Rather difficult....

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?

Originally posted by royalchicken
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 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...

Sorry, anybody need any snacks? 😛

Originally posted by ChessNut
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? 😛

Certainly Never 5!!!!!!

Got any spam?

EDIT
[i]did I spell spam right?

Originally posted by Phlabibit
Certainly Never 5!!!!!!

Got any spam?

[b]EDIT
[i]did I spell spam right?[/b]

Erhm...Phla....have you edited your post 219 times??? Or is it just my computer...😕😕😕😕

Olav

Well if its your computer, then it must be mine too. That sure is a lot of edits🙂.

As to the problem at hand. maybe in the morning, my brain doesn't like maths at the moment🙂.

James

yeah, took me a bit to get that post right......

😉

Originally posted by royalchicken
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?

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.

Originally posted by Acolyte
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.

Good going 😀. This was the same solution I found first....actually this was on a STEP exam that I downloaded at your suggestion....

Originally posted by Acolyte
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.

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) - 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.

Originally posted by Fiathahel
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.

That'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.

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.

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.