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

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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

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

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 τ:

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

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

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

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

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

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.

$P_{4}=(zC+bS)P_{2}+b^{2}P_{3}=(z^{2}+b^{2})C^{2}+mb^{2}$

$P_{5}=(mz^{2}+mzS+bC+b)P_{1}+z^{2}P_{3}$

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

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

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

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

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

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) .