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August 9

Uniform vs Pointwise Continuity

So Uniform continuity is stronger than pointwise continuity because it requires that one delta can be small enough to ensure that any two points are pointwise continuous. What is an example of a function that is pointwise continuous but not uniformly continuous? I have thought about this for quite some time but I haven't arrived at an answer; maybe I don't understand the defnintion? Thanks.

f:R→R with f(x) = x² is an example. Note that any continuous function on a compact set is automatically uniformly continuous. Rckrone (talk) 00:42, 9 August 2011 (UTC)[reply]

1/x is pointwise continuous for x>0 but not uniformly continuous. You can't guarantee the results will be within a certain distance of teach other just because the values going in are with a certain distance of each other. Dmcq (talk) 07:50, 9 August 2011 (UTC)[reply]

Is the book's answer wrong?

I have solved a problem from a book, but the book's answer is slightly different. Am I wrong or is it a typo in the book?

The question is:

You visit a golf driving range to practice your driving. The fairway (the area to which the ball is played) slopes uphill at a constant angle of $a^{\circ }$ to the horizontal from the point where the ball is played.

You play a shot with an angle of inclination $\theta$ to the horizontal with an initial velocity $U$ .

Show that if your ball lands a distance of $D$ along the fairway then $D={\frac {2U^{2}\cos \theta \sin(\theta -a)}{g\cos ^{2}a}}$

I solved this by first finding when the vertical displacement relative to the sloped fairway was equal to 0. That is

$0=Usin(\theta -a)T-{\frac {1}{2}}gT^{2}$

giving $T={\frac {2Usin(\theta -a)}{g}}$

I then use this to find the horizontal displacement (which I call $d$ at time $T$ .

$d=U\cos \theta T={\frac {2U^{2}\cos \theta \sin(\theta -a)}{g}}$

Now since the fairway slopes up at angle $a$ , the horizontal displacement $d$ can be considered the adjacent side of a right-angle triangle, and the distance alone the fairway $D$ the hypotenuse.

So $D={\frac {d}{\cos a}}$

This gives me $D={\frac {2U^{2}\cos \theta \sin(\theta -a)}{g\cos a}}$

However the solution I am supposed to get has a $\cos ^{2}a$ in the denominator instead of a $\cos a$ .

I can't see where that square would come from. Do you think it's a typo, or am I wrong? — Preceding unsigned comment added by Thorstein90 (talk • contribs) 04:52, 9 August 2011 (UTC)[reply]

I think the book answer is correct. The height of the ball is

y=UT\sin \theta -gT^{2}/2

while the height of the field is

y=d\tan a

with

d=UT\cos \theta

. We want to solve for when those two heights are equal so

UT\sin \theta -gT^{2}/2=UT\cos \theta \tan a

. Rckrone (talk) 05:24, 9 August 2011 (UTC)[reply]

(edit conflict) After a couple of pages of work, I got the book's answer. I think the place you went wrong is the equation

0=U\sin(\theta -a)T-{\frac {1}{2}}gT^{2}.

If I understand your reasoning here, you are rotating the picture by an angle of a to make the fairway level, and then solving the ballistic motion equation in this rotated frame of reference to find when the ball hits the ground. The problem is that gravity is not acting perpendicular to the fairway, so the standard ballistic motion equation doesn't apply in that rotated reference frame (in the rotated reference frame, gravity acts at an angle of a from the vertical).

What I did to solve the problem was to write ballistic motion equations describing the x- and y-coordinates of the ball in terms of time in a non-rotated reference frame, so those equations just have θ, not θ − a. Then I wrote another equation describing the elevation of the ground directly underneath the ball at time t, using the triangle with angle a and the equation for the x-coordinate of the ball at time t. Then I solved for the time when the y-coordinate of the ball equals the elevation of the ground underneath it. The

\sin(\theta -a)

factor didn't come into play until I simplified my expression for D at the end using the angle difference formula

\sin(\theta -a)=\sin \theta \cos a-\cos \theta \sin a

. —Bkell (talk) 05:31, 9 August 2011 (UTC)[reply]

I see! I thought there might have been something dodgy about the rotation, but since the answer looked similar I assumed it was how to do it. Thanks for your help. — Preceding unsigned comment added by Thorstein90 (talk • contribs) 05:50, 9 August 2011 (UTC)[reply]

Start with

x=Ut\cos \theta

,

y=(-g/2)t^{2}+Ut\sin \theta

and substitute in for t to get y as a function of x:

y(x)=-{\frac {gx^{2}}{2U^{2}\cos ^{2}\theta }}+x\tan \theta .

Now D has the property that when

x=D\cos a

,

y=D\sin a

(easy to see with a triangle). Plug those values into y(x) and re-arrange to find D. —Anonymous Dissident^Talk 10:42, 9 August 2011 (UTC)[reply]

Dirverging in all Directions

I have some more problems I just don't know how to begin with to do with flux and the Gauss divergence theorem.

Let $F=(x,y,z)$ and let S be the unit sphere, centred on $(0,0,0)$ .Verify the Divergence Theorem by computing both the flux of F through the surface, and the integral of ∇•F on the sphere.

Let $r=(x,y,z),F=r/||r^{3}||$ and let D be a simply connected solid region in ℝ³ with boundary surface S. Prove that :

∮s F • dS = 0 if (0, 0, 0) is not an element of D.

∮s F • dS = 4π if (0, 0, 0) is an element of D.

I have looked over all the notes and stuff done on this topic and I simply cannot understand where to begin. Any assistance will be gratefully received. Thank You. Chris the Russian Christopher Lilly 07:16, 9 August 2011 (UTC)[reply]

For the first problem you need to start with what exactly the divergence theorem says, which is

\int _{B}(\nabla \cdot \mathbf {F} )dV=\oint _{S}(\mathbf {F} \cdot \mathbf {n} )dA

where B is some bounded volume in R³, S is the boundary of B, and n is the unit vector normal to the surface S. The problem asks you to explicitly calculate both sides of the equation for the particular vector field F = (x,y,z) and the particular B which is the unit ball. What does ∇.F evaluate to? What is n as a function of the position on S and then what does F.n evaluate to? Rckrone (talk) 22:33, 9 August 2011 (UTC)[reply]

Thank You. I can understand that the proof lies in showing both integrals give the same value, but I was just unsure as to where to start. We had some examples of this shown to us and worked through, but a lot of this is unclear. I suspect that F•n would be (x, y, z)• (∂x/∂x,∂y/∂y,∂z/∂z), which should equal (x,y,z), and that ∇•F should be ( -2x,-2y,1)•(x,y,z) = $-2x^{2}-2y^{2}+z$ , at least I believe so, but I cannot seem to understand exactly what I am meant to be doing. This is strange, since before most of this Maths was never so difficult. Any other advice would be appreciated also. Thanks. Chris the Russian Christopher Lilly 01:06, 10 August 2011 (UTC)[reply]

The del symbol ∇ = (∂/∂x,∂/∂y,∂/∂z), so ∇•F = (∂/∂x,∂/∂y,∂/∂z)•(x,y,z) = ∂x/∂x + ∂y/∂y + ∂z/∂z = 3.

The normal vector n points perpendicular to the surface of the sphere (and by convention outward). If you think about what a sphere centered at the origin is shaped like, the vector that's perpendicular to it is one that points directly out from the origin. So at a point (x,y,z) on the sphere, the value of n (which is a function of the coordinates x,y,z) should be a vector pointing in the (x,y,z) direction. Then we need to normalize it, but since our point (x,y,z) is on the unit sphere it has length 1, so (x,y,z) is already normalized. Therefore n = (x,y,z). F•n = (x,y,z)•(x,y,z) which again we know is 1 since the points we're integrating over are all on the surface of the unit sphere. Rckrone (talk) 02:03, 10 August 2011 (UTC)[reply]

Now I get it ! Thank you, that shall assist me immensly. Chris the Russian Christopher Lilly 07:55, 12 August 2011 (UTC)[reply]

Some Basic Numerics Questions

We have a big numerics exam coming up and today we had a very lively discussion trying to settle these but to no avail and we couldn't really find a clear cut answer anywhere. Please help!
1.For a single step (Runge Kutta type) scheme to solve y'=f, is consistency enough to show convergence or do we need both consistency and stability?
2.To find the region of stability for a numerical method to solve an ODE, do we want the modulus of the stability polynomial to be strictly less than one or less than or equal to one? Is the boundary of the region included in the region? What about the origin? Is it always to be included?
Thanks! - Looking for Wisdom and Insight! (talk) 08:12, 9 August 2011 (UTC)[reply]

Limit as x approaches 0 cos(x)/x

Hi, I was doing a calculus problem and am terrible at trigonometric functions etc. The question was evaluate the following limit,Limit as x approaches 0 cos(x)/x, and the answer was

DNE (Does not exist) because the right hand limit =positive infinity while left hand limit equals negative infinity.

I dont understand how to figure out how to find the cos(x) for this problem or any problem in general and was also wondering if anyone knew a page for trigonometry that will help me with calculus. — Preceding unsigned comment added by 184.88.243.39 (talk) 23:18, 9 August 2011 (UTC)[reply]

Try to use the Taylor series of cos(x). We see that

\cos x\sim 1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+O(x^{8})\,.

Dividing by x gives us

{\frac {\cos x}{x}}\sim {\frac {1}{x}}-{\frac {x}{2!}}+{\frac {x^{3}}{4!}}-{\frac {x^{5}}{6!}}+O(x^{7})\,.

You'll be able to finish the argument from here. Think about positive x tending to zero, and then negative x tending to zero. — Fly by Night (talk) 01:04, 10 August 2011 (UTC)[reply]

Roughly speaking, in the fraction

{\frac {\cos x}{x}}

, as x approaches 0 the numerator approaches 1 (because

\cos 0=1

and

\cos x

is a continuous function), while the denominator becomes very close to 0. Think about what happens when you divide a number very close to 1 (like 0.99999) by a number very close to 0 (like 0.00001). What if the numerator becomes even closer to 1 and the denominator becomes even closer to 0? How does the answer depend on whether that number very close to 0 is positive or negative? [Note: This isn't a mathematically rigorous proof that

\lim _{x\to 0}{\frac {\cos x}{x}}

does not exist, but maybe this will help if you're having difficulty with the intuition behind it.] —Bkell (talk) 01:22, 10 August 2011 (UTC)[reply]

My understanding of the question is that the questioner does not know that cos 0 = 1, so most likely none of these answers are helpful. The most valid answer, I'm afraid, is that it is necessary to learn trig before taking calculus to have much hope of success. Trying to learn both at the same time is a recipe for failure. Looie496 (talk) 04:28, 10 August 2011 (UTC)[reply]

The first class for which I was ever instructor of record was something called Business Calculus. Although there were a few business-related "word problems", the main difference between it and regular calculus seemed to be that we didn't use trig functions.

I don't really agree that you have to know trig before doing calculus. Certainly calculus is richer if you have the trig functions around, but conceptually the two things are pretty much orthogonal. Of course if the class uses them and you don't know them, you are going to have to do some extra work. --Trovatore (talk) 21:40, 10 August 2011 (UTC)[reply]

Fair enough. If you don't use trig, you don't need to know trig -- although once you come to integral calculus you'll be seriously impaired if you don't. My point is mainly that learning calculus is hard enough in itself, and a student who needs to learn the background at the same time as learning calculus is in serious trouble. I taught a calculus-for-engineers class five or six times -- when students failed, in most cases it was due to not having a solid background. Looie496 (talk) 01:29, 11 August 2011 (UTC)[reply]