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If n is odd, then -2, if even, then 1.What is the nth-term of the sequence -2,1,-2,1,-2,.......?
Yes , but I am asking for the general term for the nth member of the sequence..btw I do have the answerIf n is odd, then -2, if even, then 1.
Then why ask the question? Testing the general knowledge of calculus II (or elementary real analysis) knowledge? Ok...btw I do have the answer
I got the answer from someone else because I couldn't see the answer. It is an interesting question and thought that some here would also find it interesting.Then why ask the question? Testing the general knowledge of calculus II (or elementary real analysis) knowledge? Ok...
Yes , but I am asking for the general term for the nth member of the sequence..btw I do have the answer
Ok..so the answer isWhat is the nth-term of the sequence -2,1,-2,1,-2,.......?
This has an exact result: -2. It is not the nth term of an oscillating sequence of the type you describe.Ok..so the answer is
(3(-1^n)/2) - 1/2
A sequence is simply a list of numbers. Thus, for example, the limit of the sequence {3., .1, .4, .1,...} is pi. The sequence -2,1,-2,1,-2,... consists of a set with only to elements: -2 & 1. It oscillates eternally/infinitively between these two values, never converging to either nor ever yielding another number. This is why most sequences don't have patterns that allow us to define them simply by listing numbers (e.g., the way the sequence that converges to pi has no pattern; this is also why we cannot generally determine the nth term of a sequence that isn't based on an expression with a variable/index term n).Obviously there is a pattern in there, but I can't remember to work out what it is. It seems to involve alternating +4 and -2....
It actually is.This has an exact result: -2. It is not the nth term of an oscillating sequence of the type you describe.
Try doing some extremely basic calculations. If that fails, use a calculator (or, if your calculator doesn't admit input such as (3(-1^n)/2) - 1/2, then simply enter the expression into Wolfram Alpha's online calculatorIt actually is.
Unless one is interested in mathematics.(3(-1^n)/2) - 1/2
may also be written as
( ( ( (-1) ^ n ) x 3 ) - 1 ) / 2
True. Now realize that your expression (the first one, not the nonsense second attempt at an equivalent expression) demands subtracting a non-integer from every infinite termThe power of (-1) will and does oscillate between (1) and (-1).
Unless one is interested in mathematics.
True. Now realize that your expression (the first one, not the nonsense second attempt at an equivalent expression) demands subtracting a non-integer from every infinite term
Now that was me being completely unfair, very tired, and unreasonably cranky. I apologize. For one thing, I was being lazy and right or wrong this was because I was too tired to do even mental computations, so I should hardly fault you even had you given an answer that made no sense at all. For another, my response was just incredibly rude and impolite. Finally, I was already short-tempered from another interaction (more legitimately, if still no less infantile) and was taking it out on you, which is one more reason NOT to be on the forums when one can't sleep.Gee, that is what I gain for assuming that you wanted a conversation?
Actually, I used Mathematica, but the syntax is simply order-of-operations. In Mathematica, you can't usually just plug in the expression; you have to write as you would in any programming language minus the fact that a CAS like Mathematica allows symbolic computations and hence one doesn't need to define n in the expression. But providing the placement of operators & parentheses is correct, the entire point of order-of-operations is to remove any ambiguity. This is particularly true when, as in Mathematica code or on the forums, one can't represent fractions and powers by placement. Thus (3(-1^n)/2) - 1/2 is not equivalent to (3((-1)^n)/2) - 1/2 (the "simple inner set of parenthesis" fix you supplied). Even rested, I'm a stickler about these things because I am frequently using MATLAB, R, or Mathematica handle complicated expressions/equations and I get the order-of-operations wrong on the first try like 99% of the time.When, before your post #12 above, were we supposed to conform to Wolfram-Alpha syntax anyway?
I don't. I do rely on statistical & mathematical software a lot, both because I suck at doing computations, and because most of the computations I do can't be done by humans.You should not rely so much on automatic parsers.
It has nothing to do with WolframAlpha or Mathematica (or computers, for that matter). This expression:Neither of us was making an effort to write in notation understandable by Wolfram Alpha in the first place, so I don't know why you are assuming otherwise anyway.
That doesn't work for the code, though (for Mathematica). But it's not the notation that's the problem. The order of operations is clearly specified. This:If you check the result of the second link above, you will see that W-A failed to understand that "x" is a multiplication symbol instead of an unknown variable.
isn't true. It's not "seeing -1^n as - (1^n)", it's correctly recognizing that exponents take precedence over addition/subtraction operators. That's how I saw it, and how I still see it, because that's what it says.As for the first expression, it is being mishandled by the automatic parser as well. It is seeing -1^n as - (1^n).
Perfect.A simple inner set of parenthesis fixes that, as seen here
The mistake was in the original formula given. You fixed it above. It simply requires making the negation/subtraction operator take precedence. Also, it was absolutely unfair of me to not correct the letter "x" with the operator. I knew what you meant and I knew the operator, I was just being an ***, for which I again apologize.Uh, I don't think that was _my_ mistake.
Apologies all around....... I should have written it with an extra parenthesis around the -1, and the whole thing in squiggly brackets to indicate a sequence like soThis has an exact result: -2. It is not the nth term of an oscillating sequence of the type you describe.
No need! It's truly my fault. Had I been less interested in behaving like an *** an more interested in the math problem, I would have done what Luis so aptly did and simply mentally added a bracket/parenthesis. I'm very sorry!Apologies all around...