hey everyone

 

this is a physics/maths ish question. to do with the michelson-moreley experiment.

 

about light travelling up the arm to the mirror kind of thing anyway, i don't know how to get the approimation part (i gather it is a first order appoximation?) can i do this using linerisation (taylor series stuff) or what? (i don't know how to do binomial expansion at the moment )

 

Thanks

 

Sarah

yes it is just linearization (of the expression (1-t)^{-1/2}) though it is quadratic in beta admittedly. it is just the first two terms of the taylor series. it isn't a binomial expansion since the exponent is not a natural number.

"it isn't a binomial expansion since the exponent is not a natural number."

 

 

The binomial expansion works for the power -1/2, and for any real or complex number.

Perhaps it is just my preference then, but I would prefer to distinguish between the exact form

 

[math](x+y)^n=\sum_{r=0}^n x^ry^{n-r}\binom{n}{r}[/math]

 

that is valid for all x and y.

 

the series expansion

 

[math] (1+x)^t = 1+ tx+\frac{t(t-1)}{2!}x^2+ \frac{t(t-1)(t-2)}{3!}x^3 + \ldots[/math]

 

[math] + \frac{t(t-1)\ldots (t-r+1)}{r!}x^r + \ldots[/math]

 

whcih is not valid unless |x|<1 (though it may converge if |x|=1) and is an analytic statement (it involves infintie sums).

 

The decision to declare the non-integer component the binomial expansion is one fraught with difficulty. Even that repository of knowledge Wolfram makes a mistake in this (that is an internal inconsistency) in that it states the binomial expansion is valid for all exponents (including complex ones) but fails to explain how to define the "binomial cofficients" in this case.

 

Perhaps the absolute is to call the first the binomial theorem, then.

sorry if this seems stupid, but i can't see exactly how to expand it using taylor series.... i end up with just 1

 

???

hold on i think i see it,

 

its not first order its second order taylor series? right?

Use matt's expansion of [math](1+x)^t[/math] with [math]x = -\beta^2[/math] and [imath]t = -\frac{1}{2}[/imath].

we are using the first order expansion of (1+x)^{-1/2} but it so happens that we are replacing x with beta^2 so it is not linear in beta. i wouldn't like to say whether that makes it not first order since it isn't clear to what's order you are referring when you say "first order".

well i mean that you have to go to f''(a) bit of the taylor series expansion....

 

is that right?

You just take the first two terms of the Taylor series.

well i mean that you have to go to f''(a) bit of the taylor series expansion....

 

is that right?

 

taylor series of what? (1-b^2)^{-1/2}? we aren't using the taylor series of that, we are using the series of (1-t)^{-1/2}. That is what i meant when i said it wasn't clear to what the "first order" referred. we are certainly using the first order estimate of the expansion of the latter. we haven't expanded the former directly.

i took the taylor series of (1-b^2)^(1/2) to the second degree (so it had f'' in there) about 0 and i came up with the answer.....???

i.e. i expanded it as if ...

 

 

f(x) = (1-x^2)^{-1/2}

then using taylor series p(n) where n = 2 for seond degree

did this about 0 (i.e. f(a) is f(0) )

 

if that helps explain what i've been on about...

Use matt's expansion of [math](1+x)^t[/math] with [math]x = -\beta^2[/math] and [imath]t = -\frac{1}{2}[/imath'].

 

 

is it linerised about 0? or some other number?

ok i see it is linerisation about 0....but then...why 0 ? and not another number?

becuase that's the easiest point to expand about or we'd have horrrible expressions flyingaround with sqrt(3/4) say if we did it about 1/2. the derivative is always going to have a 1-b^2 to some fractional power and we'd have to set b to what ever the points is we're expanding about. try it. expand about 1/2 and see what you get.

lol yeah i just did, yuck! what a mess, i see what you mean

 

Thanks matt