On 18-Jan-09 22:49, Orchid XP v8 wrote:
>>> That's pretty trivial mathematics compared to tensor calculus type 
>>> stuff.
>>> :-)  I'm not too bad at discrete math.
>>
>> Tensor calculus isn't so bad.  Among other things, I gave my nephew my 
>> tensor
>> calculus book this past summer for his baby shower.  Gotta get kids 
>> started
>> early these days, y'know.
> 
> I still can't figure out what a "tensor" is. :-P
> 
multidimensional vector.

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Orchid XP v8 <voi### [at] devnull> wrote:
> I still can't figure out what a "tensor" is. :-P

scalar = tensor of rank 0
vector = tensor of rank 1

For example, the stress tensor is a rank 2 tensor since you quantify stress in a
plane and thus need two indices to describe which plane you're talking about.
So the (1,3) element if you wrote it as a matrix would correspond to stress in
the x-z plane.  This makes for 9 elements total.  Now if we're talking
relativity and such, it gets a little more complex.  But I guess you already
looked it up on Wikipedia, so I don't know what I'm rambling on about.

 - Ricky

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Orchid XP v8 <voi### [at] devnull> wrote:
> >> That's pretty trivial mathematics compared to tensor calculus type stuff.
> >> :-)  I'm not too bad at discrete math.
> >
> > Tensor calculus isn't so bad.  Among other things, I gave my nephew my tensor
> > calculus book this past summer for his baby shower.  Gotta get kids started
> > early these days, y'know.
>
> I still can't figure out what a "tensor" is. :-P
>
> --
> http://blog.orphi.me.uk/
> http://www.zazzle.com/MathematicalOrchid*

meh, from the title thought you were talking about Final Fantasy Tactics and
perhaps some sequel... :)

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Orchid XP v8 wrote:
> However figured out that you can compute the difference from the sum and 
> the product... GENIUS!! o_O I would never have known in a million years...

Very cool. :-)

-- 
   Darren New, San Diego CA, USA (PST)
   Why is there a chainsaw in DOOM?
   There aren't any trees on Mars.

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> OK, so "everybody knows" (?!) that ax^2 + bx + c = 0 can be transformed 
> into the magical expression
>
>   x = (-b +/- Sqrt(b^2 - 4ac)) / 2a
>
> But why is that, exactly? Well, looking in my sister's A-level maths book 
> [unlike me, my sister was *taught* mathematics] I discover a long sequence 
> of algebraic manipulations that slowly transforms the one into the other.

It's not that hard or long, just involves dividing by a, completing the 
square for the first two terms, rearranging, taking the square root, and 
rearranging again for x.

But I imagine the cubic is a lot harder.

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>> But why is that, exactly? Well, looking in my sister's A-level maths 
>> book [unlike me, my sister was *taught* mathematics] I discover a long 
>> sequence of algebraic manipulations that slowly transforms the one 
>> into the other.
> 
> It's not that hard or long, just involves dividing by a, completing the 
> square for the first two terms, rearranging, taking the square root, and 
> rearranging again for x.

Yeah. As I said, a sequence of algebraic operations which "just happen" 
to lead to the answer we seek. (As in, there's no *apparent* reason why 
this particular sequence of operations does it, as opposed to some other.)

I also note in passing that I have no clue what "completing the square" 
actually means. But then, I have no formal mathematical education at 
all, so it's a miracle I even comprehend algebra in the first place!

> But I imagine the cubic is a lot harder.

http://planetmath.org/?op=getobj&from=objects&id=1407

Does it make your eyes bleed?

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Invisible wrote:

> http://planetmath.org/?op=getobj&from=objects&id=1407
> 
> Does it make your eyes bleed?

Actually... if you gather up some of those subexpressions and represent 
them as single letters, it's not that complicated.

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Invisible wrote:
> Invisible wrote:
> 
>> http://planetmath.org/?op=getobj&from=objects&id=1407
>>
>> Does it make your eyes bleed?
> 
> Actually... if you gather up some of those subexpressions and represent 
> them as single letters, it's not that complicated.

Unless my eyes deceive me, this is equivilent:

J = 2a^3 - 9ab + 26c
K = Sqrt(J^2 + 4(-a^2 + 3b)^3)
M = (1 + i Sqrt(3))/2
N = (-1 + i Sqrt(3))/2

r1 = -(a/3) +   Cbrt((-J + K)/54) +   Cbrt((-J - K)/54)
r2 = -(a/3) + M Cbrt((-J + K)/54) + N Cbrt((-J - K)/54)
r3 = -(a/3) + N Cbrt((-J + K)/54) + M Cbrt((-J - K)/54)

While hardly on the same level as the quadratic, it's not "that" hard.

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> Yeah. As I said, a sequence of algebraic operations which "just happen" to 
> lead to the answer we seek. (As in, there's no *apparent* reason why this 
> particular sequence of operations does it, as opposed to some other.)

Maybe if you've never done any equation solving before, but the steps are 
pretty obvious steps to take to rearrange an equation like that (ie to get 
all the x's in one place).

> I also note in passing that I have no clue what "completing the square" 
> actually means.

It's the key step to rearranging the quadratic equation in terms of x, and 
useful in lots of other areas too.  It's the process of rewriting terms like 
x^2+cx in the form (x+a)^2 + b.  It's not hard, and after performing a 
square root allows you to get the x by itself.

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> Unless my eyes deceive me, this is equivilent:
> 
> J = 2a^3 - 9ab + 26c
> K = Sqrt(J^2 + 4(-a^2 + 3b)^3)
> M = (1 + i Sqrt(3))/2
> N = (-1 + i Sqrt(3))/2
> 
> r1 = -(a/3) +   Cbrt((-J + K)/54) +   Cbrt((-J - K)/54)
> r2 = -(a/3) + M Cbrt((-J + K)/54) + N Cbrt((-J - K)/54)
> r3 = -(a/3) + N Cbrt((-J + K)/54) + M Cbrt((-J - K)/54)
> 
> While hardly on the same level as the quadratic, it's not "that" hard.

Did you try to derive those equations from ax^3+bx^2+cx+d=0 ?? :-)

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