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Well I never set up a system of linear equations or anything initially, I could see that they would reduce and skipped to writing (A|0) can be RE form. the rest was trying to justify using (A|0) notation to pka is all. totally sidetracked.

sorry. I'm being unclear or perhaps flat wrong. In regards to proving the set B of vectors is linearly independent, this is what I did. if B={v1,v2} where v1= <1,0,-1> and v2=<0,1,-1> then for linear independence λv1+μv2 = 0 only when λ,μ = 0 as a system of linear equations...

oh I was using the shorthand for the augmented matrix for example https://en.wikipedia.org/wiki/Augmented_matrix maybe im using it wrong the method I was shown (as I understand it) is to put the set of vectors into an augmented matrix A. If Ax=0 has a unique solution then it is linearly...

oh..., good question. I'm having trouble answering this, my head feels so jumbled on this topic. Is it not a finite subset because you can use an infinite variety of vectors to represent the plane? Ill have a go.. if the vectors from the set B form the columns of matrix A then (A|0) can be...

I'm on the basis and dimension (vector spaces) chapter of my textbook and need to prove the following: Consider the plane P in R^3 whose equation is x+y+z=0 prove P is a subspace in R^3. I suspect they don't want to us to use the method of proving subspaces by showing closure under...

Im not great with induction. How did you go from 1/(m+1) to m!0!/(m+0+1)! ? edit# ok nevermind, you multiplied by m!/m! as for finishing your last step. m!k!/(m+k+1)! - (m+1)!k!/((m+1)+k+1)! = m!k!/(m+k+1)! - (m+1)m!k!/ (m+k+2)(m+k+1)! =m!k!/(m+k+1)! * (1 - (m+1)/(m+k+2)) =(m + k + 2...

Show Imn =int( xm (1 - x)n, x=0..1) = m!n!/(m+n+1)! really having trouble with this one. first thing I tried was just integrating by parts as it is, which I couldn't make work. I tried changing the expression to: Imn = int( (xm-1) (1-x)n (1+x-1), x=0..1) which simplified to: Imn =...

yes, thanks again. I was so stuck in seeing that the input HAS to be x2-y and didn't think you could represent it a different way. duh!

well it means f(-x) = -f(x) if i remeber correctly but wouldn't that mean F(0,y) = f(0-y)= f(-y)=sin (-y) = -sin (y) ? edit: i think i got it F(x,y) = f(x2-y) F(0,y) = f(02-y)= -sin(02 - y) = sin(y) so F(x,y) = - sin (x 2 - y) = sin (y - x2 ) that look ok?

Suppose that f is a differentiable function of a single variable and F(x,y) is defined by F (x, y) = f (x2 − y). a) Show that F satisfies the partial differential equation ∂F/∂x + 2x•∂F/∂y = 0 b) Given that F(0,y) = sin y for all y, find a formula for F(x,y). for a) I take LHS...

yeah, I was getting confused because some examples dealt with polynomials of exactly 3 degrees and some 3 and lower. And some examples just state such and such is an element of P3 without a definition of what they meant by P 3. So I was trying to figure out what P3 meant when its not explicitly...

this is the class https://www.maths.unsw.edu.au/courses/math1231-mathematics-1b No, I rechecked there is no "this is what P3 is..etc" it just shows up in an example and says "let P3 be the set of polynomials of degree 3 or less...etc" So I was confused if that's what P3 always meant or...

this makes sense. I had problem show the set S={p element of P3 : p''(x)=0, for all x is an element of R} now am I interpreting this correctly p''(x)=0 means it must be p''(x)=0+0x=0 so this can only be true for the set of polynomials degree 1 or less? so the set S is the set of...

the problem is I wasn't given an explicit definition (maybe its assumed in the course) and I've been trying to piece together the definition based on examples. Which gets confusing. I will mull over this info and try to understand it. thanks

great thanks!

maybe I see it now... (-1+1) + (2(-.5)+1)x + 3(0)x2 = 0+0x +0x​2 so the zero polynomial is when a1 =-1 , a2 = -.5 and a3 = 0 However, when a3 = 0 p(x) is no longer a polynomial of degree 3? so it's not an element of S.

sorry, I got no idea. There's only one example in my book dealing with polynomial subspaces and for proving the zero polynomial it just states 'obviously it has one' and moves on, so I haven't had much to work with on this.

So is it that the zero polynomial is when a0, a1,a2,a3 = 0 ? so (0+1)+ (0+1)x + 0x 2 =/= 0 + 0 x+ 0x2 therefore S doesn't contain a zero polynomial

I'm not even sure I'm interpreting this question properly. Q: is the set S a subspace of P3 S = { p element of P3: p'(x) + x + 1 = 0 for all x is an element of R} now, first of all, I'm a bit confused with the vector space of P3 . Is it the space of polynomials of degree 3 or less or do the...

I knew I was being sloppy at the bolded but those subscripts were being a pain in the *** to type out lol thanks again