HSC 2015 MX2 Marathon ADVANCED (archive) (3 Viewers)

simpleetal · Jul 28, 2015

Re: HSC 2015 4U Marathon - Advanced Level

$If f(1)=2,$ then we get by letting y=1, that$ f(x-1)+2(1-f(x-1))-f(1-x)=0 \rightarrow 2=f(x-1)+f(1-x). $ But from equation A, we know that $f(1-x)=1-f(x-2). So$ f(x-1)-f(x-2)=1, which means that the gradient of the function is always 1, meaning that it is of the form x+b. But, f(1-x)=1-f(x-2), so 1-x+b=1-(x-2)-b, so 2b=2, and b=1. So f(x)=x+1. \\If f(1)=0, then f(x)=-f(x+1)+1. (from original equation). f(x)+f(x+1)=1. f(x)-f(-x)=0 (from equation A). So f(x)=f(-x). So,letting y=x, $f(x^2)=f(x)^2-f(2x) +1.$ But, letting y=-x and using $f(u)=f(-u), f(x^2)=f(x)^2-f(0)+1.$ So f(2x)=f(0)=1. But letting x=-1/2, f(-1)=0 which is not 1, so f(-1)=0 leads to contradiction.$ Hence, unless I’ve made sillies, (or conceptual errors) I think the only solutions are 1, or 1+x.

glittergal96 · Jul 28, 2015

Re: HSC 2015 4U Marathon - Advanced Level

You are a lot of the way there, these are most of the ideas I was looking for. But I don't think you have completed the f(1)=2 case.

I don't think f(x+1)-f(x)=1 really implies that we are a linear polynomial.
-What do you mean by "gradient" when talking about an arbitrary function?
-Why is this notion of gradient identically 1 for this function?
-Why should this fact imply f is a linear polynomial?

If gradient just means the quantity f(x+1)-f(x), this is certainly 1 everywhere, but this alone does NOT imply that we are a linear polynomial, or even continuous! We could choose an f with arbitrary values in the interval [0,1), and use f(x+1)-f(x)=1 to uniquely define a solution (to the equation f(x+1)-f(x)=1) on the whole real (well we only care about rationals, but this is no different) line.

glittergal96 · Jul 31, 2015

Re: HSC 2015 4U Marathon - Advanced Level

It's been a few days, so I will finish off your solution so we can move to something new

. You were very close, and my method started out essentially the same as yours.

Since we have f(x+1)=f(x)+1, we have f(x+m)=f(x)+m for all integers m.

Since we also have f(0)=1, we know that f(m)=m+1 for all integers m.

Now for a nonzero integer n, we have

f(n.(m/n))=f(n)f(m/n)-f(n+(m/n))+1

=> m+1=f(m)=(n+1)f(m/n)-(n+f(m/n))+1.

solving this equation for the "unknown" f(m/n) gives:

f(m/n)=(m+n)/n=(m/n)+1.

So f(x)=x+1 for all rational x.

(It is important we do things this way, because I suspect it might not be the case that every nonconstant REAL solution to the functional equation is equal to x+1).

simpleetal · Jul 31, 2015

Re: HSC 2015 4U Marathon - Advanced Level

glittergal96 said:
It's been a few days, so I will finish off your solution so we can move to something new . You were very close, and my method started out essentially the same as yours.

Since we have f(x+1)=f(x)+1, we have f(x+m)=f(x)+m for all integers m.

Since we also have f(0)=1, we know that f(m)=m+1 for all integers m.

Now for a nonzero integer n, we have

f(n.(m/n))=f(n)f(m/n)-f(n+(m/n))+1

=> m+1=f(m)=(n+1)f(m/n)-(n+f(m/n))+1.

solving this equation for the "unknown" f(m/n) gives:

f(m/n)=(m+n)/n=(m/n)+1.

So f(x)=x+1 for all rational x.

(It is important we do things this way, because I suspect it might not be the case that every nonconstant REAL solution to the functional equation is equal to x+1).

sorry looks like I didnt check BOS in time to read your previous post. (this was mainly because I had a major japanese assignment to work on, luckily it's all done now). But yes I do see the error that I previously made now that Ive read your post.

dan964 · Jul 31, 2015

Re: HSC 2015 4U Marathon - Advanced Level

glittergal96 said:
It's been a few days, so I will finish off your solution so we can move to something new . You were very close, and my method started out essentially the same as yours.

Since we have f(x+1)=f(x)+1, we have f(x+m)=f(x)+m for all integers m.

Since we also have f(0)=1, we know that f(m)=m+1 for all integers m.

Now for a nonzero integer n, we have

f(n.(m/n))=f(n)f(m/n)-f(n+(m/n))+1 (**)

=> m+1=f(m)=(n+1)f(m/n)-(n+f(m/n))+1.

solving this equation for the "unknown" f(m/n) gives:

f(m/n)=(m+n)/n=(m/n)+1.

So f(x)=x+1 for all rational x.

(It is important we do things this way, because I suspect it might not be the case that every nonconstant REAL solution to the functional equation is equal to x+1).

where did the RHS come from at (**)
nvm, it was substitution into the original question, which is on the previous page.

shervos · Aug 14, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Suppose f(x) is a real valued function such that f(xy)+f(y-x)>=f(x+y). Prove that f(x)>=0

glittergal96 · Aug 15, 2015

Re: HSC 2015 4U Marathon - Advanced Level

shervos said:
Suppose f(x) is a real valued function such that f(xy)+f(y-x)>=f(x+y). Prove that f(x)>=0

$Take $y=\frac{u+v}{2},x=\frac{u-v}{2}$ (it is easy to see this is a bijection on $\mathbb{R}^2$). \\ \\ The rearranged inequality then becomes\\ \\ $f(v)\geq f(u)-f\left(\frac{u^2-v^2}{4}\right)\quad \textrm{ for all }u,v\in\mathbb{R}.$\\ \\ But for any fixed $v\in\mathbb{R}$, we can choose $u$ such that $u=\frac{u^2-v^2}{4}$ as the discriminant of this quadratic equation is positive.\\ \\ Choosing this $u$ makes the lower bound in the inequality vanish, and we are left with $f(v)\geq 0$. As $v$ is arbitrary, we are done.$

glittergal96 · Aug 30, 2015

Re: HSC 2015 4U Marathon - Advanced Level

If $f:[0,1]\rightarrow \mathbb{R}$ is a non-negative function such that

$f(1)=1$

and

$f(x+y)\geq f(x)+f(y)\quad \textrm{for }0\leq x,y\textrm{ and }x+y\leq 1$

we say that f is a sexy function.

Find the smallest real $\alpha$ such that

$f(x)\leq \alpha x$

for all $x$ and all sexy functions $f$ .

shervos · Aug 30, 2015

Re: HSC 2015 4U Marathon - Advanced Level

What does the first line mean? Also, I think it's a good idea if symbols which are not seen in HSC exams are explained, unless they are abused here

InteGrand · Aug 30, 2015

Re: HSC 2015 4U Marathon - Advanced Level

shervos said:
What does the first line mean? Also, I think it's a good idea if symbols which are not seen in HSC exams are explained, unless they are abused here

$The first line means that $f$ is a function which has domain $[0,1]$ (i.e. the set of real numbers between $0$ and $1$ inclusive), and codomain $\mathbb{R}$ (which means that for any $x$ in the domain, $f(x)\in \mathbb{R}$).$

dan964 · Aug 30, 2015

Re: HSC 2015 4U Marathon - Advanced Level

InteGrand said:
$The first line means that $f$ is a function which has domain $[0,1]$ (i.e. the set of real numbers between $0$ and $1$ inclusive), and codomain $\mathbb{R}$ (which means that for any $x$ in the domain, $f(x)\in \mathbb{R}$).$

if it is non-negative function, am I wrong to say that y>0 or y=0 (i.e. y greater than equal to 0)
or by non-negative function are you referring to the gradient of the function?

InteGrand · Aug 30, 2015

Re: HSC 2015 4U Marathon - Advanced Level

dan964 said:
if it is non-negative function, am I wrong to say that y>0 or y=0 (i.e. y greater than equal to 0)
or by non-negative function are you referring to the gradient of the function?

$It (non-negative function) means that the function satisfies $f(x)\geq 0$ for all $x$ in the domain.$

dan964 · Aug 30, 2015

Re: HSC 2015 4U Marathon - Advanced Level

InteGrand said:
$It (non-negative function) means that the function satisfies $f(x)\geq 0$ for all $x$ in the domain.$

thanks

glittergal96 · Sep 4, 2015

Re: HSC 2015 4U Marathon - Advanced Level

glittergal96 said:
Let $f:[0,1]\rightarrow \mathbb{R}$ be a non-negative function such that

$f(1)=1$

and

$f(x+y)\geq f(x)+f(y)\quad \textrm{for }0\leq x,y\textrm{ and }x+y\leq 1.$

Find the smallest real $\alpha$ such that

$f(x)\leq \alpha x$

for all $x$ and all such $f$ .

It has been a while, so here are some hints. (I don't want to give it away just yet because I think this is a great/fun question!)

1. What must f(0) be?

2. Can you think of any relationships between convexity and the property f(x+y) >= f(x)+f(y)?
Recall that a function is said to be convex if every secant to it's graph lies above the graph itself.

3. Can you find any simple bounds for how small/large the solution alpha can be?

4. Can you then find functions which prove we cannot do better than the alpha you find in 3?

Sorry about being vague. If people don't make progress in a little more time then I will post my solution.

dan964 · Sep 4, 2015

Re: HSC 2015 4U Marathon - Advanced Level

for (1)
f(0) must be zero.
since $f(x) \geq 0$ for all x as it is non-negative.
and to satisfy the condition $f(x) \leq \alpha x$ , it cannot be anything greater than 0.
thus the only solution where both are true is when f(0)=0.

for (3)
putting x=1, f(x)=1;
gives the minimum value of alpha as 1.

but this does not take into account the equality, so its minimum value may be larger. this is just some starters.

InteGrand · Sep 4, 2015

Re: HSC 2015 4U Marathon - Advanced Level

$Well since $f(0)=0$ and $f(1)=1$, $\alpha$ would need to satisfy $\alpha \geq 1$ (otherwise $f(1)=1$ would exceed $\alpha \cdot 1=\alpha$, which we do not want), and we can achieve $\alpha=1$ by having $f(x)=x$ for all $x\in [0,1]$, and this function satisfies $f(x+y)\geq f(x)+f(y)$ (equality in fact). I'm not sure whether this is the only such function though, guessing it's not.$

glittergal96 · Sep 4, 2015

Re: HSC 2015 4U Marathon - Advanced Level

InteGrand said:
$Well since $f(0)=0$ and $f(1)=1$, $\alpha$ would need to satisfy $\alpha \geq 1$ (otherwise $f(1)=1$ would exceed $\alpha \cdot 1=\alpha$, which we do not want), and we can achieve $\alpha=1$ by having $f(x)=x$ for all $x\in [0,1]$, and this function satisfies $f(x+y)\geq f(x)+f(y)$ (equality in fact). I'm not sure whether this is the only such function though, guessing it's not.$

Yep, $\alpha \geq 1$ follows from considering the function $f(x)=x$ .

You are also right that this isn't the only function obeying the inequality in the question, so there might be solutions which are not bounded by $x$ .

See if you can find an alpha that is definitely big enough, as this is the logical next step.

glittergal96 · Sep 4, 2015

Re: HSC 2015 4U Marathon - Advanced Level

A useful fact that has not yet been mentioned/proven is that any such f must be monotonic.

InteGrand · Sep 4, 2015

Re: HSC 2015 4U Marathon - Advanced Level

glittergal96 said:
Yep, $\alpha \geq 1$ follows from considering the function $f(x)=x$ .

You are also right that this isn't the only function obeying the inequality in the question, so there might be solutions which are not bounded by $x$ .

See if you can find an alpha that is definitely big enough, as this is the logical next step.

$Wait, aren't you after the smallest possible $\alpha$ (which is $1$)? So the functions we want to find need to be bounded by $1\times x=x$?$

glittergal96 · Sep 4, 2015

Re: HSC 2015 4U Marathon - Advanced Level

InteGrand said:
$Wait, aren't you after the smallest possible $\alpha$ (which is $1$)? So the functions we want to find need to be bounded by $1\times x=x$?$

I think you might have some quantifiers the wrong way around?

We want to find a real number a such that f(x) =< ax for all f satisfying f(x+y) >= f(x) + f(y).

In particular, we want to find the least a that works for every such f.

We know a = 1 works for the function f(x) = x, but it is perfectly conceivable that there are other functions which satisfy f(x+y) >= f(x) + f(y), but for which we don't have f(x) =< 1x = x.

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