Math Challenge - March 2019

[fbs7]

wow! holy-choochoo! I didn't think of adding the exponents!

well done! thank you!

so the first term is really 2, not sqrt(2)... huh... I guess I'm not graduating to Ant-Padawan level 2 at all! oh no!

 

[fresh_42]

wow! holy-choochoo! I didn't think of adding the exponents!

This is not really true. You took the logarithm of an assumed limit and did exactly this: added the exponents - now a floor below, which made a better typeset.

 

[fbs7]

a-ha! thank you! so I'm calling myself Ant-Padawn Level 1 1/2.. no... 1 1/3

And I used what you advised about Polynomial Division... errr... Thingie.. that was very smart!

 

[fbs7]

On Yoda-level Question 4:

The question is beyond my level, but I'm trying to at least understand the item (a); I'm probably not reading the notation correctly, though:

$x*(y*z) = x*(1/2y+1/2z) = 1/2(x*y)+1/2(x*z) =$
$1/2(1/2x+3/8y+1/8z) + 1/2(1/2x+1/2z) =$
$1/2x + 3/16y+5/16z$

$(x*y)*z = (1/2x+3/8y+1/8z)*z =1/2(x*z)+3/8(y*z)+1/8(z*z) =$
$1/2(1/2x+1/2z) + 3/8(1/2y+1/2z) + 1/8(z) =$
$1/4x+3/16y+9/16z$

It doesn't seem to be associative, so I'm making some mistake. Tried as I could, still can't see my mistake.

 

[fresh_42]

On Yoda-level Question 4:

The question is beyond my level, but I'm trying to at least understand the item (a); I'm probably not reading the notation correctly, though:

$x*(y*z) = x*(1/2y+1/2z) = 1/2(x*y)+1/2(x*z) =$
$1/2(1/2x+3/8y+1/8z) + 1/2(1/2x+1/2z) =$
$1/2x + 3/16y+5/16z$

$(x*y)*z = (1/2x+3/8y+1/8z)*z =1/2(x*z)+3/8(y*z)+1/8(z*z) =$
$1/2(1/2x+1/2z) + 3/8(1/2y+1/2z) + 1/8(z) =$
$1/4x+3/16y+9/16z$

It doesn't seem to be associative, ...

Correct.

... so ...

Wrong.

... I'm making some mistake. Tried as I could, still can't see my mistake.

Me neither. Why do you expect associativity? It is an example for a non associative multiplication, i.e. a non associative algebra. Lie algebras or the Octonians are other prominent examples of non associative algebras. Both play important roles in physics, the former a bit more than the latter.

 

[fbs7]

Oh, good Lord! I was trying to find a way to prove that was associative, instead of answering if that was associative or not! I'm such an idiot.

By the way, I came with a property of an algebra A on vectors over a basis $x_i$ such that

$x_i*x_j = l_{ijk}*x_k$

that would make it associative; without taking much space, for three vectors u, v, w in Einstein's notation:

$u = u_i*x_i; v = v_i*x_i; w = w_i*x_i;$

then $u*(v*w) = (u*v)*w$ makes $u_a*x_a*(v_b*x_b*w_c*x_c) = (u_d*x_d*v_e*x_e*w_e)*w_f*x_f$

as $u_?, v_?, w_?$ are all free variables, then $d=a, e=b, f=c$, and $x_a*(x_b*x_c)=(x_a*x_b)*x_c$, that is the algebra is associative if the multiplication of the basis is associative; then, replacing with the expression for multiplication of the basis:

$x_a*(x_b*x_c) = x_a*(l_{bcg}*x_g) = l_{bcg}*x_a*x_g = l_{bcg}*l_{agh}*x_h$
$(x_a*x_b)*x_c = (l_{abi}*x_i)*x_c = l_{abi}*x_i*x_c = l_{abi}*l_{icj}*x_j$

as $x_h$ and $x_j$ are independent (that is, I assume they are independent), then $h=j$; then, eliminating $x_h$ and renaming $i=g$ and $d=h$ to make it more readable,

$l_{abi}*l_{icd} = l_{bci} * l_{aid}$ for any $a, b, c, d$

And that's where my skill ends; now, question on that... this seems like a tensor operation of some kind, is that true? That is, is $l_{ijk}$ a 3-tensor, and the expression above is a tensor operation of some kind?

 

[fresh_42]

Every bilinear multiplication can be written as a tensor. See the example of Strassen's algorithm here:
https://www.physicsforums.com/insights/what-is-a-tensor/
which is an example how matrix multiplication is written as a tensor.

As long as you do not put any additional restraints on $l_{ijk}$ as in the case of genetic algebras (or any other class of algebras), as long do you have an arbitrary algebra.

 

[fbs7]

Thank you!