Sarahisme Posted August 8, 2005 Share Posted August 8, 2005 would anyone be able to give me a hint to get me going with this problem, i am unsure as what i should start by doing... small hint? Cheers Sarah Link to comment Share on other sites More sharing options...
Sarahisme Posted August 8, 2005 Author Share Posted August 8, 2005 i have figured out how to solve it exactly (its 3/2), but 10^-6 stuff, i dunno :S Link to comment Share on other sites More sharing options...
Sarahisme Posted August 8, 2005 Author Share Posted August 8, 2005 ok, this is where i have started: Link to comment Share on other sites More sharing options...
shmoe Posted August 8, 2005 Share Posted August 8, 2005 You want an N so that: [math]\sum_{n=1}^{\infty}\frac{n^2}{3^n}-P_N<10^{-6}[/math] where [math]P_N=\sum_{n=1}^{N}\frac{n^2}{3^n}[/math] (there's no dependance on "x" here). The left side is just a sum from N+1 to infinity. Use their hint and dominate this sum with a geometric series to get an upper bound dependant on N. Choose N appropriately. Link to comment Share on other sites More sharing options...
Sarahisme Posted August 9, 2005 Author Share Posted August 9, 2005 hang on, my computer it just showing little boxes with ? for the latex stuff, damn my computer! i'll go to another one Link to comment Share on other sites More sharing options...
Sarahisme Posted August 9, 2005 Author Share Posted August 9, 2005 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... Link to comment Share on other sites More sharing options...
Sarahisme Posted August 9, 2005 Author Share Posted August 9, 2005 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 Link to comment Share on other sites More sharing options...
Sarahisme Posted August 9, 2005 Author Share Posted August 9, 2005 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) Link to comment Share on other sites More sharing options...
Sarahisme Posted August 9, 2005 Author Share Posted August 9, 2005 well i'm stuck dammit Link to comment Share on other sites More sharing options...
Sarahisme Posted August 9, 2005 Author Share Posted August 9, 2005 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] Link to comment Share on other sites More sharing options...
Sarahisme Posted August 9, 2005 Author Share Posted August 9, 2005 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] Link to comment Share on other sites More sharing options...
matt grime Posted August 9, 2005 Share Posted August 9, 2005 mmm? you want an N such that [math] \sum_{r=N+1}^{\infty} r^2/3^r[/math] is less than 10^{-6} Well, that sum is less than [math]\sum_{r=N+1}^{\infty}(2/3)^r[/math] for N sufficiently large (ie when 2^r>r^2 which is true for all r>5 certainly) and that sum equals 3(2/3)^{N+1} so pick N such that that is less than than 10^{-6} say any N greater than 100 would do Link to comment Share on other sites More sharing options...
Sarahisme Posted August 9, 2005 Author Share Posted August 9, 2005 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 Link to comment Share on other sites More sharing options...
Sarahisme Posted August 9, 2005 Author Share Posted August 9, 2005 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 Link to comment Share on other sites More sharing options...
matt grime Posted August 9, 2005 Share Posted August 9, 2005 hmm ok' date=' 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). ... i suppose yeah, ok i think what i just said above is not correct certainly it is not correct. Link to comment Share on other sites More sharing options...
matt grime Posted August 9, 2005 Share Posted August 9, 2005 yes i see now, well it works for r>=5, but thats not important sorry, but you aren't correcting me. i didnt' say that r>5 was neccesary, only sufficient. Link to comment Share on other sites More sharing options...
Sarahisme Posted August 9, 2005 Author Share Posted August 9, 2005 sorry, but you aren't correcting me. i didnt' say that r>5 was neccesary, only sufficient. lol, my dreams crushed! haha i was a fool to think i would ever be able to correct you Link to comment Share on other sites More sharing options...
matt grime Posted August 10, 2005 Share Posted August 10, 2005 but it is an important point: do you understand the difference between "if", "only if", and "if and only if"? sometimes they use sufficient, necessary, and necessary and sufficient. Link to comment Share on other sites More sharing options...
Sarahisme Posted August 10, 2005 Author Share Posted August 10, 2005 oh ok, yep, i see. Link to comment Share on other sites More sharing options...
Sarahisme Posted August 10, 2005 Author Share Posted August 10, 2005 hmmm...in the hint it says to use the comparison theorem. i don't see why we would really need to, but anyway Link to comment Share on other sites More sharing options...
matt grime Posted August 11, 2005 Share Posted August 11, 2005 because the series you can't sum r^2/3^n you replace by one you can, (2/3)^n and you compare the (easily evaluated) sum of the tail of one and which by the comparison theorem we know is more than the sum of the tail of the other. Link to comment Share on other sites More sharing options...
Sarahisme Posted August 14, 2005 Author Share Posted August 14, 2005 oh right, yep of course, sorry yeah i get it now thanks matt Link to comment Share on other sites More sharing options...
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