So, I have an exam tomorrow

Nixy

Elimi-nistrator
Staff member
and guess what I just spent an hour doing!!!!!

Shower, facial, painting toe and finger nails. :eek:

Now THAT'S what I call procrastinating!!!! :D
 

Jeslek

Banned
Nixy said:
only if you guys can teach me about n-space :circle:
For polynomials p(t) and q(t), which are part of P, define

(p(t), q(t)) = the integral from 0 to 1 of p(t)q(t) with respect to t.

This is an inner product on P. For which values of a, b, elements in R, are p(t) = 3t + 1 and q(t) = at + b orthogonal?
 

Jeslek

Banned
OK what about this then...

Let A =

[ 1 2 1 ]
[ 0 1 3 ]


Find matrix B such that AB = I sub 2 (a two by two identity matrix)
 

ipmoof

New Member
I have absolutely no idea what n-space is and I have done ALOT of maths. Maybe it's a US thing.

:confuse3:
 

Jeslek

Banned
Nullity, basis for a null space, basis, inner product spaces, dot product, ugggh.... I :love: this shit.
 

Jeslek

Banned
ipmoof said:
I have absolutely no idea what n-space is and I have done ALOT of maths. Maybe it's a US thing.

:confuse3:
2-space, 3-space, 4-space, ...? You know, two dimensions, three dimensions, etc.

x = 4 is one dimension... x + y = 5 is two dimensions, x + y + z = 3 dimensions, or 3-space.

n-space is then a function p(t) = at^n + bt^(n-1) + ... + yt + z = 0

You can find bases and all that crap for any given n-space. n-1 space is always a subspace of a given n-space. ;)
 

ipmoof

New Member
Ahh... ok, I see. So it's just real, vector-based geometry shit, right? FUN. :)

I love stuff that sounds more complicated than it is. :cool:
 

Jeslek

Banned
ipmoof said:
Euclidean space?
No, Euclidean space implies the covariant and contravariant quantities are equivalent.

Actually, if a real vector space that has an inner product defined (an inner product space) and if it is finite dimensional, its called an Euclidean space. ;)
 

Jeslek

Banned
Oh yes, whats the first thing you learn about Euclidean space? The wonderful Cauchy-Schwarz inequality

If u and v are any two vectors in an inner product space V, then

|(u, v)| <= ||u|| ||v||

I can prove it too... did it last semster. :rfap:
 
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