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

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September 18

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Continuous function

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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)[reply]

Sin(x^2) is an example of such a function. Sławomir Biały (talk) 02:27, 18 September 2011 (UTC)[reply]
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)[reply]
Yes. You can estimate the inverse Fourier transform of f by  , which is L^1. Sławomir Biały (talk) 11:10, 18 September 2011 (UTC)[reply]
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)[reply]

(Simple?) projectile motion

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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

 

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

 

but that hasn't yielded me much joy. Any advice is appreciated; is there something simpler I'm missing? —Anonymous DissidentTalk 12:47, 18 September 2011 (UTC)[reply]

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)[reply]
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)[reply]
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

 

if we make a change of variables  , this gives the family of parabolas indexed by τ:

 

A point (x,y) is on the envelope if and only if it is on a pair   of infinitely near parabolas in the family. So, for such a point,

 

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

 

Solving for y=0 gives

 

So the envelope hits the ground at

 

(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)[reply]

An algebraic approach, which may be what was requested, is this. A characteristic length is   defined by  . Introducing the dimensionless variable  , and the dimensionless height  , by substituting   and   and   into Anonymous Dissident's equation, gives
 
Dividing by  , and multiplying by   gives the simpler equation
 
This is further simplified by the trigonometric identities for the double angle  
 
Another equation is obtained by differentiation,
 
When   is maximum the differential   is zero even if   is not. So
 
Define   and   Then the system of equations is completely algebraic
 
It remains to eliminate   and   to get an algebraic equation in   alone. Bo Jacoby (talk) 02:20, 19 September 2011 (UTC).[reply]
In order to avoid making sign errors while manipulating polynomials I temporarily use the name   to signify  . So   and  . The minus sign   is then not used.
The three polynomials
 
 
 
all evaluate to zero for the wanted values of the variables   and the known values of the constants  . By construction the following polynomials also evaluate to zero.
 
 
 
 
 
 
 
Here   has been eliminated. Substitute  .
 
 
Now eliminate   from the equations  .
 
 
 
 
Now eliminate   from the equations  .
The above calculation is unfinished. I have deleted some errors. Bo Jacoby (talk) 05:12, 23 September 2011 (UTC) .[reply]