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May 26

Incredibly convulted integral

Hello, I need help solving this:

$\int _{\pi }^{\pi ^{2}}{\frac {\operatorname {erf} (x)}{{\rm {Si}}(x)}}(\cos x)^{x}(e^{\csc x})dx$

erf(x) is the error function, while Si(x) is the Sine integral.

How do I start? THANK YOU VERY MUCH.--0rrAvenger 00:59, 26 May 2007 (UTC)[reply]

Is a numerical approximation good enough? Cause I have no idea how to even begin to solve this integral analytically. nadav (talk) 02:36, 26 May 2007 (UTC)[reply]

Can you explain the origin of this integral? It looks somewhat implausible. --Lambiam ^Talk 09:02, 26 May 2007 (UTC)[reply]

High school math assignment... Just kidding. My friend made it up.--0rrAvenger 15:14, 26 May 2007 (UTC)[reply]

For π ≤ x < ³/₂π, and other parts of the range of integration, cos x is negative, and raising a negative number to an irrational value, as is required then by the factor (cos x)^x, is somewhat undefined. Our article Exponentiation ascribes the meaning exp (b·log a) to a^b, but that requires a somewhat arbitrary choice for log a when a is negative. For a = –1 and b irrational (corresponding to x = π), the possible values are dense on the unit circle in the complex plane. --Lambiam ^Talk 09:29, 26 May 2007 (UTC)[reply]

Wonky Room

Imagine a rectangular room. If you leave through one wall, you come back in on the opposite wall, rotated 180 degrees about the center of the wall - that is, walking on the ceiling. If you leave through the ceiling (or floor), you come back in through the floor (or ceiling) rotated 180 degrees about the center of each. Would this be an acceptable model of real projective 3-space? I base it on the idea of gluing together opposing boundary points on what's topologically a closed ball. Black Carrot 03:30, 26 May 2007 (UTC)[reply]

Yes, I believe the Wonky Room is an apt model of the projective space. You're basically identifying antipodal points on the boundary of a ball in a way that is analogous to the fundamental polygon for RP^2. nadav (talk) 08:23, 26 May 2007 (UTC)[reply]

Sounds a lot like the demo of Portal (computer game). -Wo o ty Woot? contribs 08:58, 26 May 2007 (UTC)[reply]

Yes the addition of "portals" to first-person shooter games so complicated the topology of the maps that companies were forced to redesign their game engines from the ground up. This was the problem 3D Realms ran into when they made Prey (video game), the first game to use portals. nadav (talk) 09:12, 26 May 2007 (UTC)[reply]

Other than Narbacular Drop. :-) Rawling4851 16:02, 30 May 2007 (UTC)[reply]

Are odd-dimensional projective spaces orientable and even-dimensional ones not? —Tamfang 03:58, 27 May 2007 (UTC)[reply]

It looks like it. In being teleported by moving through the wall acting as a stargate, vectors along all dimensions except the forward dimension, determining the orientation of the teleportee in (hyper-)space, will appear flipped to a stationary observer inside the (hyper-)box. If n is the topological dimension, n−1 vectors are flipped. By (orientation-preserving) rotations we can flip any 2m dimensions, so if n is odd the teleportee can be brought back to its previous orientation. If n is even, this is not possible; only a second teleportation can restore parity. --Lambiam ^Talk 05:58, 27 May 2007 (UTC)[reply]

*facepalm* Yah, that makes sense. —Tamfang 21:08, 27 May 2007 (UTC)[reply]

Real projective space now addresses orientability after Nbarth's recent revision. nadav (talk) 06:17, 27 May 2007 (UTC)[reply]

Could you suggest any other construction schemes (besides gluing together sides of a polyhedron) that could result in a nifty paradoxical room? I'm trying to come up with an Escher-esqe space to draw that actually follows some set of consistent topological rules. Black Carrot 01:52, 29 May 2007 (UTC)[reply]

Try getting the book The Shape of Space by Jeffrey Weeks (mathematician). It's quite on-topic here. – b_jonas 14:04, 29 May 2007 (UTC)[reply]

The Bot Has Failed!

More discussion at the reference desk discussion page. Root⁴(one) 04:25, 26 May 2007 (UTC)[reply]

Factor Theorem

Is the factor theorem true for expressions (polynomials) of multiple variables? For example to factorise a² + ab - ab² - b³, Can I treat it as a function of a - ie. f(a) = a² + ab - ab² - b³ and then say that since f(b²) = 0, then (a - b²) is a factor of this expression. I know it's true in this case but I want to know whether the proof is valid. --AMorris (talk)●(contribs) 09:16, 26 May 2007 (UTC)[reply]

The short answer is: yes. If the expression can be viewed as a polynomial P[V] in some variable V, and vanishes when some expression E is substituted for V; that is, P[E] can be simplified to 0 using the normal rules of algebra, then P[V] can be rewritten in the form (V−E) × Q[V], in which Q[V] is again a polynomial in V. The proof applies without change. --Lambiam ^Talk 09:58, 26 May 2007 (UTC)[reply]

Something I don't understand about integration

We'll be studying Integration in Additional Mathematics next term. We have already studied Differentation, and I understand Integration is the reverse process of differentiation. So if you differentiate a function and then integrate the result, you get the original function again. Right?

But there's something I don't understand about Integration. It's difficult to explain, so I'll use an example.

Let y = x^2 + 3x + 4

So dy/dx = 2x + 3. Right?

Now, let's say I am asked to integrate 2x + 3.

(My keyboard doesn't have an integration sign.)

When 2x is integrated, I get x^2 (I did this mentally, not using any formula).

When 3 is integrated, I get 3x.

So when I integrate 2x + 3, I get x^2 + 3x. Am I correct?

Now my question is, what happened to the 4? When I'm asked to integrate 2x + 3, how do I get back the 4? How do I even know that the last number's 4? Change the 4 in the original function to 5, and dy/dx is still 2x + 3.

--Kaypoh 11:16, 26 May 2007 (UTC)[reply]

For this reason, you always have

+C

at the end of any integral without limits, where C is a constant (the 4 in your example). When the integral has limits, you don't need the

+C

.

(e/c) An excellent question. The answer is that when you perform the integration (that is, when you find the antiderivative), you must not forget to add the Arbitrary constant of integration. As you point out, you do lose the specific constant term when you differentiate and then antidifferentiate. nadav (talk) 11:37, 26 May 2007 (UTC)[reply]

Thanks for the quick response! Does that mean that there's no way to find out whether the nunmber is 4, 5 or something else, so I use c to represent the number? --Kaypoh 11:50, 26 May 2007 (UTC)[reply]

Yes, that's exactly right, unless you have additional information that tells you which value of C to choose. If all you know is that you are looking for a function F whose derivative is some function f that you are given, then you must include the + C. Sometimes, you are given additional information about this function F you are seeking, for example you might be told that F(2)=3. In those cases (called Initial value problems), you can solve for the exact value of C that's required for the equation to hold. nadav (talk) 12:04, 26 May 2007 (UTC)[reply]

I would say that you have misunderstood the fundamental theorem of calculus in a common way. Differentiation loses information; the derivative of a function tells us how it is turning at each point, but not an absolute position. Integration starts at a given position and accumulates all the twists and turns to give an absolute position everywhere. So we can integrate and then differentiate without loss, but if we differentiate then integrate we necessarily introduce uncertainty. A traditional way to denote that uncertainty is with a "constant of integration",

\int f(x)\,dx=F(x)+C.\,\!

It is easy to forget the constant, because it is irrelevant to definite integrals. Want a vivid mental picture? Suppose I tell you to go one block north and turn left. That's derivative-style information. But are we in New York or Hong Kong or Moscow? That's the constant of integration. And notice that no matter which city we are in, from the path we travel (the integral) we can recover the directions (the derivative). --KSmrq^T 12:44, 26 May 2007 (UTC)[reply]

This is a very useful way to think about the general problem, but I imagine it would be difficult to understand the concept of accumulating the twists and turns taken by the function without having studied definite integrals first. nadav (talk) 13:05, 26 May 2007 (UTC)[reply]

The easiest way to see that differentiation loses information is to consider what the derivative of 4 is, in your example. It's zero. It disappears. Now ask yourself what the integral of 0 is. Remember that the integral of a function f(x) that when differentiated gets you f back. Let f(x) = 0. But the derivative of any constant function is 0. We know that some constant must have been in the integral, so we have to put it back (hence the +C that you see).

Now let's try a real world example to help explain this. Say you have an equation that describes your distance from, say Tokyo, as a function of time. From this equation you can find the derivative to determine your velocity as a function of time. Note that your starting distance from Tokyo doesn't affect the results, though. Then, if you reverse the process, and integrate the equation for your velocity as a function of time, you don't quite have the original equation back, as the starting distance from Tokyo is lost. Here it is with numbers:

Distance from Tokyo: s = 2t + 5, Velocity: v = 2

Distance from Tokyo: s = 2t + 9, Velocity: v = 2

Now we integrate v = 2 and get s = 2t + C, where C is the unknown starting distance from Tokyo (perhaps s₀ would be a better name in this case). If we had taken the derivative of the velocity equation, we would have determined that there is no acceleration: a = 0. Integrating from this wouldn't tell us the starting velocity or starting position. If you wanted to keep all the information, you could record that a = 0, v_o = 2, and s_o = 5 (or 9). From that, you could rebuild the orginal displacement equations. StuRat 18:40, 26 May 2007 (UTC)[reply]

Irrelevant note: this example works perfectly on a line (where Tokyo is one point and your position is another). But on a plane (a map of Japan, say), looking at your distance from Tokyo and the derivative thereof only tells you about the radial component of your velocity—you could run in circles around Tokyo all day without changing your distance from the city center. [The same is true on a globe (or oblate spheroid), except that you need to be a little more careful when determining which component of velocity you have information about.] Tesseran 18:48, 27 May 2007 (UTC)[reply]

x₀

How do you pronounce x₀? My professor says "x not", but I have no clue why he says it like this. Thanks in advance, --Abdull 15:06, 26 May 2007 (UTC)[reply]

x naught.--0rrAvenger 15:12, 26 May 2007 (UTC)[reply]

Definitions can be found at Wiktionary and Dictionary.com. nadav (talk) 15:19, 26 May 2007 (UTC)[reply]

Thanks to both of you! --Abdull 15:28, 26 May 2007 (UTC)[reply]

Some emphasize that the 0 is a subscript by saying "x sub zero" or "x sub naught". -- Meni Rosenfeld (talk) 17:09, 26 May 2007 (UTC)[reply]

I've also heard "x initial", especially in physics class where it usually indicates some starting point you have to fill in. Black Carrot 01:56, 29 May 2007 (UTC)[reply]

Klein bottle

If and how can i graph a klien bottle on my ti-84 plus silver edition?

A Klein bottle is a three-dimensional figure which requires an extra dimension to be comprehended fully. What do you mean by "graph", since the ti-84 can only graph 2 dimensions.--0rrAvenger 17:30, 26 May 2007 (UTC)[reply]

That's not entirely correct. A Klein bottle is a two-dimensional surface, which can be embedded in 3D Euclidean space in a way which is not completely satisfactory (has self intersections). I think it can be properly embedded in 4D space. Either way, it is probably too much for a TI-84 Plus series calculator to handle. -- Meni Rosenfeld (talk) 17:57, 26 May 2007 (UTC)[reply]

I haven't used one in years, but I think TI-89 series should be able to handle graphing the Klein bottle. The lower level TI calculators such as the TI-84 don't have built in 3d graphing functionality, I believe. They perhaps could be programmed in C or assembly to do this (I figure a TI-BASIC program would be too slow), but that sounds difficult. nadav (talk) 18:09, 26 May 2007 (UTC)[reply]

I'm not sure what is sought for here. Is it to start with a two-dimensional surface embedded in a four-dimensional space and project it down onto the two-dimensional screen? Heh, I would probably just draw the contour by hand from an original like the one to the right. ;-) Anyway, I actually wrote a program (in TI-BASIC) on my TI-83 once that renders a 3d scene consisting of a single sphere illuminated by a single, distant light source, seen from a distant observer. I.e., it was essentially the Moon seen from Earth. Greyscale was achieved with dithering. It took like 10 minutes to render, but it looks nice. :-) —Bromskloss 18:21, 26 May 2007 (UTC)[reply]

I couldn't resist sharing it. —Bromskloss 19:27, 26 May 2007 (UTC)[reply]

Meni Rosenfeld (and anyone else who is interested): That "not completely satisfactory embedding" you are talking about is called an "immersion". --Tugbug 23:22, 26 May 2007 (UTC)[reply]