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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

$x=Vt\cos \theta ,y=-\frac{g}{2}{t}^2+Vt\sin \theta +h$

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.

Template:Od 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 τ:

$P_{\tau }:y=\frac{-g}{2V^2}(1+{\tau }^2)x^2+\tau x+h$

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,

$0=\frac{-g}{V^2}\tau d\tau x^2+d\tau x⟹x=V^2/(g\tau )$

(the nontrivial zero). Plugging into the equation for y(x) gives

$y(V^2/(g\tau ))=h+\frac{V^2({\tau }^2-1)}{2g{\tau }^2}$

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

$0=-z^2+2z\sin \theta \cos \theta +2b{\cos }^2\theta .$

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,

$0=-2zdz+\sin \alpha dz+z\cos \alpha d\alpha -b\sin \alpha d\alpha .$

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

$P_1=m{z}^2+zS+bC+b$

$P_2=zC+mbS$

$P_3=C^2+S^2+m$

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.

$P_4=(zC+bS)P_2+b^2{P}_3=(z^2+b^2)C^2+m{b}^2$

$P_5=(m{z}^2+mzS+bC+b)P_1+z^2{P}_3$

$=(m{z}^2+mzS+bC+b)(m{z}^2+zS+bC+b)+z^2(C^2+S^2+m)$

$=(m{z}^2+bC+b{)}^2+m{z}^2{S}^2+z^2{C}^2+z^2{S}^2+z^{2}m$

$=(m{z}^2+bC+b{)}^2+z^2{C}^2+z^{2}m$

$=b^2{C}^2+(m{z}^2+b{)}^2+2bC(m{z}^2+b)+z^2{C}^2+z^{2}m$

$=(z^2+b^2)C^2+2b(m{z}^2+b)C+(m{z}^2+b{)}^2+z^{2}m$

Here $S$ has been eliminated. Substitute $w=z^2$.

$P_4=(w+b^2)C^2+m{b}^2$

$p_5=(w+b^2)C^2+2b(mw+b)C+(mw+b{)}^2+mw$

Now eliminate $C^2$ from the equations $0=P_4=P_5$.

$P_6=m{P}_4+P_5$

$=m((w+b^2)C^2+m{b}^2)+(w+b^2)C^2+2b(mw+b)C+(mw+b{)}^2+mw$

$=b^2+2b(mw+b)C+(mw+b{)}^2+mw$

$=2b(mw+b)C+w^2+m(2b+1)w+2b^2$

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) .