HSC 2016 MX2 Marathon (archive) (5 Viewers)

leehuan · Nov 15, 2015

Re: HSC 2016 4U Marathon

Is that second part really manageable by current 2016ers?

InteGrand · Nov 15, 2015

Re: HSC 2016 4U Marathon

leehuan said:
Is that second part really manageable by current 2016ers?

Yes, just need to provide a counter-example.

leehuan · Nov 15, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
Yes, just need to provide a counter-example.

Oh right.
But how would you go about proving a general case?

Also, in the context of maths what do we regard as convex and concave

InteGrand · Nov 15, 2015

Re: HSC 2016 4U Marathon

leehuan said:
Oh right.
But how would you go about proving a general case?

Also, in the context of maths what do we regard as convex and concave

$Convex means ``concave up'' and concave means ``concave down''. You can find formal definitions on the Wikipedia pages for them:$

• https://en.wikipedia.org/wiki/Convex_function

• https://en.wikipedia.org/wiki/Concave_function

InteGrand · Nov 15, 2015

Re: HSC 2016 4U Marathon

leehuan said:
Oh right.
But how would you go about proving a general case?

To disprove the statement's converse, one only needs a counter-example. What do you mean general case?

Paradoxica · Nov 15, 2015

Re: HSC 2016 4U Marathon

leehuan said:
Oh right.
But how would you go about proving a general case?

Also, in the context of maths what do we regard as convex and concave

There is no general case for the converse. All you have to show is that the converse does not hold true in general, as the converse statement is already generalised to all convex exponentials.

i.e the converse states:

$If $e^{f(x)}$ is convex, then $f(x)$ is convex$

which is not true, as counterexamples exist.

leehuan · Nov 15, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
$Convex means ``concave up'' and concave means ``concave down''. You can find formal definitions on the Wikipedia pages for them:$

• https://en.wikipedia.org/wiki/Convex_function

• https://en.wikipedia.org/wiki/Concave_function

Great, that's what I was thinking but I needed confirmation, since for some reason during the HSC I never came across that language.
------------------------------
What I meant was, if you had g(x)=e^(f(x)) and you knew g"(x)>0 would it be possible to then work backwards and realise "but f"(x) isn't always > 0"

If the counterexample answer will suffice then I won't comment further.

glittergal96 · Nov 15, 2015

Re: HSC 2016 4U Marathon

i) Let g=e^f

g(tx+(1-t)y) =< e^(tf(x)+(1-t)f(y)) (convexity of f)
= (e^f(x))^t(e^f(y))^(1-t)
= g(x)^tg(y)^(1-t)
=< tg(x)+(1-t)g(t) (weighted AM-GM in two variables *)

for all 0 =< t =< 1.

ii) f(x)=log(x) is a counterexample, it is strictly concave as it's second derivative is negative, and e^f(x)=x is trivially convex.

Proof of weighted AM-GM in two variables:

log is a concave function because its second derivative is negative.

This implies that
tlog(x)+(1-t)log(y)=< log(tx+(1-t)y)
x^t y^(1-t) =< tx + (1-t)y (exponentiating both sides.)

Note that the proof of i) makes no assumption on the regularity of f. Convex functions need not be very smooth!

glittergal96 · Nov 15, 2015

Re: HSC 2016 4U Marathon

A followup question based on the second derivative test for convexity:

Given that f''(x)>0 for all a =< x =< b, show that f is convex on the interval [a,b].

That is, show that f(tx+(1-t)y) =< tf(x)+(1-t)f(y) for all x and y in [a,b], and all t in [0,1].

(This result is of course common sense, but should really be proved rigorously. Thankfully it is not too hard to do so.)

Paradoxica · Nov 15, 2015

Re: HSC 2016 4U Marathon

glittergal96 said:
i) Let g=e^f

g(tx+(1-t)y) =< e^(tf(x)+(1-t)f(y)) (convexity of f)
= (e^f(x))^t(e^f(y))^(1-t)
= g(x)^tg(y)^(1-t)
=< tg(x)+(1-t)g(t) (weighted AM-GM in two variables *)

for all 0 =< t =< 1.

ii) f(x)=log(x) is a counterexample, it is strictly concave as it's second derivative is negative, and e^f(x)=x is trivially convex.

Proof of weighted AM-GM in two variables:

log is a concave function because its second derivative is negative.

This implies that
tlog(x)+(1-t)log(y)=< log(tx+(1-t)y)
x^t y^(1-t) =< tx + (1-t)y (exponentiating both sides.)

Note that the proof of i) makes no assumption on the regularity of f. Convex functions need not be very smooth!

Wrong, linear functions are technically ambiguous in their convexity and concavity. A more rigorous counter-example would be f(x) = 2log(x)

glittergal96 · Nov 15, 2015

Re: HSC 2016 4U Marathon

Paradoxica said:
Wrong, linear functions are technically ambiguous in their convexity and concavity. A more rigorous counter-example would be f(x) = 2log(x)

It is literally a matter of definition. My use of the word "convex" is exactly that in https://en.wikipedia.org/wiki/Convex_function (as well as most textbooks I have ever seen it in).

Using this terminology, x is a convex function, and log(x) is a strictly concave function. Strictly concave functions are not convex. Hence log(x) is a counterexample to your claim. If you meant something else by your use of the word "convex" you should have made that clear in the original question.

Sy123 · Nov 15, 2015

Re: HSC 2016 4U Marathon

glittergal96 said:
A followup question based on the second derivative test for convexity:

Given that f''(x)>0 for all a =< x =< b, show that f is convex on the interval [a,b].

That is, show that f(tx+(1-t)y) =< tf(x)+(1-t)f(y) for all x and y in [a,b], and all t in [0,1].

(This result is of course common sense, but should really be proved rigorously. Thankfully it is not too hard to do so.)

$\\ $Given$ \ f''(x) > 0 \ $for all$ \ a \leq x \leq b$

$\\ $Then$ \ f'(x) \ $is a monotone increasing function on$ \ a \leq x \leq b$

$\\ $Then, for$ \ a \leq x, p \leq b \ $we have$ \ x \geq p \Rightarrow f'(x) \geq f'(p) $, and we have$ \ x \leq p \Rightarrow f'(x) \leq f'(p)$

$\\ $Considering the function$ \ g_p(x) = f(x) - xf'(p) \ $so that$ \ g_p'(x) = f'(x) - f'(p)$

$\\ $Then$ \ x \geq p \Rightarrow \ g_p'(x) \geq 0 \Rightarrow \ g_p \ $is monotone increasing on$ \ p \leq x \leq b \\ \Rightarrow g_p(x) \geq g_p(p) \Rightarrow f(x) - xf'(p) \geq f(p) - pf'(p) \Rightarrow f(x) \geq f(p) + (x-p)f'(p)$

$\\ $On the other case$ \ x \leq p \Rightarrow \ g_p'(x) \leq 0 \Rightarrow \ g_p \ $is monotone decreasing on$ \ a \leq x \leq p \\ \Rightarrow g_p(x) \geq g_p(x) \Rightarrow \ f(x) \geq f(p) + (x-p)f'(p)$

$\\ $Thus, independent of$ \ p \ $we have$ \ f(x) \geq f(p) + (x-p)f'(p)$

$\\ $So,$ \ tf(x) + (1-t)f(y) \geq f(p) + f'(p)( tx + (1-t)y - p)$

$\\ $Since$ \ a \leq p \leq b \ $set$ \ p = tx + (1-t)y \ $which lies in the interval, as$ \ p \ $lies in the interval between$ \ x \ $and$ \ y \ $which is itself in the interval between$ \ a \ $and$ \ b$

$\\ \Rightarrow \ tf(x) + (1-t)f(y) \geq f(tx + (1-t)y)$

Sy123 · Nov 15, 2015

Re: HSC 2016 4U Marathon

Bringing it back to more familiar ground:

$\\ $By plotting a rhombus onto the complex plane and using complex numbers, prove that their diagonals are perpendicular to each other$$

Drsoccerball · Nov 19, 2015

Re: HSC 2016 4U Marathon

$(i)After serving his time in prison, Integrand decides to quit university and join the delivery company known as UPS. On that specific day Integrand’s boss tells him, ''I need you to unload this heavy and fragile box.'' So Integrand being the genius he is, comes up with a mathematical way of unloading the box.$

$Integrands height at which he holds the box is R meters above a tire.$

$$ His boss looked at Integrands set up. He has one big tire which had a thickness of W meters. It is Given that air resistance is $ mv^2k $ and the resistance of the tire is $ k^3mv^2.$

$$ By taking the floor as the origin prove $ :$

$e^w(\frac{ge^{2k^3+2k}}{k^3+k})+ \frac{g}{ke^{4kw}} +(\frac{g}{k}-\frac{g}{k^3 +k})=0$

$(ii) $ Assuming that the max speed the box can reach before shattering is 100 m/s prove that the minimum thickness of W is: $$

$W= \frac{ln(\frac{k}{10000k-g})}{4k}$

$(iii) $ Hence or otherwise calculate the value for k at this thickness $$

InteGrand · Nov 19, 2015

Re: HSC 2016 4U Marathon

Drsoccerball said:
$$ By taking the floor as the origin prove $ :$

$e^w(\frac{ge^{2k^3+2k}}{k^3+k})+ \frac{g}{ke^{4kw}} +(\frac{g}{k}-\frac{g}{k^3 +k}) = 0$

$How can this be 0? Since $g,k>0$ and $\frac{g}{k}> \frac{g}{k^3 +k}$, the L.H.S. is positive. The L.H.S. is dimensionally dodgy too.$

Drsoccerball · Nov 19, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
$How can this be 0? Since $g,k>0$ and $\frac{g}{k}> \frac{g}{k^3 +k}$, the L.H.S. is positive. The L.H.S. is dimensionally dodgy too.$

Why can g be negative?

InteGrand · Nov 19, 2015

Re: HSC 2016 4U Marathon

Drsoccerball said:
Why can g be negative?

$Conventionally $g$ refers to the positive quantity, like 9.8 m/s$^2$.$

Drsoccerball · Nov 19, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
$Conventionally $g$ refers to the positive quantity, like 9.8 m/s$^2$.$

Im pretty sure I didnt make a mistake in my working tho...

porcupinetree · Nov 23, 2015

Re: HSC 2016 4U Marathon

Next Q:

$\textrm{A solid has a base that is the region enclosed by the curve y=lnx, the x axis and the line x=3.}$
$\textrm{Cross-sections taken perpendicular to the x axis are regular hexagons, with one side in the base.}$
$\textrm{Find the volume of the solid.}$

Ekman · Nov 23, 2015

Re: HSC 2016 4U Marathon

InteGrand said:
$Conventionally $g$ refers to the positive quantity, like 9.8 m/s$^2$.$

Key word there is conventionally, however this isn't a conventional question, since when did integrand go to prison, come out, go to uni to only quit later and get hired by UPS, and still have time to answer questions on bos

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