Recent content by Mmmm

Wooooooooooooohooooooooooooooooooooooooooooo... I've figured it out... At last... My mistake was in thinking that A=\left[A\right]_{\beta}^{\beta}. The map A:X\rightarrow X maps a vector in the vector space X to a new vector in X. Wheras \left[A\right]_{\beta}^{\beta} maps the...

No problem, glad it helped... Now all you have to do is prove your answer for (c) using induction.

Thanks, that helps a lot. Did you notice that in my first post I mentioned that if you choose the second basis as \beta A rather than A\beta then you get the desired result? \begin{aligned} & \beta x=(\beta A)y\\ \Rightarrow & A^{-1}\beta^{-1}\beta x=y\\ \Rightarrow &...

OK... but [b1 b2 ... bn]x = A[b1 b2 ... bn]y multiplying by A-1 gives A-1[b1 b2 ... bn]x = [b1 b2 ... bn]y but [b1 b2 ... bn]y is meaningless right? it's the components in Ab paired with the b basis!

but even then you would have the matrix being [b1 b2 ... bn]-1A-1

so using your notation: [b1 b2 ... bn]x = [Ab1 Ab2 ... Abn]y = A[b1 b2 ... bn]y where x is in \beta coords and y is in A \beta coords taking the first and last of these equalities: [b_1 b_2 ... b_n]x = A[b_1 b_2 ... b_n]y \Rightarrow A^{-1}[b_1 b_2 ... b_n]x = [b_1 b_2 ... b_n]y...

Right! which is what I did and got \text{conversion matrix from }\beta \text{ to } A\beta=\beta^{-1}A^{-1}\beta rather than the A^{-1} which the book claims. Am I right?

In this case I am changing bases, so the identity map leaves the vector unchanged but changes the components...if you see what i mean... I'm beginning to get the feeling that the terminology used in this book isn't standard!

I know A\beta is definitely a basis, the question is whether the identity map between the two bases is equal to A^{-1}. ie does A^{-1}\mathbf{x^{\beta}}=\mathbf{x^{A\beta}} My reasoning says it should be A^{-1}\mathbf{x^{\beta}}=\mathbf{x^{\beta A}} notation: \mathbf{x^{\beta}} meaning the...

Yes, it's linear. So you would say I'm right then? or are you saying it's straightforward to get the required result?

(c)=(b)-(a) with 2n replaced with n. also, for your previous proofs for which you said you used lots of algebra, consider this: you have \sum_{r=1}^{n}r^{2}=\frac{1}{6}n(n+1)(2n+1)\] (a) find \sum_{r=1}^{n}(2r)^{2}\] (write this in terms of the first sum) (b) find \sum_{r=1}^{n}(2r+1)\]...

The fact that your equation involves dx\dy rather than dy\dx means that just by adding xtany to both sides you get the required form! and you can find the integrating factor in terms of y. ie use x' + Px = Q Where P and Q are functions of y

Homework Statement I have posted this problem on another website (mathhelpforum) but have received no replies. I don't know whether this is because no one knows what I am talking about or if it's just that no one can find a fault with my reasoning. Please please please could you post a reply...

ah, yes so a simple substitution X=x'-x will shift the ball to the origin. when x'=x X=0. dx'=dX so there is no difference. Thanks for your time Lanedance

Ahh... Having had a bit of time to mull it over, p is not dependant on t (or r), so integrating wrt r first \frac{\partial}{\partial t} \int_0^ t \int_0^{\pi}\int_0^{2 \pi} r^2 sin\phi \nabla^2 p(x') d \phi d \theta dr =\frac{\partial}{\partial t} \int_0^{\pi}\int_0^{2 \pi}...