Sarahisme

can you do the same thing with taylor series thingos?

and for the period of the clock (relative to the observer): T = 2Lc/({gamma}*(c^{2}-v^{2})) but then this does equal what its meant to.... (i.e. T = 2L{gamma}/c )

and between reflection and return (relative to the observer): t = L/({gamma}*(c+v))

the time between the emission and reflection events relative to the observer ....i got: L/{gamma} + vt = ct therefore t = L/({gamma}*(c-v))

i know the period should be the same as a clock whose length is perpendicular to the motion but i just can't get the equation out...

hey peoples would someone please be able to help me a bit with this relativity problem? Cheers Sarah

i can't quite work this one out... i think the first step is to say that we want to find complex root of this: z^{n} = a but yeah...i dunno...can someone give me a few pointers please Sarah

can someone please explain to me how it goes from the first line to the second? (from a normal equation to an infinite series)?? Thanks Sarah

yay! i might have actually got this one right hopefully, how'd i go... i reckon R = 2 , where R is the radius of convergence of the power series....

ok here is another proof of mine for this question.... i am a little unsure about the inequality bit because well...i sorta made that up somewhat...anyways here it is.....:

how about this.... use the limit comparison test... so let |a(n)^{2}| = c(n) lim,n->infinity, (c(n))/(a(n)) = +infinity well ok....maybe not, but at least i'm trying right? :S

hey guys could i get a hint at how to do this problem please? i just can't quite work out what my first step should be,.... Thanks -Sarah

lol yeah thats what i was thinking is that sometime if you know the measurements are biased, say you know that you know they are below what the true value is, then that will give you a lower bound on the value....that sort of thing...so yeah... well what about the second question then?

oh and this was another question i was a little unsure about.... the choices are as shown: nominal interval ratio ordinal well yep did i do i good job? -Sarah

hey i was trying this basic science quiz and well you can see the answer i gave.... is it correct? or i mean should there be more boxes ticked or anything? i am pretty sure its right....thought i'd double check though -Sarah

no no i have done the work, but as it takes ages to fill in that bit (to type into here i mean) i havent put it up.

lol yeah i see well it would have taken ages to type out, but if i explained what the working was,.... so the proof is still good right?

or is that both observers agree that the reading on the clock at even B is L/(gamma*v) but the time they think it takes to get to event B is different ???

one other related question.... what if there was another clock which was attached to the rod (i.e. in the rods inertial reference frame) then what would it read at event B?

would both observers agree that the time taken for event A then B to occur be L/(gamma*v) i am pretty sure the answer is no. and since this is the time as mesured by the observer on the rod, then how can the observer on the clock use this time to calcualte the length of the rod?

hi i can't quite get my head around how this works... or well the way this book does it anyways.... the observer on the rod reads the time of the clock to be 5:(L/(gamma*v)) and yes i agree with this but then the book calucaltes the length of the rod (from the point of view of another observer situated on the clock) by using the time seen by the first observer (who is situated on the rod) that is the observer on the clock calcuates the length of the rod by doing this: length of rod = v * L/(gamma*v) but this is uses the value of the clock as seen by the observer on the rod. but wouldnt the oberver on the clock see a different time to the one the observer on the rod sees? and yes i know the rod is seen to be shorter (from the point of view of the observer on the clock) but still...if someone could explain this too me, that'd be great! Thanks Sarah

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

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

is it linerised about 0? or some other number?

"In fact, between "substituting" and "eventually" is hardly one step." what do you mean?