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Another maths question
- Thread starter Jedster
- Start date
If n is odd, then -2, if even, then 1.
Jedster
Well-Known Member
Yes , but I am asking for the general term for the nth member of the sequence..btw I do have the answer
Then why ask the question? Testing the general knowledge of calculus II (or elementary real analysis) knowledge? Ok...
Jedster
Well-Known Member
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.
I'll put up the answer ina few days if no one answers.
Rick O'Shez
Irishman bouncing off walls
I'm a bit rusty but here is my working so far:
1st term: n-3
2nd: n-1
3rd: n-5
4th: n-3
5th: n-7
6th: n-5
7th: n-9
8th: n-7
9th: n-11
10th: n-9
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....
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@Spiny Norman
The pattern seems to be that odd-indexed terms are four units lower than their immediate predecessors, while even-indexed terms are two units lower instead.
Since that works out to an average of decreasing the term in two units for each two increases of the index, the actual terms remain stable indefinitely.
@Jedster
We can of course simply describe it as -2 for odd index numbers and +1 for even index numbers.
But if we want to instead describe it as a single calculation in relation to the value of "n", it can be done in several other, functionally equivalent ways.
Are we allowed to use the binary modulo function? "n mod 2" results in 0 for an even value of n, and in1 for an odd value of n.
If it is allowed, then the n-th term is calculated as
1 - 3 * (n mod 2)
The pattern seems to be that odd-indexed terms are four units lower than their immediate predecessors, while even-indexed terms are two units lower instead.
Since that works out to an average of decreasing the term in two units for each two increases of the index, the actual terms remain stable indefinitely.
@Jedster
We can of course simply describe it as -2 for odd index numbers and +1 for even index numbers.
But if we want to instead describe it as a single calculation in relation to the value of "n", it can be done in several other, functionally equivalent ways.
Are we allowed to use the binary modulo function? "n mod 2" results in 0 for an even value of n, and in1 for an odd value of n.
If it is allowed, then the n-th term is calculated as
1 - 3 * (n mod 2)
Jedster
Well-Known Member
Ok..so the answer is
(3(-1^n)/2) - 1/2
This has an exact result: -2. It is not the nth term of an oscillating sequence of the type you describe.
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).
It actually is.
You may be having trouble with the notation. See if this helps:
(3(-1^n)/2) - 1/2
may also be written as
( ( ( (-1) ^ n ) x 3 ) - 1 ) / 2
The power of (-1) will and does oscillate between (1) and (-1). Therefore, each individual term of the sequence will be either -1 or 2, as desired.
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 calculator
Unless one is interested in mathematics.
Your first expression
And the second
Try balancing the parentheses next time. Or, "you may be having trouble with" some absolutely BASIC notation. I would suggest something to help, but as you've already tried to help me by suggesting I fundamentally misinterpret notation, I am not sure how to proceed.
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
Rick O'Shez
Irishman bouncing off walls
Could we have an easy one next time, please. 
Gee, that is what I gain for assuming that you wanted a conversation?
When, before your post #12 above, were we supposed to conform to Wolfram-Alpha syntax anyway?
You should not rely so much on automatic parsers.
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.
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. As for the first expression, it is being mishandled by the automatic parser as well. It is seeing -1^n as - (1^n).
A simple inner set of parenthesis fixes that, as seen here:
http://www.wolframalpha.com/input/?i=(3((-1)^n)/2)+-+1/2
Uh, I don't think that was _my_ mistake... but if it is so important to you that I prove that I can handle Wolfram-Alpha with my apparently "nonsense" efforts, what about this? The only change was using an asterisk to indicate multiplication instead of the previous "x" letter.
http://www.wolframalpha.com/input/?i=(+(+++(++(-1)+^+n+)++*+3+)+-+1+)+/+2
Next time, try establishing the goal before blaming people for failing to accidentally fulfill it, ok?
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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.
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.
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.
It has nothing to do with WolframAlpha or Mathematica (or computers, for that matter). This expression:
(3(-1^n)/2) - 1/2
is unambiguous (or is supposed to be). It says I'm taking -1/2 to the n power, multiplying it by 3, and subtracting 1/2.
If I plug the terms of the sequence into the formula, I always get -2 (because I am always subtracting 1/2 from -1.5).
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:
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.
Perfect.
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.
Jedster
Well-Known Member
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 so
{ (3((-1)^n)/2) - 1/2 }
I would have lost a mark or 2 on an exam
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!