Sarahisme

ok would i also be able to do it by induction? (as below...) show for n = 1 2*3^{1} - 3*2^{1} = 6 - 6 = 0 TRUE show for n = 2 2*3^{2} - 3*2^{2} = 18 - 12 = 6 TRUE assume true for n = k & n = k - 1 i.e. a(k) = 2*3^{k} - 3*2^{k} a(k-1) = 2*3^{k-1} - 3*2^{k-1} by defn. a(n+1) = 5*a(n) - 6*a(n-1) or a(k+1) = 5*a(k) - 6*a(k-1) .....then substitute in the assumption from above and eventually it should give : a(k+1) = 2*3^{k+1} - 3*2^{k+1} hows that? is that correct now? that method made more sense to me Cheers Sarah

"Are you sure you have not assumed what you were going to prove?" that is what i was afraid of...but i thought of it as if i assumed that was true then it would have to work when put into the given formula for a(n+1)... is that kind of logic incorrect? "Anyway, a method that could be used is induction: First show that formula is correct for n = 1 and for n = 2, then show that if the formula is true for n = k and for n = k + 1, then it also is true for n = k + 2." yeah thats induction but i thought the step you assume in induction (true for n=k) is what we are trying to prove.... anyways i dunno, i am most likely wrong

ok i have come up with a solution....but it seems very iffy to me....anyways here it is.... _______________________________________________________________________ a(n) = 2*3^n - 3*2^n a(n+1) = 2*3^{n+1) - 3*2^{n+1) a(n-1) = 2*3^{n-1) - 3*2^{n-1) 6*a(n-1) = 12*3^{n-1} - 18*2^{n-1} using the formula given (for a(n+1) ), rearrange that in terms of a(n): i.e. a(n) = (1/5)(a(n+1) + 6*a(n-1)) = (1/5)(2*3^{n+1} - 3*2^{n+1} + 12*3^{n-1} - 18*2^{n-1}) = (1/5)(2*3^{n+1} + 12*3^{n-1} - (3*2^{n+1} + 18*2^{n-1})) = (1/5)(6*3^n+4*3^n - (6*2^n + 9*2^n)) = (1/5)(10*3^n - 15*2^n) = 2*3^n - 3*2^n = a(n) Q.E.D is that completely wrong? lol i don't see how that would explain for n>=1 or n>=2 though.... anyways thats it -Sarah

Hi all! i can't work out the method to solving this problem or proving it i suppose i guess induction doesnt work but i am stumped for ideas, if anyone has any hints they could give that would be great! i don't to be told exactly how to do it at the moment, but i do need a few hints to get me started. Cheers Sarah

ok yep i got it now, thanks DQW

ok, but the distance from P to Q must be very very very small then?

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...

"PQ = vt" "LQ = ct" i don't see how these are the same times (t's)??

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.....???

"L/(1-v^2/c^2)^.5" was that right though?

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

hi this answer seems too simple to be true.... i reckon its L/(1-v^2/c^2)^.5 and yeah i don't know what it means by "primary evemts occurring in this measurement"?? Thanks for you help ") Sarah

well actually i just get Sin(theta)/Beta

hey i have another question if someone is willining to help i can't work out how to use the c and v in this diagram ?? Thanks guys Sarah

hold on i think i see it, its not first order its second order taylor series? right?

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

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

yep, ok , i see it now, thanks guys

oh lol, so its just by definition ?

can someone cleverer than me point out how this problem goes from the second last line to the last? Cheers Sarah

Hi all again! I'm sure there is a simple answer/solution to this problem...or a way to do it anyway, i just can't seem to get it. why does this = 1/e? Thanks Sarah

ok thanks swansont and everyone else

oh right cancelling! man i feel like an idiot! *shame* lol thanks mezarashi

i don't see why L'Hop's rule doesnt work by just straight forward application in this limit? the solution i have subs in a value so that the limit has m->0 ? Thanks guys, sorry if this seems like another inane question

thanks matt, sorry bout that guys