Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[da_nang]

The coordinates of the matrix are the points that would yield the polynomial x^2, no? The computing of the determinant probably gives you some sort of existence and/or uniqueness condition, but that's just a guess. If the three points are not distinct, then your polynomial is definitely not unique, and that determinant is definitely zero. No idea though

The value of the determinant should tell you about the existence of solutions.

Not quite. If det(A) =  0, then Ax = b has either no solution or infinitely many. If det(A) != 0, then it has exactly one solution.

Spoiler: Details (click to show/hide)

To properly distinguish these, you need the rank. For a n x n matrix A: The equation Ax = b has exactly one solution if rank(A) = rank(A|b) = n, where A|b is the augmented matrix. If rank(A) = rank(A|b) < n, then there are infinitely many solutions. If rank(A) != rank(A|b), then there are no solutions.

In other words, existence is guaranteed if rank(A) = rank(A|b) and unique if rank(A) = rank(A|b) = n.

[Helgoland]

Rank isn't an easily computed invariant like determinant though, and mere existence of a solution isn't all that hard to establish: Just check that the points given don't map one number to two different outputs, and you can construct a solution geometrically. Use the constant part to drag the first point to the x axis, drag the second point to the x axis with the linear term*, and then apply the intermediate value theorem to establish that you can hit the third point using the remaining term.

*This is where we'd hit a snag if we had two outputs above the same input, since that would require a flip by 90°.

[Arx]

I'm completely stumped. I'm reading over a nonlinear dynamical systems paper, and at one point they go from



to



And I can't for the life of me figure out how that's correct. The best I can manage is



But I can't see how that extra factor in the denominator reduces the numerator to the form they have. I feel like I'm just being dense, anyone see what my mistake is?

All the constants are nonzero and positive.

[askovdk]

But I can't see how that extra factor in the denominator reduces the numerator to the form they have. I feel like I'm just being dense, anyone see what my mistake is?

All the constants are nonzero and positive.

Do we know anything about z?
If it's a constant like a-f, then I can't see how the we can keep a linear dependence in the numerator for the derivative but somehow reduce the resulting quadratic dependence in the denominator.

[Arx]

z is a constant, yes. e is also an arbitrary constant, not Euler's number. The only relationship between the constants is

b > d + z/f

I'm starting to wonder if I've missed something from context or something strange like that.

At this point I'd settle for proof that psi'(0) is negative iff z/(f^2) > e, which is the conclusion the authors draw, and as long as that's correct I think everything else hangs together.

I'm just thoroughly confused, because it seems highly unlikely that the authors would be wrong, but I can't see how they could be right either...

[askovdk]

z is a constant, yes. e is also an arbitrary constant, not Euler's number. The only relationship between the constants is

b > d + z/f

I'm starting to wonder if I've missed something from context or something strange like that.

At this point I'd settle for proof that psi'(0) is negative iff z/(f^2) > e, which is the conclusion the authors draw, and as long as that's correct I think everything else hangs together.

I'm just thoroughly confused, because it seems highly unlikely that the authors would be wrong, but I can't see how they could be right either...

Cool, then I think we can get it 
Inserting y=0 in the denominator makes it: f(b-d)-z = f( b - d - z/f), but b > d+z/f  =>  b - d - z/f > 0, i.e. it can't be zero and (especially squared) it will be positive and therefore not affect the sign of the derivative.

So we just have to look at the nominator of the fraction derivative: n'(y) * d(y) - d'(y)*n(y)
I get
n'(y) = (d+ey) + (f+y)*e
d'(y) = (b-d-ey) +(f+y)*(-e)

Inserting 0 as y in these and slowly lifting the (negative) parentheses ends me with a lot of cancelling parts, so only bef² - bz  is left

I.e.  Phi'(0) = a*(bef² - bz) / f²* 'something positive' = -ab*(z/f²-e) / 'something positive'.

I.e. The sign of the derivative at y=0 is indeed negative iff z/f²-e > 0   <=>  z/f² > e

   Let me know if you want to compare how I lift the parentheses.


(It would seem that there is an typo in the book, as the equation for phi'(0) would be correct if the denominator was squared.)

[Arx]

That's what I got as well, yes. Which is good, because it means the rest of the results should be fine. If anyone has an explanation as to the derivatives, though, I'd love to see it - I'm still baffled.

[askovdk]

...  If anyone has an explanation as to the derivatives, though, I'd love to see it - I'm still baffled.

? I don't understand, - i.e. would you like to see how I get the long list of mostly cancelling terms?

[Arx]

No, that's easy. What I'd like to know is exactly how the authors of the paper got



As the derivative of psi at 0.

[askovdk]

No, that's easy. What I'd like to know is exactly how the authors of the paper got



As the derivative of psi at 0.

Hmm, I'm sure I see the problem, - beside that I truly believe there is a typo in the book.

I.e. Phi(y) = a* n(y)/d(y)
Phi'(y)=a*(n'(y)*d(y) - d'(y)*n(y) ) / d²(y)
= a*( (1*(d+ey)+(f+y)*e)*((f+y)*(b-d-ey)-z) - (((b-d-ey)+(f+y)*(-e))*((f+y)*(d+ey)+z)) ) / ((f+y)*(b-d-ey)-z)²

Inserting 0 as y

= a*( (1*(d+ey)+(f+y)*e)*((f+y)*(b-d-ey)-z) - (((b-d-ey)+(f+y)*(-e))*((f+y)*(d+ey)+z)) ) / ((f+y)*(b-d-ey)-z)²
= a*( (d+f*e)*(f*(b-d)-z) - (((b-d)-f*e))*((f*d+z)) ) / (f*(b-d)-z)²

... reducing the numerator ...

= a* (bef² - bz) / (f*(b-d)-z)²

pulling a b out of the numerator, and an f² out of numerator and denominator

= a*b* ( f²*(e-z/f²) ) / ( f*((b-d) - z/f ) )²
= a*b* ( f²*(e-z/f²) ) /  f²*(b-d - z/f )²

and for some reason wanting a '-' in front

= (-1)*(-1) a*b* ((e-z/f²) ) /  (b-d - z/f )²
= -ab* (  -1*(e-z/f²) ) /  (b-d - z/f )²
= -ab* (  -e +z/f²) ) /  (b-d - z/f )²
= -ab* (   z/f² -e ) /  (b-d - z/f )²

Q.E.D.      (Still assuming a typo in the book.)

Or are we talking past each others?

[da_nang]

Definitely looking like a typo, if my own algebra is anything to go by.

[Arx]

The assumption that there's a typo in the paper is kinda suboptimal, but it looks correct. Oddly, WolframAlpha thinks that's not right, but there are now three different people contradicting it so ¯\_(ツ)_/¯.

Thank you guys.

[da_nang]

I mean, it makes sense that there is a typo.

Let u be the numerator and v be the denominator. Both u and v can be seen as first-order polynomials in terms of z: p(z, y, parameters) = (some sign value)z + (some function)(y, parameters).

Calculating d(u/v)/dy = (u'v - uv')/v² gives us at most a first-order polynomial in the resulting numerator (as p' = (some function)' with respect to y) and exactly second-order denominator. _{[Pardon my programmer-ish notation]}

That cannot be achieved with the author's suggested solution.

[Parsely]

Why is this expression true? (That "Pi" represents the pi symbol. I thought the alt code for it looked too much like an "n".)

sin(2*Pi/3) = √(3)/2

I don't understand how that conversion works.

[Arx]

Geometric construction using a hexagon, if I recall properly.

Easier than that, even. Sin(2pi/3) is just sin(pi/3) in the 2nd quadrant (if you prefer, you can think of it as sin(pi-pi/3)), which is (√3)/2. (Since the sign of sin (best phrase) doesn't change in the 2nd quadrant because it's a function only of y and r, which are both still positive.)