View Full Version : Help with sigma notation where the top value is x?
rkell48
Aug18-11, 08:33 PM
1. The problem statement, all variables and given/known data
In each case, x is an integer between -6 and 6 inclusive.
2. Relevant equations
x
Σ 2i =12
i=1
3. The attempt at a solution
2x1 = 12 + 2x2 = 12 +......+2x(x) = 12
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
NascentOxygen
Aug18-11, 09:01 PM
I surmise you to be saying x is an integer and lies somewhere between -6 and +6. Correct?
So I'd read it as the sum of all terms, 2*i
for all integer values i starting from 1 and stepping through to x
but stopping when that sum equals 12.
Sure, you don't initially know the value of x, but it's the only unknown in the puzzle, so you can discover what value it must have, readily enough.
BTW, it is not a good idea to use the character 'x' for both multiplication and the unknown in the same line!! There are plenty of alternatives to choose from. 2(1) + 2(2) + ....
is as good as any. But if you can find a typeface with a large dot that sits well above where decimal points get positioned, then (preferably where there are no decimal points) you can use that large dot to denote multiplication. Or on the web, you can simply use an
asterisk.
rkell48
Aug18-11, 09:04 PM
I surmise you to be saying x is an integer and lies somewhere between -6 and +6. Correct?
So I'd read it as the sum of all terms, 2*i
for all integer values i starting from 1 and stepping through to x
but stopping when that sum equals 12.
hmm? i genuinely have no idea of what you said..
Ray Vickson
Aug18-11, 09:37 PM
1. The problem statement, all variables and given/known data
In each case, x is an integer between -6 and 6 inclusive.
2. Relevant equations
x
Σ 2i =12
i=1
3. The attempt at a solution
2x1 = 12 + 2x2 = 12 +......+2x(x) = 12
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
When x = 0 we are summing 2*i for i from 1 to zero, so get 2*1 + 2*0 = 2.
When x = 4 we are summing 2*i for i from 1 to 4, so we get 2*1 + 2*2 + 2*3 + 2*4 = 20, etc. When x = -5 we are summing 2*i for i going from 1 to -5, so we get 2*1 + 2*0 + 2*(-1) + 2*(-2) + 2*(-3) + 2*(-4) + 2*(-5). So, for any integer x between -6 and 6 you can do the sum, and you want to find out which x (if any) leads to a sum = 12. You could do it by a brute-force approach, trying every x, but there are quicker ways. However, the first step is to understand what the question is asking.
RGV
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