23 Jul 03
1 edit
Originally posted by royalchickenx is an integer that when used within the equation in question the calculated result would come out as '3' and not '2' and not '1' unless you move onto '2' when you shall not stop there as you shall move immediately to '3'. Once you get to '3' you shall toss the Holy Handgrenade and all of your enemies shall perish...
This is not mine. Sorry for the failure of narrative. Suppose:
3 = 2/x(1) - x(1) + 2/x(2) - x(2) + ... +2/x(n) -x(n)+....
What is x(n) in terms of n?
Sorry, anybody need any snacks? 😛
Originally posted by ChessNutCertainly Never 5!!!!!!
x is an integer that when used within the equation in question the calculated result would come out as '3' and not '2' and not '1' unless you move onto '2' when you shall not stop there as you shall move immediately to '3'. Once ...[text shortened]... our enemies shall perish...
Sorry, anybody need any snacks? 😛
Got any spam?
EDIT
[i]did I spell spam right?
01 Aug 03
1 edit
Originally posted by royalchickenHow about:
This is not mine. Sorry for the failure of narrative. Suppose:
3 = 2/x(1) - x(1) + 2/x(2) - x(2) + ... +2/x(n) -x(n)+....
What is x(n) in terms of n?
x(n) = 1 if n=1,2 or 3
x(n) = sqrt(2) otherwise
NOTE: There are clearly an infinite number of solutions, so I assume you're just looking for one.
02 Aug 03
1 edit
Originally posted by AcolyteGood going 😀. This was the same solution I found first....actually this was on a STEP exam that I downloaded at your suggestion....
How about:
x(n) = 1 if n=1,2 or 3
x(n) = sqrt(2) otherwise
NOTE: There are clearly an infinite number of solutions, so I assume you're just looking for one.
14 Aug 03
Originally posted by Acolytesorry to bother you, but
How about:
x(n) = 1 if n=1,2 or 3
x(n) = sqrt(2) otherwise
NOTE: There are clearly an infinite number of solutions, so I assume you're just looking for one.
sqrt(2) - sqrt(2) + sqrt(2) - sqrt(2) + sqrt(2) - sqrt(2) + ...
aint summable (if that is the correct english word for 'sommeerbaar'😉
Cause it would be the same as
[sqrt(2) - sqrt(2)] + [sqrt(2) - sqrt(2)] + [sqrt(2) - sqrt(2)] + ...
= 0 + 0 + 0 ... = 0
and as
sqrt(2) + [-sqrt(2) + sqrt(2)] + [-sqrt(2) + sqrt(2)] + ....
= sqrt(2) + 0 + 0 + ... = sqrt(2)
If my thinking is rigth there aint a solution to this question cause therefor the sequence
2/x(1), -x(1), 2/x(2), -x(2), 2/x(3), -x(3), ...
should converge to 0. But that can't be done cause 2/x(n) or x(n) is greater then 1 for all n.
14 Aug 03
1 edit
Originally posted by FiathahelThat's a good point, and it emphasises the importance of specifying exactly what is tending to infinity. I read the expression in the question as 'the sum from t=1 to n of 2/x(t) - x(t) tends to 3 as n tends to infinity.' In this case, my example is equal to 3 for all n >= 3 and so the series tends to 3.
sorry to bother you, but
sqrt(2) - sqrt(2) + sqrt(2) - sqrt(2) + sqrt(2) - sqrt(2) + ...
aint summable (if that is the correct english word for 'sommeerbaar'😉
Cause it would be the same as
[sqrt(2) - sqrt(2)] + [sqrt(2) - sqr ...[text shortened]... can't be done cause 2/x(n) or x(n) is greater then 1 for all n.
If, on the other hand, you interpret the '=' as 'the sum tends to 3 as the number of terms tends to infinity', where a term is either 2/x(n) or -x(n) for some n, then the expression is indeed impossible.
14 Aug 03
The first interpretaion is the one I assume the STEP examiner intended. What Acolyte's answer basically says is that lim (n->infinity) 1-1 = 0, and Fiathahel quite rightly says that 1-1+1-1+1-1+1... is meaningless unless the nth term is defined. The same is true of any conditionally convergent series in that rearrangement of terms is not allowed. Leibniz wrote on this one.