IB Maths Tutor - Sydney (2 Viewers)

perfect6768 · Apr 19, 2013

bump for great results

Drongoski · Apr 21, 2013

perfect6768 said:
bump for great results

I'm glad you think highly of my estimated ATAR of > 64.95. Thank you.

nerdasdasd · Apr 24, 2013

Magical Kebab said:
What's IB?

Bumppp.

It is a way of getting into Uni, but it is the standard across global universities (you can apply everywhere in the world ).

It is still just a number .

Capt Rifle · May 10, 2013

bump

Very diverse tuition...

nerdasdasd · May 10, 2013

Bump ^^

Drongoski · May 12, 2013

Thank you Capt & Nerd. Very kind of you both.

Drongoski · Jun 2, 2013

Capt Rifle said:
bump

Very diverse tuition...

haha - have to diversify to survive, with so much competition in the HSC tuition market from freshly-minted hotshots each year far abler than I and often far far cheaper than I. It is a bruisingly tough business!

Drongoski · Jun 17, 2013

SpiralFlex · Jul 17, 2013

Bump!

Drongoski · Jul 21, 2013

Ah thanks Spiral - you are always so kind.

Streetboy · Jul 29, 2013

heay man, I trials next week, how much do you charge.

PM, to let me know, keep in mind that I have financial problems,

can you help me!

hsclover · Aug 27, 2013

bump for a great tutor

hsclover · Aug 29, 2013

bump

Drongoski · Sep 3, 2013

Oh thanks for the bump, hsclover.

nerdasdasd · Sep 9, 2013

Bump

.

Drongoski · Sep 28, 2013

IB Maths (Higher Level)
Paper 3 - Series & Differential Equations
May 2008

Q5(b): Determine whether the series is convergent or divergent:

$\sum ^{\infty}_{n=0} (\root \3 \of {n^3+1}- n)$

Solution (using difference of 2 cubes)

$0 < \root \3 \of {n^3+1} - n = \frac { {(\root \3 \of {n^3 + 1} - n) }[(n^3 + 1)^{\frac {2}{3}} + n \root \3 \of {n^3 + 1} + n^2]} {[(n^3 + 1)^{\frac{2}{3}} + n \root \3 \of {n^3 + 1} + n^2]} \\ \\$
$\\ = \frac {(n^3 +1) - n^3}{ \cdots} = \frac {1}{\cdots} < \frac {1}{n^2} $ and $ \sum ^{\infty}_{n=0} \frac {1}{n^2} $ converges (p-series)$$

Therefore, by the Comparison Test, the given infinite series converges.

funnytomato · Sep 28, 2013

Drongoski said:
IB Maths (Higher Level)
Paper 3 - Series & Differential Equations
May 2008

Q5(b): Determine whether the series is convergent or divergent:

$\sum ^{\infty}_{n=0} (\root \3 \of {n^3+1}- n)$

Solution (using difference of 2 cubes)

$0 < \root \3 \of {n^3+1} - n = \frac {(\root \3 \of {n^3 + 1} - n)[(n^3 + 1)^{\frac {2}{3}} + n \root \3 \of {n^3 + 1} + n^2]}{[(n^3 + 1)^{\frac{2}{3}} + n \root \3 \of {n^3 + 1} + n^2]} \\ \\ \\ = \frac {(n^3 +1) - n^3}{ \cdots} = \frac {1}{\cdots} < \frac {1}{n^2} $ and $ \sum ^{\infty}_{n=0} \frac {1}{n^2} $ converges $$

Therefore, by the Comparison Test, the given infinite series converges.

what do you think would be the major difference between IB and HSC for maths?

Drongoski · Sep 29, 2013

The IB (Higher Level) has a Options part: You choose 1. Students do the one the school opts for. The Options is essentially a 1st Year Uni Topic. Quite impressive & demanding. Questions asked are designed to really test your understanding.

obliviousninja · Oct 25, 2013

Thanks Drongoski for the paper clip trick, although I was pretty stupid and didn't recognise the circle geometry proof...sigh.

obliviousninja · Oct 25, 2013

Drongoski said:
Ah! I hope you at least drew the 2 circles & get some marks.

Proving collinearity is often tricky. You must first assume QT and TP are not parts of the same straight line.

Nup, I was pretty stupid, I went to prove similar triangles. I should at least be able to get an E4 though.

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