Wikipedia:Reference desk/Archives/Mathematics/2011 September 18

= September 18 =

Continuous function
Is there a non-uniformly continuous bounded real valued function defined on all real numbers? Money is tight (talk) 02:06, 18 September 2011 (UTC)
 * Sin(x^2) is an example of such a function. Sławomir Biały  (talk) 02:27, 18 September 2011 (UTC)
 * Thanks! I have another question. If f is a continuously twice differentiable function on the real line with compact support, is it the fourier transform of some L^1 function? Money is tight (talk) 06:41, 18 September 2011 (UTC)
 * Yes. You can estimate the inverse Fourier transform of f by $$C(1+x^2)^{-1}$$, which is L^1.  Sławomir Biały  (talk) 11:10, 18 September 2011 (UTC)
 * I don't fully understand what you mean. By C do you mean a constant? How and in what sense is the estimation (uniform pointwise L^1 L^2 norm etc.)? Is the inverse transform in L^1? Money is tight (talk) 01:25, 19 September 2011 (UTC)

(Simple?) projectile motion
This is an apparently simple problem which seems to involve algebra of frustrating complexity. A particle is projected from the top of a cliff of height h metres with a velocity of V m/s into the sea. What is its maximum horizontal range? The equations of motion are
 * $$x = Vt\cos\theta, \qquad y = -\frac{g}{2} t^2 + Vt\sin\theta + h$$

where θ is the angle of projection and the variable in question. There are two obvious approaches: Find the larger root T of y(t) = 0 and maximise x(T) in terms of θ; or rewrite y as a function of x and maximise the larger root X of y(x) = 0 in terms of θ. Both are very algebra-heavy and I keep getting lost. A craftier approach might be to maximise the roots of y(x) = 0 with product and sum examination rather than actual computation, where
 * $$y(x) = \frac{-g}{2V^2\cos^2 \theta} x^2 + x\tan \theta + h,$$

but that hasn't yielded me much joy. Any advice is appreciated; is there something simpler I'm missing? — Anonymous Dissident  Talk 12:47, 18 September 2011 (UTC)
 * The problem is basically an exercise in seeing things through to the end. Solve y=0 for t, keeping the positive root only, and substitute back into x.  You then get x as a function of theta, which can be maximized by methods of one variable calculus.  Details are in the article Range of a projectile.  Sławomir Biały  (talk) 12:59, 18 September 2011 (UTC)
 * I think the problem can be simplified a bit by noticing that the point of maximum distance would be on the envelope of the parabolas which are the various trajectories. To get the envelope, set (dx/dt)(dy/dθ)=(dx/dθ)(dy/dt) to get an additional equation, I get v=g t sin θ. This can be plugged in to y=0 to get a linear equation for sin θ, and both of these can be plugged into the equation for x to get the maximum distance. Or you can just find the equation of the envelope, which is another parabola, and solve for y=0.--RDBury (talk) 13:54, 18 September 2011 (UTC)
 * Slight correction, I should have said "linear equation for sin2 θ". Also, since the envelope is a parabola, it has a focus which, interestingly imo, is the location of the cannon.

That's pretty clever. Starting from Anonymous Dissident's equation
 * $$y(x) = \frac{-g}{2V^2\cos^2 \theta} x^2 + x\tan \theta + h,$$

if we make a change of variables $$\tau=\tan\theta$$, this gives the family of parabolas indexed by &tau;:
 * $$P_\tau:\quad y = \frac{-g}{2V^2}(1+\tau^2) x^2 + \tau x + h$$

A point (x,y) is on the envelope if and only if it is on a pair $$P_\tau, P_{\tau+d\tau}$$ of infinitely near parabolas in the family. So, for such a point,
 * $$0 = \frac{-g}{V^2}\tau d\tau x^2 + d\tau x \implies x=V^2/(g\tau)$$

(the nontrivial zero). Plugging into the equation for y(x) gives
 * $$y(V^2/(g\tau))= h+\frac{V^2 (\tau^2-1)}{2 g \tau^2}$$

Solving for y=0 gives
 * $$\tau=\frac{V}{\sqrt{2gh+V^2}}.$$

So the envelope hits the ground at
 * $$x=\frac{V^2}{g\tau} = \frac{V}{g}\sqrt{2gh+V^2}.$$

(This doesn't seem to agree with the answer Range of a projectile gives. I think there must be an error in the article somewhere.)   Sławomir Biały  (talk) 19:05, 18 September 2011 (UTC)
 * An algebraic approach, which may be what was requested, is this. A characteristic length is $$L$$ defined by $$V^2=Lg$$. Introducing the dimensionless variable $$z$$, and the dimensionless height $$b$$, by substituting $$x=Lz$$ and $$h=Lb$$ and $$y=0$$ into Anonymous Dissident's equation, gives
 * $$0 = \frac{-g(Lz)^2}{2Lg\cos^2\theta} + Lz\,\tan \theta  + Lb.$$
 * Dividing by $$L$$, and multiplying by $$2 \cos^2 \theta$$ gives the simpler equation
 * $$0 = -z^2+2\,z\,\sin\theta\cos\theta+2\,b\,\cos^2\theta.$$
 * This is further simplified by the trigonometric identities for the double angle $$\alpha =2\, \theta$$
 * $$0 = -z^2 + z\, \sin\alpha  + b\, \cos\alpha+b.$$
 * Another equation is obtained by differentiation,
 * $$0 = -2z\, dz+\sin\alpha \, dz+z \, \cos\alpha \,d\alpha-b\, \sin\alpha\, d\alpha .$$
 * When $$z$$ is maximum the differential $$dz$$ is zero even if $$d\alpha$$ is not. So
 * $$0 = z\,\cos\alpha - b\,\sin\alpha .$$
 * Define $$C = \cos \alpha$$ and $$S = \sin \alpha .$$ Then the system of equations is completely algebraic
 * $$0=-z^2+zS+bC+b=zC-bS=C^2+S^2-1\, .$$
 * It remains to eliminate $$C$$ and $$S$$ to get an algebraic equation in $$z$$ alone. Bo Jacoby (talk) 02:20, 19 September 2011 (UTC).
 * In order to avoid making sign errors while manipulating polynomials I temporarily use the name $$m$$ to signify $$-1$$. So $$m+1=0$$ and $$m^2=1$$. The minus sign $$-$$ is then not used.
 * The three polynomials
 * $$P_1=mz^2+zS+bC+b$$
 * $$P_2=zC+mbS$$
 * $$P_3=C^2+S^2+m$$
 * all evaluate to zero for the wanted values of the variables $$z,C,S$$ and the known values of the constants $$b,m$$. By construction the following polynomials also evaluate to zero.
 * $$P_4=(zC+bS)P_2+b^2P_3=(z^2+b^2)C^2+mb^2$$
 * $$ P_5 = (mz^2+mzS+bC+b)P_1+z^2P_3 $$
 * $$ = (mz^2+mzS+bC+b)(mz^2+zS+bC+b)+z^2(C^2+S^2+m)$$
 * $$ =(mz^2+bC+b)^2+mz^2S^2+z^2C^2+z^2S^2+z^2m $$
 * $$ =(mz^2+bC+b)^2+z^2C^2+z^2m $$
 * $$ =b^2C^2 +(mz^2+b)^2+2bC(mz^2+b)+z^2C^2+z^2m $$
 * $$ =(z^2+b^2)C^2 +2b(mz^2+b)C +(mz^2+b)^2+z^2m $$
 * Here $$S$$ has been eliminated. Substitute $$w=z^2$$.
 * $$P_4=(w+b^2)C^2+mb^2$$
 * $$p_5=(w+b^2)C^2 +2b(mw+b)C +(mw+b)^2+mw $$
 * Now eliminate $$C^2$$ from the equations $$0=P_4=P_5$$.
 * $$P_6=mP_4+P_5$$
 * $$=m((w+b^2)C^2+mb^2) +(w+b^2)C^2 +2b(mw+b)C +(mw+b)^2+mw$$
 * $$=b^2 +2b(mw+b)C +(mw+b)^2 +mw$$
 * $$=2b(mw+b)C + w^2+m(2b +1)w+2b^2$$
 * Now eliminate $$C$$ from the equations $$0=P_4=P_6$$.
 * The above calculation is unfinished. I have deleted some errors. Bo Jacoby (talk) 05:12, 23 September 2011 (UTC).