HSC 2014 MX2 Marathon ADVANCED (archive) (1 Viewer)

[seanieg89]

Re: HSC 2014 4U Marathon - Advanced Level

Can we define g(x) = f(x) for x >=0 and g(x) = f(-x) for x < 0?

No. Counterexample, f(x) = x => your g(x) is |x|, which is not differentiable at 0.

(This question is actually easier than I originally hoped. It is possible to sidestep the approach I was hoping for and use an easier argument in dimension 1.)

 

[seanieg89]

Re: HSC 2014 4U Marathon - Advanced Level

Note the edit: we also need to assume that the one sided second derivative at 0 of f exists, and is equal to the limit of f''(y) as y -> 0 from above.

 

[mathing]

Re: HSC 2014 4U Marathon - Advanced Level

Note the edit: we also need to assume that the one sided second derivative at 0 of f exists, and is equal to the limit of f''(y) as y -> 0 from above.

I guess we can define g(x) = -f(-x) + f(0) for x < 0 then the derivatives will be symmetrical around the x = 0 and it will make sure g is continuous at 0.

 

[seanieg89]

Re: HSC 2014 4U Marathon - Advanced Level

I guess we can define g(x) = -f(-x) + f(0) for x < 0 then the derivatives will be symmetrical around the x = 0 and it will make sure g is continuous at 0.

The first derivatives will "match up" but the second derivatives won't.

E.g. f(x)=x^2. If we do this odd reflection, then the resulting function will have second derivative 2 for positive x and -2 for negative x. (And we need g to have a continuous second derivative.)

Using reflection based arguments is definitely a good way to go though!

 

[mathing]

Re: HSC 2014 4U Marathon - Advanced Level

The first derivatives will "match up" but the second derivatives won't.

E.g. f(x)=x^2. If we do this odd reflection, then the resulting function will have second derivative 2 for positive x and -2 for negative x. (And we need g to have a continuous second derivative.)

Using reflection based arguments is definitely a good way to go though!

Good counter examples

I think we only have to worry above the 2nd derivative changing signs if the first derivative is zero at x = 0. Can we take the union of the two approaches? So if f'(x+) = 0 mirror it around x = 0 otherwise take the negative mirror to keep the 2nd derivative equal at x = 0.

 

[seanieg89]

Re: HSC 2014 4U Marathon - Advanced Level

Good counter examples

I think we only have to worry above the 2nd derivative changing signs if the first derivative is zero at x = 0. Can we take the union of the two approaches? So if f'(x+) = 0 mirror it around x = 0 otherwise take the negative mirror to keep the 2nd derivative equal at x = 0.

This can happen even if the first derivative is nonzero, eg f(x)=x^2+x.

If you want to use a reflection argument, then I don't think simply using even, odd, or a combination of the two will work.

You need to either construct a more complicated way of "reflecting", or else think of simpler ways to extend a function in a way that "matches up".

The latter approach I am thinking of only works in dimension 1, but is quite simple. The former is what I am hoping someone will think of.

 

[mathing]

Re: HSC 2014 4U Marathon - Advanced Level

This can happen even if the first derivative is nonzero, eg f(x)=x^2+x.

If you want to use a reflection argument, then I don't think simply using even, odd, or a combination of the two will work.

You need to either construct a more complicated way of "reflecting", or else think of simpler ways to extend a function in a way that "matches up".

The latter approach I am thinking of only works in dimension 1, but is quite simple. The former is what I am hoping someone will think of.

I think I see the extending way. We have the differential equation y''(0) = f''(0+) and y'(0) = f'(0+) and y(0) = f(0) so we can solve for some quadratic.

 

[mathing]

Re: HSC 2014 4U Marathon - Advanced Level

For the more complicated reflection, can we look at f'(0+) and f(0) then shift our origin up f(0) units on the y axis and apply a rotation by f'(0+) so it's now 0. With the derivative 0 in these co-ordinates can we now do the negative reflection?

*Should be regular mirror not negative.

 

[seanieg89]

Re: HSC 2014 4U Marathon - Advanced Level

I think I see the extending way. We have the differential equation y''(0) = f''(0+) and y'(0) = f'(0+) and y(0) = f(0) so we can solve for some quadratic.

Yep, exactly. Do you see why this fails though, if instead we are considering a function of two variables defined in the half-plane {x >= 0}? We could try the same construction on each line y = c, but the y-dependence of our extension would not be as nice as the y-dependence of our original function.

 

[seanieg89]

Re: HSC 2014 4U Marathon - Advanced Level

For the more complicated reflection, can we look at f'(0+) and f(0) then shift our origin up f(0) units on the y axis and apply a rotation by f'(0+) so it's now 0. With the derivative 0 in these co-ordinates can we now do the negative reflection?

*Should be regular mirror not negative.

Not entirely sure what you are saying here, it sounds fairly different to what I did. You should try to prove it carefully if it seems to work visually.

 

[mathing]

Re: HSC 2014 4U Marathon - Advanced Level

Yep, exactly. Do you see why this fails though, if instead we are considering a function of two variables defined in the half-plane {x >= 0}? We could try the same construction on each line y = c, but the y-dependence of our extension would not be as nice as the y-dependence of our original function.

In the 2D case do we assume all the right hand directional derivatives agree? If that's the case wouldn't it work.

 

[seanieg89]

Re: HSC 2014 4U Marathon - Advanced Level

In the 2D case do we assume all the right hand directional derivatives agree? If that's the case wouldn't it work.

No, if we assume the function is continuously differentiable (but not necessarily twice differentiable) in the right half plane, then for positive x, the "linear polynomial extension" gives:



This of course makes the left hand x-derivatives on the y axis exist and they are exactly what they should be.

If we try to calculate the y-partial derivative at a point in this left half-plane (like (-1,0)) though, then the expression involves a term like:



which we cannot know exists without further assumptions on f.


This reasoning shows that most approaches based on boundary derivatives are doomed to failure in higher dimensions, you need to just use knowledge of f inside the right half plane rather than knowledge of its derivatives on the boundary.

 

[mathing]

Re: HSC 2014 4U Marathon - Advanced Level

No, if we assume the function is continuously differentiable (but not necessarily twice differentiable) in the right half plane, then for positive x, the "linear polynomial extension" gives:



This of course makes the left hand x-derivatives on the y axis exist and they are exactly what they should be.

If we try to calculate the y-partial derivative at a point in this left half-plane (like (-1,0)) though, then the expression involves a term like:



which we cannot know exists without further assumptions on f.


This reasoning shows that most approaches based on boundary derivatives are doomed to failure in higher dimensions, you need to just use knowledge of f inside the right half plane rather than knowledge of its derivatives on the boundary.

I see. In the original question it says the 2nd derivative is continuous, so I wasn't sure what the assumptions were for the 2D case. Is the 2nd derivative existing still an assumption for the 1D case?

 

[seanieg89]

Re: HSC 2014 4U Marathon - Advanced Level

I see. In the original question it says the 2nd derivative is continuous, so I wasn't sure what the assumptions were for the 2D case. Is the 2nd derivative existing still an assumption for the 1D case?

Yep, in saying that the second derivative is continuous I am of course asserting that it exists.

Hint: The trick is to evaluate f at several points in your definition of reflection. Eg something like f(-x) = f(x) + f(x/2), but obviously not exactly that.

 

[Sy123]

Re: HSC 2014 4U Marathon - Advanced Level

 

[RealiseNothing]

Re: HSC 2014 4U Marathon - Advanced Level

Here is a bit of general calculus (for the uni students here, this idea is important in extending elements of Sobolev spaces beyond the boundary of a domain in a way that's pretty smooth).

(Recall that "f is bounded" means there is a positive M such that |f(x)| =< M for all x in the domain.)

Let f be a real valued function defined on the non-negative reals which is bounded and continuous, and has continuous second derivative over the positive reals. (And this continuity extends to the second order one-sided derivative at 0, which we assume exists).

Prove that there exists a function g satisfying the same properties that is defined on the whole real line and agrees with f at non-negative reals.

So is my interpretation correct:

For we want to define piecewise . Then show that if is defined over all reals, we can have some which is also bounded and continuous, and all those second derivative properties also hold?

 

[seanieg89]

Re: HSC 2014 4U Marathon - Advanced Level

So is my interpretation correct:

For we want to define piecewise . Then show that if is defined over all reals, we can have some which is also bounded and continuous, and all those second derivative properties also hold?

Yep.

 

[RealiseNothing]

Re: HSC 2014 4U Marathon - Advanced Level

Let the angles of the triangle be

Use cosine rule on the smaller triangles made out of one side of the large triangle, half a side of the large triangle, and a median:



Do this for all three smaller triangles within the larger triangle to get and

Then use cosine rule on the large triangle for all three sides to get:



Then you make the subject:



Do this for all three equations you have now.

Now you just substitute in these values to your three equations from the smaller triangles and add all three equations to get the result.

 

[Sy123]

Re: HSC 2014 4U Marathon - Advanced Level

 

[RealiseNothing]

Re: HSC 2014 4U Marathon - Advanced Level

I'm getting: