Mathematics Thread

[TheNextFeliks]

Ok thanks.

 

[brandbest1]

@A Leman yeah I just learned l'hopitals rule just recently (as well as maclaurin and Taylor series).

EDIT: Mathematica... I tried the free trial just now... 4 GB lol.

 

[CarlBrannen]

Well I'm starting to work on my thesis. It's going to be in gravitation. The calculations are messy but easy so I'm using the computer algebra software "MAXIMA".

Some of you might find it handy. It does some of the things Mathematica does but has the advantage of being free:
http://maxima.sourceforge.net/

 

[CarlBrannen]

Cmowla, last time I was in college I got a free or cheap version of mathematica. But I'd rather use the free product.

 

[CarlBrannen]

At one time I had high hopes for Mathematica and other computer algebra systems. Let me pose here a problem that I've not seen one of these systems solve, but which is fairly easy for a human with a little time. This is a generalization of quadratic equations, that is, instead of one quadratic equation in one unknown you begin with several quadratic equations in several unknowns. Here's the version with 6 unknowns. (By the way, this has applications to elementary particle theory.):

There are six complex numbers, I, J, K, R, G, and B and they satisfy the following six equations:

I = I*I + J*K + K*J + R*R + G*G + B*B,
J = I*J + J*I + K*K +R*G + G*B + B*R,
K = I*K + J*J + K*I + R*B + G*R + B*G,
R = I*R + J*G + K*B + R*I + G*K + B*J,
G = I*G + J*B + K*R + R*J + G*I + B*K,
B = I*B + J*R + K*G + R*K + G*J + B*I.

Two easy solutions are (I,J,K,R,G,B) = (0,0,0,0,0,0) and (1,0,0,0,0,0).

Mathematical note:
If you look at the quadratic terms, all of the 36 possible ways of multiplying one of {I,J,K,R,G,B} by one of the same set show up. Which element things end up with follows the multiplication rules of a finite group, that is, the group of permutations on 3 elements. Then "I" is the identity permutation, "J" and "K" are the permutations that rotate all three elements and "R", "G" and "B" are the three remaining permutations that swap two elements and leave the remaining one unchanged.

The relationship to quantum field theory is that this is a calculation in "Hopf algebra" and from the physics point of view what's being done here is a search for the propagators of the (finite) field theory. That is, you're solving the equation P*P = P for P an element of the Hopf algebra.

The math guys call this multiplication (i.e. the P*P = P) the "group algebra over the complex numbers of the finite group G", where in this case G is the set of permutations on three elements. This is defined in Jacobson's book "Basic Algebra I" in section 7.1 "Definition and Examples of Associative Algebras". In my copy (copyright 1974), it's on page 388.

So the math theory is somewhat involved, but the actual calculation is straight quadratic equations. My experience is that the computer algrebra systems can't handle it. Maybe they've improved since the last time I tried it.

 

[CarlBrannen]

At one time I had high hopes for Mathematica and other computer algebra systems. Let me pose here a problem that I've not seen one of these systems solve, but which is fairly easy for a human with a little time. This is a generalization of quadratic equations, that is, instead of one quadratic equation in one unknown you begin with several quadratic equations in several unknowns. Here's the version with 6 unknowns. (By the way, this has applications to elementary particle theory.):

There are six complex numbers, I, J, K, R, G, and B and they satisfy the following six equations:

I = I*I + J*K + K*J + R*R + G*G + B*B,
J = I*J + J*I + K*K +R*G + G*B + B*R,
K = I*K + J*J + K*I + R*B + G*R + B*G,
R = I*R + J*G + K*B + R*I + G*K + B*J,
G = I*G + J*B + K*R + R*J + G*I + B*K,
B = I*B + J*R + K*G + R*K + G*J + B*I.

Two easy solutions are (I,J,K,R,G,B) = (0,0,0,0,0,0) and (1,0,0,0,0,0).

Mathematical note:
If you look at the quadratic terms, all of the 36 possible ways of multiplying one of {I,J,K,R,G,B} by one of the same set show up. Which element things end up with follows the multiplication rules of a finite group, that is, the group of permutations on 3 elements. Then "I" is the identity permutation, "J" and "K" are the permutations that rotate all three elements and "R", "G" and "B" are the three remaining permutations that swap two elements and leave the remaining one unchanged.

The relationship to quantum field theory is that this is a calculation in "Hopf algebra" and from the physics point of view what's being done here is a search for the propagators of the (finite) field theory. That is, you're solving the equation P*P = P for P an element of the Hopf algebra.

The math guys call this multiplication (i.e. the P*P = P) the "group algebra over the complex numbers of the finite group G", where in this case G is the set of permutations on three elements. This is defined in Jacobson's book "Basic Algebra I" in section 7.1 "Definition and Examples of Associative Algebras". In my copy (copyright 1974), it's on page 388.

So the math theory is somewhat involved, but the actual calculation is straight quadratic equations. My experience is that the computer algrebra systems can't handle it. Maybe they've improved since the last time I tried it. If Mathematica can solve this now, I'd be very motivated to get it. I'm pretty sure that MAXIMA hasn't improved to the point where it can solve this.

 

[vcuber13]

Solve for real values of x.

\( Log_2 (2^{x-1} + 3^{x+1}) = 2x - Log_2 (3^x)
\)

 

[ben1996123]

Solve for real values of x.

\( log_2(2^{x-1}+3^{x+1})=2x-log_2(3^x)
\)

wulfram says its not possible symbolically

 

[5BLD]

wulfram says its not possible symbolically

It is... there's an answer. I know it but it requires fudging so im still trying to solve it properly

Spoiler: answer

x = -(log(2))/(log(3/2)) = (log(2))/(log(2)-log(3))

 

[vcuber13]

wulfram says its not possible symbolically

wolfram says there is an exact elementary answer

 

[5BLD]

yay

Spoiler

(log(2^(x-1)+3^(x+1)+(log(3^x))))/(log(2))= 2 x
log(3^x (2^(x-1)+3^(x+1))) = 2 x log(2)
3^x (2^(x-1)+3^(x+1)) = 4^x
and then it gets too long to type out...

 

[Ross The Boss]

this thread makes me feel like such an *****. lol

edit: seriously, i d i o t is censored? wow.

 

[ben1996123]

this thread makes me feel like such an *****. lol

edit: seriously, i d i o t is censored? wow.

ino lol tis funee

ïdiot

hear's my question for somepony two solve

pruf that \( \int_{-\infty}^{\infty} e^{\frac{-x^2}{2}}dx=\sqrt{2\pi} \)

 

[TheNextFeliks]

Lol. I have proof:
Given circle with points (-1,0) and (1,0)
Prove that equation of circle is x^2-2yk+y^2=1 with center (0,k)
I know k must be 0 but I can't figure out how to prove it.

 

[vcuber13]

k need not be 0
circles have equations in the form (x-a)^2 + (y-b)^2 = r^2, with centre (a,b) and raduis r
x^2+(y-k)^2=r^2
r^2=k^2+1
x^2+y^2-2yk+k^2=k^2+1
x^2-2yk+y^2=1


i dont know if youll like this ben:

Spoiler

d/dx erf(x)=2*Pi^(-0.5)*e^(-x^2)
d/dx erf(2^(-0.5)*x)=(2/Pi)^(0.5)*e^(-0.5*x^2)
erf(2^(-0.5)*t)=(2/Pi)^(0.5)* integral (0,t) [e^(-0.5*x^2)]dx
2*sqrt(Pi/2)* erf(2^(-0.5)*t)=2* integral (0,t) [e^(-0.5*x^2)]dx
sqrt(2*Pi)* erf(inf)=2* integral (0,inf) [e^(-0.5*x^2)]dx
sqrt(2*Pi)=integral (-inf,inf) [e^(-0.5*x^2)]dx

 

[A Leman]

hear's my question for somepony two solve

pruf that \( \int_{-\infty}^{\infty} e^{\frac{-x^2}{2}}dx=\sqrt{2\pi} \)

I am too lazy to try to type a complete proof on SS because mine is a bit long so I will skip obvious steps. Anyway, the easiest way to solve some problems it to think of a similar problem in an extra demension. In this case, z=e^(-x^2-y^2). If you find the volume under that surface at the verticies (±b,±b). it is
int([-b,b] e^(-x^2)dx)*int([-b,b] e^(-y^2)dy)=4*(int([0,b] e^(-x^2) dx))^2
V over the whole xy plane is V=lim(b->∞) 4(int([0,b] e^(-x^2) dx)^2= 4(int([0,∞] e^(-x^2) dx)^2
this can also be found in polar coordinates as V=lim(a->∞) int[0,2pi]int[0,a] (e^(-r^2)*r dr*dθ)=lim(a->∞) pi*(1-e^(-a^2))=pi
so 4(int([0,∞] e^(-x^2) dx)^2=pi
int([0,∞] e^(-x^2) dx=sqrt(pi)/2
int([-∞,∞] e^(-x^2) dx=sqrt(pi)
now u substitute u=x/sqrt(2) so
\( \int_{-\infty}^{\infty} e^{\frac{-x^2}{2}}dx=\sqrt{2\pi} \)

btw, how did you type your integral on SS. I'm curious.

 

[ben1996123]

Lol. I have proof:
Given circle with points (-1,0) and (1,0)
Prove that equation of circle is x^2-2yk+y^2=1 with center (0,k)
I know k must be 0 but I can't figure out how to prove it.

Spoiler

(-1,0) and (1,0) centre (0,k) => radius = 1

x² + y² - 2yk = 1
x² + (y-k)² - k² = 1
x² + (y-k)² = 1+k²

1+k² = 1 => k = 0

i dont know if youll like this ben:

Spoiler

d/dx erf(x)=2*Pi^(-0.5)*e^(-x^2)
d/dx erf(2^(-0.5)*x)=(2/Pi)^(0.5)*e^(-0.5*x^2)
erf(2^(-0.5)*t)=(2/Pi)^(0.5)* integral (0,t) [e^(-0.5*x^2)]dx
2*sqrt(Pi/2)* erf(2^(-0.5)*t)=2* integral (0,t) [e^(-0.5*x^2)]dx
sqrt(2*Pi)* erf(inf)=2* integral (0,inf) [e^(-0.5*x^2)]dx
sqrt(2*Pi)=integral (-inf,inf) [e^(-0.5*x^2)]dx

ew erf(x) dunrike it

I am too lazy to try to type a complete proof on SS because mine is a bit long so I will skip obvious steps. Anyway, the easiest way to solve some problems it to think of a similar problem in an extra demension. In this case, z=e^(-x^2-y^2). If you find the volume under that surface at the verticies (±b,±b). it is
int([-b,b] e^(-x^2)dx)*int([-b,b] e^(-y^2)dy)=4*(int([0,b] e^(-x^2) dx))^2
V over the whole xy plane is V=lim(b->∞) 4(int([0,b] e^(-x^2) dx)^2= 4(int([0,∞] e^(-x^2) dx)^2
this can also be found in polar coordinates as V=lim(a->∞) int[0,2pi]int[0,a] (e^(-r^2)*r dr*dθ)=lim(a->∞) pi*(1-e^(-a^2))=pi
so 4(int([0,∞] e^(-x^2) dx)^2=pi
int([0,∞] e^(-x^2) dx=sqrt(pi)/2
int([-∞,∞] e^(-x^2) dx=sqrt(pi)
now u substitute u=x/sqrt(2) so
\( \int_{-\infty}^{\infty} e^{\frac{-x^2}{2}}dx=\sqrt{2\pi} \)

btw, how did you type your integral on SS. I'm curious.

thats pretty much what i did

[noparse]\( \int_{-\infty}^{\infty} e^{-\frac{x^2}{2}}dx \)
\( \int \)[/noparse]

 

[vcuber13]

Spoiler

(-1,0) and (1,0) centre (0,k) => radius = 1

x² + y² - 2yk = 1
x² + (y-k)² - k² = 1
x² + (y-k)² = 1+k²

1+k² = 1 => k = 0

what is this showing?

 

[TheNextFeliks]

what is this showing?

k=>0. You got it.

 

[vcuber13]

so, k cannot equal -3?