HSC 2016 Maths Marathon (archive) (2 Viewers)

Paradoxica · Apr 13, 2016

Re: HSC 2016 2U Marathon

Is TeX broken?

I don't see any TeX posts right now.

Nailgun · Apr 13, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
Rewritten in 2U style:

$\noindent Let $f(t)$ be a function with domain $0\leq t \leq 1$. Let $a<b$ be fixed real numbers. Let $f(t) = a(1-t) + bt$ for $0\leq t \leq 1$. Find the range of $f(t)$.$

lel rewritten in 2U style, yet shorter than using the notation

InteGrand · Apr 13, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
lel rewritten in 2U style, yet shorter than using the notation

Shorter but less accurate.

parad0xica · Apr 13, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
Shorter but less accurate.

Sadly... slight derail but what's the reason HSC doesn't want to teach students the beautiful maths notations and proper way of defining functions?

Paradoxica · Apr 13, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
Shorter but less accurate.

pfft technicalities

rigour is what turns a lot of people off from mathematics...

so many talented people out there who were/are shut off by the "necessary" technicalities of mathematics. (bookwork can always be done afterwards, the rough outline of the logic is what matters most, is it not?)

Yes I am an Intuitionist. No, I will not change my mind about that.

Nailgun · Apr 14, 2016

Re: HSC 2016 2U Marathon

InteGrand said:
Rewritten in more 2U style:

$\noindent Let $f(t)$ be a function with domain $0\leq t \leq 1$. Let $a<b$ be fixed real numbers. Let $f(t)$ be defined as $f(t) = a(1-t) + bt$ for $0\leq t \leq 1$. Find the range of $f(t)$.$

$$\noindent Observe that $f(t)$ is a linear function of $t$,\\as it can be rearranged to $f\left( t \right) =(b-a)t+a$\\f\left( 0 \right) =a$ $ and $ $f\left( 1 \right) =b$\\$Hence, it is evident by inspection that in the domain of $0\leq t \leq 1$ $f(t)$ is a linear function from [$(t,f(t)$] $(0,a)$ and $(1,b)$ with a positive gradient as $a<b$. Therefore the range is $a\le f(t)\le b$.$

leehuan · Apr 14, 2016

Re: HSC 2016 2U Marathon

Paradoxica said:
pfft technicalities

rigour is what turns a lot of people off from mathematics...

so many talented people out there who were/are shut off by the "necessary" technicalities of mathematics. (bookwork can always be done afterwards, the rough outline of the logic is what matters most, is it not?)

Yes I am an Intuitionist. No, I will not change my mind about that.

Because the average student will get lost in the symbols

Nailgun · Apr 15, 2016

Re: HSC 2016 2U Marathon

leehuan said:
Because the average student will get lost in the symbols

lol you realise he's in favor of less technical symbols?

leehuan · Apr 16, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
lol you realise he's in favor of less technical symbols?

Oh, I quoted the wrong thing. My bad.

This was what I was trying to quote

parad0xica said:
Sadly... slight derail but what's the reason HSC doesn't want to teach students the beautiful maths notations and proper way of defining functions?

si2136 · Apr 20, 2016

Re: HSC 2016 2U Marathon

Paradoxica said:
$xe^x + \mathcal{C}$

I don't get it, shouldn't it be (e^x+1)/x

InteGrand · Apr 20, 2016

Re: HSC 2016 2U Marathon

si2136 said:
I don't get it, shouldn't it be (e^x+1)/x

Paradoxica intended it as a mathematical joke.

The actual answer is of course e^x + c.

leehuan · Apr 23, 2016

Re: HSC 2016 2U Marathon

Instructive question

$$Use the definition of the derivative to find $\frac{d}{dx}{x}^{\frac{1}{3}}$

Paradoxica · Apr 23, 2016

Re: HSC 2016 2U Marathon

leehuan said:
Instructive question

$$Use the definition of the derivative to find $\frac{d}{dx}{x}^{\frac{1}{3}}$

"THE" definition?

There are two definitions of the derivative.

leehuan · Apr 23, 2016

Re: HSC 2016 2U Marathon

Paradoxica said:
"THE" definition?

There are two definitions of the derivative.

...and what is this second definition?

Nailgun · Apr 23, 2016

Re: HSC 2016 2U Marathon

leehuan said:
Instructive question

$$Use the definition of the derivative to find $\frac{d}{dx}{x}^{\frac{1}{3}}$

im confused how is this a question

Paradoxica · Apr 23, 2016

Re: HSC 2016 2U Marathon

leehuan said:
...and what is this second definition?

Symmetric Derivative

leehuan · Apr 23, 2016

Re: HSC 2016 2U Marathon

Paradoxica said:
Symmetric Derivative

Ok, I meant the ordinary derivative if we need to be pedantic.

Nailgun · Apr 23, 2016

Re: HSC 2016 2U Marathon

leehuan said:
Ok, I meant the ordinary derivative if we need to be pedantic.

I still don't understand the question

isnt it just 1/3x^(-2/3)

leehuan · Apr 23, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
I still don't understand the question

isnt it just 1/3x^(-2/3)

The definition is lim h->0 ( f(x+h)-f(x) ) /h

But I haven't done the question yet to say for sure if it works

Nailgun · Apr 23, 2016

Re: HSC 2016 2U Marathon

leehuan said:
The definition is lim h->0 ( f(x+h)-f(x) ) /h

But I haven't done the question yet to say for sure if it works

how do you simplify [cuberoot(x+h)-cuberoot(x)]/h lelel

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