Let's discuss Mathematics

[Robert FIN]

I shure hope this is understandable

Spoiler :

Hey, thanks a lot for your time and effort

The calculation worked when I removed the D of the inner function (the one that was not needed). The answer happened to be the same when I did it the wrong way or the right way so that is why I was so confused.

Cheers!

 

[Lohrenswald]

Now I shall ask for help

an integral
the integrand is:
(x-x0) * e^(-mw/h * (x-x0)^2) * (1+sqrt(2mw/h)*(x-x0) + 2mw/h * (x-x0)^2)
x is the variable, everything else is constants
the limits are from minus to plus infinite

how does one solve this (by hand)?

 

[Kyriakos]

(disclaimer: non-helpful post)

So with the integrand being (x-x0)^2*[(e^stuff)*1+sqrt(other stuff)+2mw/h]?

I hate west calculus, not going to help here (can't), but after the stuff e is raised in are turned to log, and assuming x0 is one of x's variations, can't you do the antiderivative stuff using the fundamental theorem of west calculus?

I haven't yet felt like approaching that again (disliked it in the end of highschool as well)

 

[Lohrenswald]

here is a more clear representation

this doesn't look like the fundamental theorem can be used easily

 

[Atticus]

1. Get rid of the parentheses inside the integral.
2. You have then an integral of sum of three things. You can integrate the things separately.
3. For the first one the multiplier of e^stuff is the derivative of [stuff], almost, so you know how to integrate it.
4. For the others this doesn't work. Use partial integration instead to get the power of multiplier what you want. You may have to do that many times.
5. First though, substitute x for x-x_0. It makes calculations easier and doesn't require any extra hassle, since you're integrating from -inf to inf.

 

[stormtrooper412]

yeah introduce a new variable t = x-x0 and for the sake of simplicity, put C = mw/h since it's repetitive. If you have issues, I can write it on paper and scan it, but it won't be until tonight

 

[Robert Can't]

Does anyone know if the notation L^2(G) is?
Its coming up with regards to Class functions and I'm not sure if its just a different notation for Class(G) or if it means something else

 

[Atticus]

I'm not sure what you mean by class, but if G is a set, that means the set of measurable functions whose square has finite integral.

For example (suppose G=[0,1]):

f is in L^2 ([0,1])
if and only if
f : [0,1] -> R is measurable and \int f^2 < infinity.

 

[Robert FIN]

A football player scores when performing a penalty kick with the probability of 80% (0.8). How many times he has to try until he will get a goal with probability of 99% (0.99) or more?

Answer:

Spoiler :

At least 3 times

I don't get it... I tried to use log but that went wrong. It just doesn't work to power 0.8 (like 0.8^2) because the answer will be less than 0.8....

Doing revision for a math test. Let's see if someone is that fast, lol

 

[Harv]

A football player scores when performing a penalty kick with the probability of 80% (0.8). How many times he has to try until he will get a goal with probability of 99% (0.99) or more?

Answer:

Spoiler :

At least 3 times

I don't get it... I tried to use log but that went wrong. It just doesn't work to power 0.8 (like 0.8^2) because the answer will be less than 0.8....

Doing revision for a math test. Let's see if someone is that fast, lol

80% probability of scoring a goal with one kick:
20% probability of missing

P = 1 - 0.2^1 = 0.8

Three kicks

P = 1 - 0.2^3 = 0.992
99.2%

 

[Robert FIN]

80% probability of scoring a goal with one kick:
20% probability of missing

P = 1 - 0.2^1 = 0.8

Three kicks

P = 1 - 0.2^3 = 0.992
99.2%

Thanks, I did not read this in time and kinda failed the test but at least I know how to do it now ^_^

 

[Atticus]

Here's two puzzles:

1. You want to cut a piece of Brie cheese in two equal pieces. The piece is a segment of a circle with the angle \alpha. Because of the preferences of the cheese eaters the other piece should contain the tip and the other the arc. Where do you cut?

2. What is the average distance of two points in a square?

 

[Mise]

I worked it out in Excel but got no useful insight at all. You can put that on my tombstone

 

[Atticus]

I'm not sure if there's any useful insight. I've just come across with these two questions, but never gave them thought myself. So, to force myself to do that I posted them here.

I'm not entirely sure how you'd even do the latter with Excel. Take a very fine grid and calculate?

 

[Mise]

Just a whole bunch of simulations with pairs of coordinates. See where the average tends to...

 

[Atticus]

For fun's sake I did the same with python:

Spoiler :

from math import sqrt
from random import random

def d(x,y):
  dx = (x[0] - y[0])**2
  dy = (x[1] - y[1])**2
  return sqrt(dx+dy)

def randomDist(pointCount):
  sum = 0.0
  for i in range(0,pointCount):
    x = [random(),random()]	
    y = [random(),random()]	
    sum += d(x,y)
  return ( sum / pointCount )

# MAIN:

print(randomDist(1000000))

A few runs puts it in the interval [0.521, 0.522]. Gonna think about how to do it properly later.

 

[dutchfire]

Here's two puzzles:

1. You want to cut a piece of Brie cheese in two equal pieces. The piece is a segment of a circle with the angle \alpha. Because of the preferences of the cheese eaters the other piece should contain the tip and the other the arc. Where do you cut?

2. What is the average distance of two points in a square?

1) Based on mathematical intuition, I believe it doesn't depend on \alpha. (If you had a quarter circle and had to divide it in half, you could just divide the full circle in an inner and outer half, and then cut it into 4 pieces later).
Assuming you make a 'round' cut at constant radius, it has to be at 1/2 Sqrt(2) of the radius of the full Brie cheese.

 

[Robert Can't]

I've managed to get the answer to the second problem in the form of an integral (a horrible looking integral but an integral all the same). Now I just need to integrate it. Will get back to you as soon as I've found the answer.

 

[Atticus]

I was thinking you'd make one straight cut, not a circular one.

As for puzzle two, I'm hoping there would be some easier way than the straight up integration of distances.

 

[uppi]

I was thinking you'd make one straight cut, not a circular one.

In that case I get x = sqrt(alpha/(2*sin(alpha))), where x is the distance from the tip (as a fraction of the radius) at both edges for the two points that define the cut and alpha is in radians.

For alpha << 1 this reduces to dutchfire's answer.