Testwiki:Reference desk/Archives/Mathematics/2011 February 15

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February 15

Unusual description of the Klein bottle

Hi all,

I'm trying to show that the klein bottle K may be described as $(S^1\vee S^1){\cup }_f{B}^2$ for a suitable $f:S^1\to S^1\vee S^1$, where $\vee$ is the wedge union (identifying a single point in each circle) and ${\cup }_f$ is the gluing of $B^2$ along the equivalence relation which equates $x\in S^1$ with $f(x)\in S^1\vee S^1$ (as in Hatcher's book chapter 0). Use this to give a presentation of ${\mathrm{\Pi }}_1(K)$ with 2 generators and 1 relation.

I asked a more basic question a while back - http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Mathematics/2011_January_30#Explanation_of_a_few_topology_concents - and i'm trying to think along related lines. Given the identifications of the edges of the square of the klein bottle, do we take the map which, while x moves around the circle, goes around 1 loop of $S^1\vee S^1$, then around the other loop, then the first loop in the same direction, then the second loop in the opposite direction? This corresponds to the edges of the klein bottle square, with 2 sides having the same orientation and 2 sides opposite orientation, I think. Does that work okay for part (i)? Thanks very much, help greatly appreciated. Typeships17 (talk) 02:18, 15 February 2011 (UTC)

Yes. Algebraist 02:27, 15 February 2011 (UTC)

A derivatives question

A problem I invented, but am unable to solve: A vehicle is at a standstill. It then begins to travel with a derivative of speed which uses the unit m/s⁸, which, for the sake of the question, we'll call it "hurtle", let's say. The vehicle has a constant "hurtle" of 8m/s⁸. How far away, in metres or kilometres, will the vehicle be after 8 seconds, in relation to its starting point? Brambleclawx 03:33, 15 February 2011 (UTC)

Assume your vehicle is on a perfectly flat surface, with starting position x₀. By "It then begins to travel with a derivative of speed which uses the unit m/s⁸" I assume you mean the eighth derivative of the vehicle's position with respect to time is a constant 8, or in symbols f⁽⁸⁾(t)=8. To solve this you would just compute a multiple integral from x₀ to 8 (assuming x₀<8, i.e. the vehicle is moving forward and not backward), specifically, integrating eight times. This would look like:

$\underset{0}{\overset{8}{\int }}...\int 8dt...dt$. Notice the assumption that x₀, v₀, a₀, etc. are all 0, which is the same as saying all the constants of integration are 0. [Sorry, I hit enter before I was ready to save] 72.128.95.0 (talk) 04:22, 15 February 2011 (UTC)

You're integrating over time from 0 to 8, not over location from $x_0$ to 8. It's completely irrelevant whether $x_0<8$. Also, if you're writing it as an integral you can't use the same variable letter. The full expression is

$\underset{0}{\overset{8}{\int }}\underset{0}{\overset{t_1}{\int }}\underset{0}{\overset{t_2}{\int }}\underset{0}{\overset{t_3}{\int }}\underset{0}{\overset{t_4}{\int }}\underset{0}{\overset{t_5}{\int }}\underset{0}{\overset{t_6}{\int }}\underset{0}{\overset{t_7}{\int }}8d{t}_{8}d{t}_{7}d{t}_{6}d{t}_{5}d{t}_{4}d{t}_{3}d{t}_{2}d{t}_1$

-- Meni Rosenfeld (talk) 11:29, 15 February 2011 (UTC)

The short answer is $8\frac{m}{s^8}\frac{(8s{)}^8}{8!}\approx 3329m$.

For a longer answer you may want to start with Taylor series. -- Meni Rosenfeld (talk) 11:08, 15 February 2011 (UTC)

..and you really wouldn't want to be sitting in that vehicle for 8 seconds. It starts off very gently - after 1 second, its speed is about 1.5 mm/s and acceleration is about 1 cm/s². After 4 seconds, performance is more respectable - speed is 26 m/s, so almost 60 mph, and acceleration is about 45 m/s², which is about 4.5 g. But after 8 seconds it is travelling at 3.3 km/s and its acceleration is a whopping 2912 m/s², which is nearly 300g ! It reaches escape velocity about 1.5 seconds later, and approaches 1/10 light speed within 30 seconds. Gandalf61 (talk) 12:43, 15 February 2011 (UTC)

If you want to state the problem completely, you should probably clarify that by "standstill" you do indeed mean that the velocity and all other derivatives are zero at t=0. The word "standstill" by itself could just indicate that the velocity is zero, which in this problem is not enough to solve for the other integration constants. Staecker (talk) 12:44, 15 February 2011 (UTC)

I think you've all answered my question very well... and yes, by standstill, I meant that the car was not moving at all, and all other derivatives were 0. Thank you! By the way, Gandalf61's points really were quite amusing. Brambleclawx 04:05, 16 February 2011 (UTC)