Let's discuss Mathematics

A nice little theorem I had to prove in a topology course a couple of days ago:

Given a set X and two metrics d1 and d2 on this set X. Prove that if sequences a_n that converge in (X, d1) if and only if they converge in (X, d2) then the limit to which they converge is identical in d1 and d2.
 
Is d3 = max (d1, d2) a metric as well? Cos it would be easy then ;)
 
It might fail the triangle inequality.
 
Yeah that's the only one it would fail.

I'm not going to try and prove it on Friday night though ;)
 
My proof:

Given a sequence a_n that converges to x in the d1 metric -> take a look at the sequence

b_n = a_(n/2) if n is even
= x if n is not even

Clearly this sequence also converges to x in the d1 metric. By assumption, it must converge to some y in the d2 metric. But y must be equal to x, otherwise the sequence wouldn't converge. So the sequence b_n converges to x in d2, that means a_n also converges to x in d2.
The argument is symmetric, so a sequence a_n that converges to y in d2 also converges to y in d1.
 
I have a potentially stupid question. If I have a simple curve (y=mx+c), and I want to divide the curve into sections of equal area (trapeziums, basically), how do I find out what values of x satisfy that?

E.g. I have a line that starts at y=750 and ends at y=540, and I want to divide them into 100 individual trapeziums of equal area. How do I do that? Am I being stupid????
 
Integrate the curve.

Divide area by 100.

That's the area each trapezium needs to be. The start from one side and just allocate appropriate amount on x axis to get the area of the trapezium to be correct (area = (ystart + yend) * deltaX / 2, solve for deltaX and yend).

EDIT: ystart and yend are in terms of x too remember.
 
So there's no way of working out the x's without going through each step? There's no function that will magically generate them for me? :( Oh well.
 
I'm sure there is, it's just algebra.
 
Step away from the computer and use a pencil and a bit of paper then. Usually works wonders.

Simplify it first, use 2 trapeziums, then see what happens when you increase it to 3, etc.

I'm not going to do it for you, lazy sod ;)
 
I would, but I don't know the formula for the area of a trapezium... so instead I'll pose an integral riddle.

How, using 2 steps, would you solve this integral?

[tex]\int \frac{e^{x^2}}{x} dx[/tex]

If the LaTEX doesn't go through, here's an image.

Well, I'll edit the image in.

EDIT:
texclip20100913025721.png
 
I posted the formula for the area of the trapezium ;) It's easy because it has the x axis as one of the sides and 2 of the sides are parallel to the y axis.

Hint: It's the area of a rectangle + the area of a triangle.

the rectangle has are deltaX * ymin. The triangle has area deltaX * (ymax - ymin ) / 2

So the area is

deltaX * (ymin + (ymax - ymin) / 2)

= deltaX * ( ymin - ymin/2 + ymax/2 )

= deltaX * ( ymin + ymax ) / 2
 
I did that, Para, and I couldn't do it! That's why I'm asking here!!

This is what happened:

63c20c1f.jpg
 
Where is y = mx + c in that calculation? I think that might help ;)
 
IdiotsOpposite: meh, integration, I hate it.

I suppose it is an integration by parts thing?
 
You can't hate integration! I love it!

Spoiler :


STEP 1:

Substitute u=x^2

integral 1/2 e^(u)/u du

STEP 2:

Integrate:

1/2 Ei(u) + C

Unsubstitute (not a step):

1/2 Ei(x^2) + C


EDIT: The whole message has been changed.
 
Did you do the substitution correct?

Don't you have to write dx in terms of du as well? It's been a long time since I did any integration.
 
Well, what I did is...

e^(x^2)/x dx

If u = x^2

then du = 2x dx

So we have e^(u)/x * du/2x

Which is...

1/2 e^(u)/x^2 du

Which is 1/2 e^(u)/u du

And integrating that gives you Ei(u)/2

Which is Ei(x^2)/2 + C.

YAY! It worked.

Also, integration is one of my favorite areas and it will stick with me at least through differential equations so...
 
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