Recent content by fourier jr

Hi all can I go ahead with spoilers now that the movie is out? So there's that scene in The Last Jedi where Holdo jumps to hyperspace straight into Snoke's flagship, cuts it in half and also destroys about a dozen star destroyers, so I wondered if F=ma what would the force of that collision be...

notice the pattern ##1+q = \frac{1-q^{2}}{1-q}## ##1+q^{2} = \frac{1-q^{4}}{1-q^{2}}## ##1+q^{4} = \frac{1-q^{8}}{1-q^{4}}## etc

that's how I would have done it

I think it would be helpful to know that f is even if f(x) = f(-x)

& taking things out of context

another one which I thought was hard is lang's algebra. there's a lot of stuff incorporating other subjects so it helps to know a bit about them, and a lot of the problems are good too I remember reading about that one too now that you mention it. it would be good to give that one a go...

algebraic geometry by hartshorne has a reputation for being extremely hard but I haven't seen much of it. it's supposedly a shorter version of EGA.

if you write (3/4)^(n+1) as ##e^{(n+1)\ln{(\frac{3}{4})}}## maybe it will be easier to see why it goes to zero since the ln is negative & you don't need l'hopital's rule either.

I don't think that shows that 0 is in W though, since you need to have L(0) = M(0) for that to be true. I would start L(0) = L (x - x) = ... & I would also skip the lines that say ∴ L(x+y) + (-M(x+y)) = 0_v because the previous lines are enough

to show that it is closed you need to show that if x, y ∈ W and a ∈ F then x + y ∈ W and ax ∈W also but you've already done this by showing L(x + y) = M(x + y) etc above.

if you can show that f is a homomorphism & find its kernel then you'll have your isomorphism by the first isomorphism theorem. the range of f is actually {an ∈ G | n ∈ Z}, not just an. the kernel of f is {m ∈ Z | am = e}, not nZ (but you're not far off). you might need to show that (Z, +) is a...

did you enter the right thing into your calculator with brackets in the right place, etc? I don't know how that y got left behind, you know your calculator better than I do.I got the same thing wolfram did.

think ##\frac{a}{1-r}##

looks ok to me. actually maybe you should change < to ≤ for when cos2 is zero

I had the same proof that you got in post #11 & I don't think the invertibility had anything to do with the second variable, I think it just made it a bit more complicated than necessary.