Recent content by CGM

Uncertainty in (1 + σf/f) = √((δσf/σf)2 + (δf/f)2) ? or uncertainty in f + σf =√(δσf)2 + (δf)2

1. The problem statement, all variables and given/known dat Show that if you have a signal f and noise σf for a stellar flux measurement, then the uncertainty on the observed magnitude is given by σm = 1.0875 x σf / f Homework Equations (you should use the fact that when ε<<1 then log(1±ε) ≈...

Subbing in k to be an even number: = 1/pi { -2/k+1 + 2/k-1 } = 2/pi { -1/k+1 + 1/k-1 } = -2/pi(k2-1) Got it I think EDIT: I see now where I lost the half so I don't know how I've ended up with the right answer

I can't find the 1/2 and I can't figure out how to simplify it even with cos(kpi) = (-1)k substituted in = 1/pi { [ (-(-1)k+1 - 1)/(k+1) ] - [ (-(-1)k-1 - 1)/(k-1) ] }

Not sure what I've done actually

The last line has the limits put in but I don't know where to go from there EDIT: Just realized I put 'x's in instead of 'pi's So, it should be: = 1/pi { (-cos((k+1)pi)-1)/(k+1)) - (-cos((k-1)pi)-1)/(k-1)) }

ak = 1/pi [ ∫sin((k+1)x) - ∫sin((k-1)x) ] = 1/pi { [(-1/k+1)cos((k+1)x)] - [(-1/k-1)cos((k-1)x)] } = 1/pi { (-cos((k+1)x)-1)/(k+1)) - (-cos((k-1)x)-1)/(k-1)) } Is this even close?

Okay so, that gives me the ao value Then ak = 1/pi ( ∫sin(x)cos(kx) ) My tutor told me to use: sin((k+1)x) = sin(kx)cos(x) + cos(kx)sin(x) sin((k-1)x) = sin(kx)cos(x) - cos(kx)sin(x) to make the integration easier but I can't see what to do with this

a0 = 1/pi [ ∫(0) + ∫(sinx)] (0 between 0 and -pi and sinx between pi and 0) a0 = 1/pi (0 + [-cosx]) = 1/pi (0 + (-1+1)) = 0 Sorry I'm not sure about how to do the notation properly on a computer

Homework Statement I have attached a screenshot of the question. I know how to use Fourier's theorem for one function but have no idea how to attempt it with a discontinuous function like this. I tried working out a0 by integrating both functions with the limits shown, adding them and...

Ah yes, I see my mistake. Thank you. t'=t/y

Yes that's what I said, maybe my notation is confusing. t'=yt-((v/c)^2)t t'=t(y-(v/c)^2) Is this the answer or is there more simplification to be done?

Never mind I did something silly. So 0=y(x-vt) => x=vt t'=y(t-(v/t)^2 . t)

So, t'=yt?