PJelly wrote: » Don't sob about yesterday! This is the paper 2 thread, this is the future! Be positive. I kept thinking on how badly I did on paper one, even the simple bits, but it's over and no amount of moping will change it.
jamesr1775 wrote: » pretty cool thing for circle. appeard in 2004 i think but you can use if an axis is a tangent to circle. if y axis is tangent to circle then f^2 = c if x axis is a tangent to circle then g^2 = c
Shayanzadeh wrote: » It's -A/B not -B/A. So in your 2x-y=0 you would get 2 as slope. Easy to mix up with -B/A which is from alpha + betha.
jamesr1775 wrote: » yep and then u can get the C variable by subbing (0,f) in to equation of circle with the 2gx and 2fy
GV_NRG wrote: » can someone please explain this -A/B methoo please?
PJelly wrote: » Take the three parts of a line like A B and C Ax+By+C you could say. Or Ay+Bx+c. To get the slope, put -A/B Sooo. 4x-6y+3 would have a slope of 4/6 = 2/3 (Minus's cancel) And 7x+13y-9 would have a slope of -7/13 Geddit?
canister94 wrote: » eq of tangent of circle x1x+y1y=r squared difference equations un=la power of n +mb power n
kpac wrote: » I was going to say the difference equation wouldn't come up because it appeared last year, but after yesterday I wouldn't rule out anything.
mtb_kng wrote: » Yea, the examiners might ask us where they're going on holidays, I know it's not on the curriculum, but sure that didn't stop them yesterday did it? :pac: "Q6a (i) There are loads of people complaining about paper one. What's the probability you passed? "
PJelly wrote: » There's one part of McLaurin series that I struggle to grasp. When you need to find the value of X that you sub in for say, the 1997 example. You need root ten and you have root 1 + x. But you cant let X=9. So you need to rejig the equation around. There was a similar question like it somewhere where you need to get root seventeen. I always struggle with the actual re-jigging of the equation.
partyatmygaff wrote: » First of all, you have to realise the use of a Maclaurin series. All it does is convert a function like sqrt(1+x) in to a simple series that can be summed arithmetically. The reason this is important is because calculators can only add or subtract. The more terms you get in a Maclaurin series the more accurate it becomes. This is why some calculators advertise "8 decimal point precision" e.t.c. Now, look at this 1997 question. You're being asked to find the value of sqrt(10) using the Maclaurin series of sqrt(1+x). Essentially you're being asked to do the job your calculator would normally do. Now let's make this extremely easy. sqrt(10) can be written as sqrt(9+1). You want to get one of those numbers down to 1 and the other between -1 and 1. So what you do is factorise. You then get sqrt(9(1+1/9)) which can also be rewritten as sqrt(9) x sqrt(1+1/9). Simplifying that you get 3sqrt(1+1/9). The question is asking you to evaluate sqrt(10) using the series expansion. You now have 3.sqrt(1+1/9) and 1.sqrt(1+x). What you now do is sub x = 1/9 in to your expansion and then multiply the answer of that by three.
partyatmygaff wrote: » Now, look at this 1997 question. You're being asked to find the value of sqrt(10) using the Maclaurin series of sqrt(1+x). Essentially you're being asked to do the job your calculator would normally do. Now let's make this extremely easy. sqrt(10) can be written as sqrt(9+1). You want to get one of those numbers down to 1 and the other between -1 and 1. So what you do is factorise. You then get sqrt(9(1+1/9)) which can also be rewritten as sqrt(9) x sqrt(1+1/9). Simplifying that you get 3sqrt(1+1/9). The question is asking you to evaluate sqrt(10) using the series expansion. You now have 3.sqrt(1+1/9) and 1.sqrt(1+x). What you now do is sub x = 1/9 in to your expansion and then multiply the answer of that by three.
LilMissCiara wrote: » Can you not then equate 9+1 to X+1 and have X = 9 and put that in to what you worked out for sqrt(1+x). Why has it to be between -1 and 1.. I probably sound really stupid right now!
partyatmygaff wrote: » That's because the series only converges when x is between -1 and 1. The Maclaurin series only works when the series is convergent. A convergent series is one which has a limit. It always converges on to a definite number. A divergent series does not converge on to a definite number, it goes on to infinity which isn't of much use when you're trying to evaluate a function.
partyatmygaff wrote: » That's because the series only converges when x is between -1 and 1. The Maclaurin series only works when the series is convergent. A convergent series is one which has a limit. It always converges on to a definite number. A divergent series does not converge on to a definite number, it goes on to infinity which isn't of much use when you're trying to evaluate a function. This is the reason why we do the ratio test. We want to find the range of values of x where the series is convergent so that we can make use of the series to evaluate functions. This mightn't be the best of explanations as it's the result of me trawling through the theory behind it on the odd website and Wikipedia article but for the purposes of Q8, I think it's good enough.
john1741 wrote: » So say if it looks like x=6 would you use 1/6 instead? Putting a one over whichever number making it a fraction
luciemc wrote: » anyone doing question 9?
LilMissCiara wrote: » Ok thanks, it makes sense... But.. Just after looking at the 2009 question and it had something very similar. I understand how to get it to -1>X<1 but I'm confused by something.. You get the maclaurin series for Root(1+X) which is grand. Then you want to approximate Root(17) so you break that down and it becomes 4Root(1/16 + 1) and you use 1/16 for X. But that's 4Root(1+X) and you are then putting X in to just Root(1+X). I would multiply the answer by 4 but the solution says it's wrong.. Can anybody spread some light on that! Solution = http://www.studentxpress.ie/papers/optionsoln2009.pdf
Michael_E wrote: » I know all the proofs now, I think. The Perpendicular distance one is so easy to make mistakes in, though.
