{n-1}{5^{n-1}}+\frac{n}{5^n}$
$5x=1+\frac{2}{5}+\frac{3}{5^2}...+\frac{n}{5^{n-1}}$

-----

$5x-x=1+\frac{2}{5}-\frac{1}{5}+\frac{3-2}{5^2}+\frac{4-3}{5^3}+...+\frac{n-n+1}{5^{n-1}}-\frac{n}{5^n}$
$4x=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{n-1}}-\frac{n}{5^n}$

as n goes to infinity,

$4x=1+\frac{1}{4}-0$
$x=\frac{5}{16}$

hup · Sep 2, 2011

thoth1 said:
try this question:
$I_{n}=\int_{0}^{\pi }\frac{1-\cos nx}{1-\cos x}dx$
and $I_{3}=3\pi$
show that:
$\int_{0}^{\frac{\pi }{2}}\frac{\sin ^{2}3\theta }{\sin ^{2}\theta }d\theta = \frac{3\pi }{2}$

$I_{n}=\int_{0}^{\pi }\frac{1-\cos nx}{1-\cos x} \\ I_{n}=\int_{0}^{\pi }\frac{1-(1-2\sin ^{2}\frac{nx}2{})}{1-(1-2\sin ^{2}\frac{x}{2})} \\ =\int_{0}^{\pi }\frac{\sin ^{2}\frac{nx}{2}}{\sin ^{2}\frac{x}{2}} \\ Let\ \theta =\frac{x}{2}$

result follows not good at writing it

thoth1 · Sep 2, 2011

try this one:

For the series S(x) = $1+2x+3x^{2}+....+(n+1)x^{n}$

find (1-x)S(x) and hence find S(x).

this question and the question solved above were in my 4U trial.

Drongoski · Sep 2, 2011

thoth1 said:
try this one:

For the series S(x) = $1+2x+3x^{2}+....+(n+1)x^{n}$

find (1-x)S(x) and hence find S(x).

$S(x) = 1 + 2x + 3x^2 + 4x^3 + \dots + nx^{n-1} + (n+1)x^n$

$xS(x) = x + 2x^2 + 3x^3 + \dots + nx^n + (n+1)x^{n+1}$

$\therefore S(x) - xS(x) = 1+x+x^2+x^3 + \dots + x^n - (n+1)x^{n+1}$

$\therefore (1-x)S(x) = \frac {1(x^{n+1} - 1)}{x-1} - (n+1) x^{n+1}$

$\therefore S(x) = \frac {1-x^{n+1}}{(1-x)^2} - \frac {(n+1)x^{n+1}}{1-x}$

thoth1 · Sep 2, 2011

Drongoski said:
$S(x) = 1 + 2x + 3x^2 + 4x^3 + \dots + nx^{n-1} + (n+1)x^n$

$xS(x) = x + 2x^2 + 3x^3 + \dots + nx^n + (n+1)x^{n+1}$

$\therefore S(x) - xS(x) = 1+x+x^2+x^3 + \dots + x^n - (n+1)x^{n+1}$

$\therefore (1-x)S(x) = \frac {1(x^{n+1} - 1)}{x-1} - (n+1) x^{n+1}$

$\therefore S(x) = \frac {1-x^{n+1}}{(1-x)^2} - \frac {(n+1)x^{n+1}}{1-x}$

thats right.

Drongoski · Sep 2, 2011

thoth1 said:
thats right.

Oh thanks. Was wondering what answer was expected.

thoth1 · Sep 3, 2011

Drongoski said:
Oh thanks. Was wondering what answer was expected.

yeh the only difference was in the solutions the school gave us they wrote the answer as one fraction.

ZombieApocalypse · Sep 3, 2011

Drongoski said:
ohex is a very clever boy.

Yes he certainly is <333333

OE FAN4LYF <3 <4 <5 <6

ohexploitable · Sep 3, 2011

ZombieApocalypse said:
Yes he certainly is <333333

OE FAN4LYF <3 <4 <5 <6

go to ur maffs coordinator/head teacher and ask to be put back in the 4u class

plz

ZombieApocalypse · Sep 3, 2011

ohexploitable said:
go to ur maffs coordinator/head teacher and ask to be put back in the 4u class

plz

I think if I asked nicely enough, they'd consider letting me come back, as classes haven't "officially" start.

But 4u maffs is too hard for me

I'm going to suffer and probably fail+drag all my other subjects down with 4u.

The only girl who's doing 4u in our school this year who is 905348439058035 times better than me dropped it after trials, so if she can't do it, I have no hope

SpiralFlex · Sep 3, 2011

ZombieApocalypse said:
I think if I asked nicely enough, they'd consider letting me come back, as classes haven't "officially" start.

But 4u maffs is too hard for me
I'm going to suffer and probably fail+drag all my other subjects down with 4u.

The only girl who's doing 4u in our school this year who is 905348439058035 times better than me dropped it after trials, so if she can't do it, I have no hope

When there is a will, there is a way. What's so hard about Complex numbers?

cutemouse · Sep 4, 2011

SpiralFlex said:
What's so hard about Complex numbers?

[maths]i^2=-1[/maths]

Wtf, this shiz just got real...

ZombieApocalypse · Sep 4, 2011

SpiralFlex said:
When there is a will, there is a way. What's so hard about Complex numbers?

Complex didn't scare me from 4u. It's the other topics. I looked through the book and I don't get it

No point doing well in one topic and then failing the 7 others when you know you could do well in all topics in 2u

SpiralFlex · Sep 4, 2011

ZombieApocalypse said:
Complex didn't scare me from 4u. It's the other topics. I looked through the book and I don't get it
No point doing well in one topic and then failing the 7 others when you know you could do well in all topics in 2u

You haven't even learnt it yet...How would you know?

ZombieApocalypse · Sep 4, 2011

SpiralFlex said:
You haven't even learnt it yet...How would you know?

But in 2u, I could just read the textbook and get it. I can't do that for 4u. I'll be completely reliant on the teacher, and I don't like that. I like to read things myself and the understand it... if you get me.

Besides, i'm not smart enough for 4u.

SpiralFlex · Sep 4, 2011

ZombieApocalypse said:
But in 2u, I could just read the textbook and get it. I can't do that for 4u. I'll be completely reliant on the teacher, and I don't like that. I like to read things myself and the understand it... if you get me.

Besides, i'm not smart enough for 4u.

...

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