[Wildlife]

I really have nooo idea wat ur talkin about...........

[Rampage]

Originally posted by CP5670
(have not been able to get an exact indefinite integral...)

Can you get an exact integral using regular calculus?

I mean:  [INT] f(x) dx = F(x) + C

"C" could be ANYTHING!  You need to know the derivative of the function to find "C".

[CP5670]

Well, the problem with that equation is that it is an implicit function and cannot be solved algebraically for either variable. (at least, I have not been able to do so yet ) I am not sure if this works, but it might be possible to treat it as a function of another variable, say something like z=f(x,y) and find a double integral with respect to x and y.

Does anyone know if there is a way to get integrals for implicit functions? (a chain rule for integrals maybe?)

There is something called Euler's method for solving equations of the form f(x,y)=dy/dx numerically, but that only give approximate results, and I like things to be precise.

Finding C once you have the antiderivative is pretty easy; you just find a point on the original equation, substitute it into the new one and solve for C. It is the antiderivative that's the hard part here.

[Galemp]

Originally posted by Wildlife
I really have nooo idea wat ur talkin about...........

Don't worry, you're not missing anything. It's just really advanced mathematics, taught in secondary schools but mostly universities.

[Vortex]

Yeh I'm just getting to this in my HL class, don't scare me yet CP!

[Setekh]

Mmmmmvectors.

CP, you would be in heaven in my school. Maths this, maths that... how old are you again?!!

[CP5670]

I'm 15.4 or so, but hey the stuff is not that advanced or anything... :D

This is a bit harder material, but that makes it lots of fun:

http://mathworld.wolfram.com/Euler-LagrangeDifferentialEquation.html
http://mathworld.wolfram.com/LegendreDuplicationFormula.html

Cool stuff, huh? I read this stuff constantly in my spare time; it's simply great...

[Setekh]

Sounds like you're about my age. What's your birthday again?

Oh, and I gotta agree. Maths is just plain cool. PS. if fractal functions were in there, I'd vote for that definitely...

[wEvil]

How about subtracting Guinness from your glass?

dont tell me none of you are training for St.Patricks' Day????



Maths on the other hand...I like the theory but I havent done it since i was 16.  four years..yipes.

[Rampage]

This is saaaad:

Six votes for "Subtracting alcohol from glass functions"

[Thorn]

*tries to make sense of whats being said*
This is making my head hurt...
Well, back to subtraction for me
*wanders off to the fridge*

[edit] actually, I kinda like the Irish function...

This coffee isnt Irish enough!

[/edit]

[Setekh]

Originally posted by wEvil
Maths on the other hand...I like the theory but I havent done it since i was 16.  four years..yipes.

I hear that most of the people around me don't use maths on even a monthly basis after their high-school years. Except add and subtract, maybe.

I do know, though, two people who are doing courses in Pure Mathematics at university....

[Vortex]

I'm 15.4

OMG only a math person would say that... :D:D (I like math too so don't think I'm insulting you)

[Vortex]

Hehe, my favorite proof (well actually it doesn't work but it's funny to show to people who don't see the flaw in it).

The 2=1 proof

Let a=b

a = b
a^2 = ab
(a^2) - (b^2) = ab - (b^2)
(a+b)(a-b) = b(a-b)
a+b = b

Since a=b

b+b = b
2b = b
2=1

[CP5670]

That was actually sort of inaccurate and had been calculated a few months ago; my real age is closer to

15.51780821 +
 ¥
 S (91789021/100000000ª)
a->1

a = b
a^2 = ab
(a^2) - (b^2) = ab - (b^2)
(a+b)(a-b) = b(a-b)
a+b = b

Since a=b

b+b = b
2b = b
2=1

uh...how did you get b+b=b from a=b? also, the fifth step should actually be (a-b)(a+b-b)=0 to ensure that no solutions are left out. :D So either a=b or a=0 could be the solution.

[Vortex]

Ok I'll do it again:

Let a = b

a = b
a^2 = ab (multiply both sides by a)
a^2 - b^2 = ab - b^2 (subtract both sides by b squared)
(a+b)(a-b) = b(a-b) (factor the left side, factor a b out on the right side)
a+b = b (divide both sides by (a-b))
b+b = b (since a=b has been stated at the beginning, sub b in for a)
2b = b (just some cleanup)
2=1 (divide both sides by b)

[aldo_14]

Originally posted by Setekh


I hear that most of the people around me don't use maths on even a monthly basis after their high-school years. Except add and subtract, maybe.

I do know, though, two people who are doing courses in Pure Mathematics at university....

Haven't used it since last year at Uni.... well, a little big of order n calculation, but that's really just programming.

[CP5670]

I am probably going into theoretical physics, which is pretty much all math (especially diff equations), so I would be using it all the time.

[Setekh]

Lol, then you can wear even bigger boots when there's a science discussion 'round here.

[Zazicle]

Originally posted by Vortex
Ok I'll do it again:

Let a = b

a = b
a^2 = ab (multiply both sides by a)
a^2 - b^2 = ab - b^2 (subtract both sides by b squared)
(a+b)(a-b) = b(a-b) (factor the left side, factor a b out on the right side)
a+b = b (divide both sides by (a-b))
b+b = b (since a=b has been stated at the beginning, sub b in for a)
2b = b (just some cleanup)
2=1 (divide both sides by b)

You can't divide through by (a-b) because (a-b) = 0.