Recent content by toni

no, the 5-hour and 7-hours door will lead him back to where he was

can anyone give me some advice?

oh I am in trouble now...the answer is given by tutor ="=

A miner is trapped in mine containing 3 doors The first door will take him out of the tunnel in 2 hours the second door will take hin out of the tunnel in 5 hours and the third door will take him out of the tunnel in 7 hours. Assuming that the miner at all times is equally likely to choose any...

I see! Thank you soooo much!

I got this from my Economics lecture notes... How come 1 to (n-1) can be split up into [(1 to infinity) minus (n to infinity)]? What confuses me is the former one has (n-2) terms, but the latter one has only (n-1) terms... And it doesn't make sense to me graphically...

can anyone tell me... If xf"+f'+xf = 0 Is d^n/dx^n (xf"+f'+xf) = 0 ?? cos the solution says d^n/dx^n [x(xf"+f'+f)-f] = 0...that puzzles me. --'

this is what I've done...still far from the destination Sum(0 to infinity) (-1)^n [(1)(3)...(2n-1) x^(n+1/2) / (2n)!] donno how to transform x^(n+1/2) to x^(n)?...thank you for your help!

i donno...seems i still not getting the point at all ==; it's like if I am given a question asking me to find a taylor series of a function, i will definitely lost. and the "center of convergence" how am i going to find it? using convergence test to find the interval of x and pick the middle...

I stuck at the second "="...i know it goes like this because the formula...i just someone explain to me why it works like that. thank you soooo much!

Hmmm..it's actually on a solution sheet. I stuck at that point when i tried to go through it again before midterm... the question is here, just in case. Thank you soooo much!

Sorry, i shouldve made things clearer. Here is everything before this. r is the radius of convergence.

is "a" arbitrary? "a" is also said to be the "center" right?

so can we say taylor series represents function in a more precise way than power series does?

I just don't understand why it goes like this...why 'therefore'? i know (x^2)f''(x)+xf'(x)+((x^2)-1)f(x)) = x(xf''(x)+f'(x)+xf(x)) - f(x), but why it goes to zero when taking (d^n/dx^n)?