Sarahisme

hi all, here is the question, i will post my answer in the next post: Sarah

yeah thats what i thought, but that would be rather complex , hmm ... oh well i thought the problem was supposed to be simple guess not, huh?

lol, my dreams crushed! haha i was a fool to think i would ever be able to correct you

thanks for you help everyone

thanks for you help matt

thanks for all your help guys (especially swansont of course )

yes i see now, well it works for r>=5, but thats not important (my only ever chance to correct you! ) hmm ok now, we want: [math] 3(\frac{2}{3})^{N+1}<10^{-6} [/math] and so any value of N>=36 will do yep i think i understand this now, i knew what it was asking, i just didnt quite know how to implement their hint, hehe i think i've got the hang of this latex thing now thanks matt

hmm ok, i thought of another method, which sort of makes sense. we need to find a N s.t [math] \sum_{r=N+1}^{\infty} r^2/3^r<10^{-6} [/math] since each term is <= the last one in this series, then if you can make the first term <10^{-6} then you can easily come up with a value of N. (i got 17 or 18). hmm but then i suppose you could have a series which goes 1, 0.8, 0.5, 0.1,0.1,0.1... and so yeah if you kept adding those all up i suppose yeah, ok i think what i just said above is not correct

ok this is the way i think to go is [math] \sum_{n=1}^{N}\frac{n^2}{3^n}>\sum_{n=1}^{\infty}\frac{n^2}{3^n}-10^{-6} [/math] then you somehow find out what that N value is, because you know that [math] \sum_{n=1}^{\infty}\frac{n^2}{3^n}=\frac{3}{2} [/math]

ok so, so far this is whats happenin: [math] \sum_{n=1}^{\infty}\frac{n^2}{3^n}-\sum_{n=1}^{N}\frac{n^2}{3^n}<10^{-6} =\sum_{n=N+1}^{\infty}\frac{n^2}{3^n} [/math]

well i'm stuck dammit

ok now i've got this, (after using the formula for the partial sum of a geometric series applied to the series i mentioned in my previous post)

ok so i will bound it with the series 1+2+4+8+16+.... which is geometric series with a=1 and a comon ratio r=2

hmm ok so far it makes sense sort of, i have tryed manually trying to find a value for N, but obviously that is not going to work, but i can see that its heading in the right direction...

lol i have a mac which doesnt explain why stuff on this website has stopped loading properly laterly :S (like latex and such )

well i am pretty sure this method is correct, as it gives a feesible answer

i think it is correct this time, although the orginal question itself(as has already been mentioned) was quite dodgey

hang on, my computer it just showing little boxes with ? for the latex stuff, damn my computer! i'll go to another one

i guess i am just a nervous sort of person as well as the stuff you said, so yeah i am never confident, even when i have the answer right the first time

oh ok well i am not worried about the math being right (thats my fault if its wrong), but the method is right, the thinking is right, in that you add the two different methods of heat loss together....?

oops sorry yeah, lol just me rushing things so what did you think of my updated answer (in post #10) ?

ok, this is where i have started:

ok here is my solution now, i really hope its right , well here goes....how is it swansont? (and others )

lol i didnt quite understand that, sorry, but what i said in post #16 was correct?

i have figured out how to solve it exactly (its 3/2), but 10^-6 stuff, i dunno :S