An extra credit assignment in my class involved determining the next line in this sequence:
1
1 2
2 1
1 2 1 1
1 1 1 2 2 1
the next line is:
3 1 2 2 1 1
you say the previous line outloud, and that is the next line. "one 1" on the first line, so the second line is 1 1 etc etc etc
my argument is that there are actually an infinite number of answers to this problem. I believe that there are an infinite number of answers to any (increasing?) pattern that only gives you a finite number of numbers. There are an infinite number of functions that when solved, give you the sequence of numbers, and any number after that (depending on which equation you choose).
f(x) = (x-1)*(x-11)*(x-21)*(x-1211)*(x-111221)*(x-12345676)* .......
I am arguing that any function is a pattern. My professor (non math) says that this is not true. So far, I have argued that patterns do not need to be infinite in nature (ex: 3 -3 2 -2 1 -1 0 ), thus any function that predicts just one answer after a given pattern is valid. I have also argued that every function is a pattern, but all patterns are not functions.
A very simplified example would be 1, 2. Off the top of your head there are at least 3 very simple (non function)patterns that you can think of. Add one each time, increment the powers of 1 (1^n), or multiply by 2.
Even if you take a very long sequence such as 1 3 6 9 12 15 18 21 24, there are still an infinite number of answers to this problem due to you being able to insert any number into B and have the equation, when solved still equal all of those numbers in a series. (x-1)(x-3)(x-6)(x-9)(x-12)(x-15)(x-18)(x-21)(x-24)(x-B) you can then subsequently add on any (x-???) to the end as many times as you want, and it will still, when solved, give you the original sequence of numbers, or any subset of them that you want to select.
thoughts?
1
1 2
2 1
1 2 1 1
1 1 1 2 2 1
the next line is:
3 1 2 2 1 1
you say the previous line outloud, and that is the next line. "one 1" on the first line, so the second line is 1 1 etc etc etc
my argument is that there are actually an infinite number of answers to this problem. I believe that there are an infinite number of answers to any (increasing?) pattern that only gives you a finite number of numbers. There are an infinite number of functions that when solved, give you the sequence of numbers, and any number after that (depending on which equation you choose).
f(x) = (x-1)*(x-11)*(x-21)*(x-1211)*(x-111221)*(x-12345676)* .......
I am arguing that any function is a pattern. My professor (non math) says that this is not true. So far, I have argued that patterns do not need to be infinite in nature (ex: 3 -3 2 -2 1 -1 0 ), thus any function that predicts just one answer after a given pattern is valid. I have also argued that every function is a pattern, but all patterns are not functions.
A very simplified example would be 1, 2. Off the top of your head there are at least 3 very simple (non function)patterns that you can think of. Add one each time, increment the powers of 1 (1^n), or multiply by 2.
Even if you take a very long sequence such as 1 3 6 9 12 15 18 21 24, there are still an infinite number of answers to this problem due to you being able to insert any number into B and have the equation, when solved still equal all of those numbers in a series. (x-1)(x-3)(x-6)(x-9)(x-12)(x-15)(x-18)(x-21)(x-24)(x-B) you can then subsequently add on any (x-???) to the end as many times as you want, and it will still, when solved, give you the original sequence of numbers, or any subset of them that you want to select.
thoughts?
It's been some years since I bent my mind to pure maths, but I was a little confused by your use of language - I'm not sure if different terms are use in different locations, which could confuse things.
I would have expressed it as "a unique solution that may contain an infinite number of terms". I think there is a difference.
It's like saying there are an infinite number of answers to the value of the square root of two, depending on how many significant figures you choose to consider.
If you allow for infinite series, the McLaurin and/or Taylor series might come close. Although I seem to recall there are certain functions for which you cannot find a Taylor or McLaurin series.
I think again, its a matter of language that's obscuring what you are asking - or my understanding. You have above a polynomial of order 10 (i.e, x appears at most to the 10th power). It has 10, always 10, and only 10 roots [values for which f(x,B)=0]. Of course the values of the 10 roots depend on B as well, hence it must be written as f(x,B), not just f(x).
When you add an extra term (x-C) it becomes an 11th order polynomial f(x,B,C), with 11 roots. And so on.
So I don't think you can say it has an infinite number of solutions. It has n solutions, where n is the order of the polynomial.
and f(x,B) is not equal to f(x,B,C) although they share common roots.
If you graph one these functions, it is going to look like a wavey line, vaguely similar to a section of a sinusoid, although very assymmetric and irregular. The roots (solutions) correspond to where the line crosses the horizontal axis, where f(x,B,c)=0.
As you can see there can be an infinite number of such wavey lines that cross the horzontal axis at exactly the same points, but may do wildly different things between those points.
Yes, maybe I'm agreeing with you - an infinite number of solutions is possible.... f(x), g(x), h(x) that all share a common subset of roots, but are different functions.
It's easy to DRAW suich a curve, but it might be quite impossible to precisely reduce the drawn curve to a polynomial of the form
(x-a)(x-b)(x-c)...(x-n)
I would have expressed it as "a unique solution that may contain an infinite number of terms". I think there is a difference.
It's like saying there are an infinite number of answers to the value of the square root of two, depending on how many significant figures you choose to consider.
If you allow for infinite series, the McLaurin and/or Taylor series might come close. Although I seem to recall there are certain functions for which you cannot find a Taylor or McLaurin series.
I think again, its a matter of language that's obscuring what you are asking - or my understanding. You have above a polynomial of order 10 (i.e, x appears at most to the 10th power). It has 10, always 10, and only 10 roots [values for which f(x,B)=0]. Of course the values of the 10 roots depend on B as well, hence it must be written as f(x,B), not just f(x).
When you add an extra term (x-C) it becomes an 11th order polynomial f(x,B,C), with 11 roots. And so on.
So I don't think you can say it has an infinite number of solutions. It has n solutions, where n is the order of the polynomial.
and f(x,B) is not equal to f(x,B,C) although they share common roots.
If you graph one these functions, it is going to look like a wavey line, vaguely similar to a section of a sinusoid, although very assymmetric and irregular. The roots (solutions) correspond to where the line crosses the horizontal axis, where f(x,B,c)=0.
As you can see there can be an infinite number of such wavey lines that cross the horzontal axis at exactly the same points, but may do wildly different things between those points.
Yes, maybe I'm agreeing with you - an infinite number of solutions is possible.... f(x), g(x), h(x) that all share a common subset of roots, but are different functions.
It's easy to DRAW suich a curve, but it might be quite impossible to precisely reduce the drawn curve to a polynomial of the form
(x-a)(x-b)(x-c)...(x-n)
It took me awhile to see it; there's a mistake though.... it ough to be
1
1 1
2 1
1 2 1 1
1 1 1 2 2 1
Read it as:
start with "1"
above is one "1", i.e., 1 1
above is two "1's", i.e., 2 1
above is one "2", one "1", i.e., 1 2 1 1
above is one "1", one "2", two "1's", i.e., 1 1 1 2 2 1
above is three "1", two "2", one "1", i.e., 3 1 2 2 1 1
Then you read these as 1; 11; 21; 1211
i.e., one, eleven, twenty-one, a-thousand-two-hundred-and-eleven.
which are then regarded as the N roots of the polynomial
f(x) = (x-1)(x-11)(x-21)(x-1211)...(x-N)
solved for f(x)=0
The mathematical expression that actually computes each subsequent number in the sequence based on its place in the sequence is going to be some rather nasty expression, that i don't even want to start figuring out at this time of night!
These sort of things always pop up in the MENSA tests!
1
1 1
2 1
1 2 1 1
1 1 1 2 2 1
Read it as:
start with "1"
above is one "1", i.e., 1 1
above is two "1's", i.e., 2 1
above is one "2", one "1", i.e., 1 2 1 1
above is one "1", one "2", two "1's", i.e., 1 1 1 2 2 1
above is three "1", two "2", one "1", i.e., 3 1 2 2 1 1
Then you read these as 1; 11; 21; 1211
i.e., one, eleven, twenty-one, a-thousand-two-hundred-and-eleven.
which are then regarded as the N roots of the polynomial
f(x) = (x-1)(x-11)(x-21)(x-1211)...(x-N)
solved for f(x)=0
The mathematical expression that actually computes each subsequent number in the sequence based on its place in the sequence is going to be some rather nasty expression, that i don't even want to start figuring out at this time of night!
These sort of things always pop up in the MENSA tests!
One might question if this was actually a mathematical series because without the "english" to explain it, it doesn't make sense. In other words, I doubt that a mathematical "expression" or function could be written that would solve to this set of sequences. Hence, mathematical skills are of no help.
I think it's quite possible - indeed I'm certain of it, since mathematics is just a more concise and elegant way of describing number relationships than can be done in conversational English.
The fact that it can be described in a long drawn out post implies it follows specific unambiguous rules, therefore can be written as a mathematical formula. Although quite likely an 'ugly' one. I'll give it a shot when I'm more awake (it's 1:35am here).
And it would have to be presented here as a GIF, this text editor couldn't handle it.
This is what was given:
"
What is the next line in this pattern:
1
11
21
1211
111221
"
I should have said that: I believe that there are an infinite number of equations that will give the given sequence of numbers from the problem, and subsequently there are also an infinite number of numbers that are the correct next line in the pattern.
"
What is the next line in this pattern:
1
11
21
1211
111221
"
I should have said that: I believe that there are an infinite number of equations that will give the given sequence of numbers from the problem, and subsequently there are also an infinite number of numbers that are the correct next line in the pattern.
infinia said:
In absolute terms yes but IQ is a relative expression of your placement within a bell curve like distribution against people of your age. So as the average "smarts" increase or decrease with age, you normally stay fairly well in the same relative location and hence keep more or less the same IQ.
This thread gave me nightmares - litterally!
Can I accuse you of deploying a Weapon of Math Disruption ?
I don't think so. But it might be because I am interpreting your statement differently.
If the rule is established on how to find the next number in the sequence, based on the current number, then that next number is unique. And hence the sequence - as a whole - is unique.
*IF* you limit the sequence to say 3 terms, then you may be able to find many (or infinite) sequences that share the same first three terms.
But that is not the same thing a saying there are an infinite number of sequences that satisfy 'the rule'. If that were true, then the 'rule' (i.e., your word description of how to find the next line) contains some ambiguity.
I was re-thinking this:
It actually doesn't matter whether there exists an expression to find the nth term, or not. The sequence is defined as
Xn ={1, 11, 21, 1211, 111221, ... }
A set doesn't HAVE to have a formula that makes the next term easy to calculate; it can just be defined by a list of it's elements - it is still a valid sequence.
There are sublte differences among the definitions of 'series' ,' sequence', and 'progression', that I now forget, so in in that truest sense maybe you can't call it a 'series'.
Can I accuse you of deploying a Weapon of Math Disruption ?
I don't think so. But it might be because I am interpreting your statement differently.
If the rule is established on how to find the next number in the sequence, based on the current number, then that next number is unique. And hence the sequence - as a whole - is unique.
*IF* you limit the sequence to say 3 terms, then you may be able to find many (or infinite) sequences that share the same first three terms.
But that is not the same thing a saying there are an infinite number of sequences that satisfy 'the rule'. If that were true, then the 'rule' (i.e., your word description of how to find the next line) contains some ambiguity.
I was re-thinking this:
It actually doesn't matter whether there exists an expression to find the nth term, or not. The sequence is defined as
Xn ={1, 11, 21, 1211, 111221, ... }
A set doesn't HAVE to have a formula that makes the next term easy to calculate; it can just be defined by a list of it's elements - it is still a valid sequence.
There are sublte differences among the definitions of 'series' ,' sequence', and 'progression', that I now forget, so in in that truest sense maybe you can't call it a 'series'.
AudioFreak said:
So if a person is near the median and tested early before a significant education, say in scientific problem solving, then there could be relatively good chance of an increase in IQ. But OTH if that same person was in the top 2%, likely they would remain in that percentile grouping.
Steerpike said:
Agreed, one can have a "sequence" of totally unrelated numbers or a sequence of letters, but this is not mathematical. To me, to be definable as a "math" problem, it has to be a "series" where an arbitrary term can be defined by its location. If this is not true then it cannot be looked at mathematicaly.
Holy Mother of Gauss! How did 9 get in there?
I think I must have mistyped '19' as '9', and then, with brain switched off, seen that 9 was out-of-sequence & just moved it.
Has anyone here read "The man who loved only numbers", about Hungarian mathematician Paul Erdos?
I think I must have mistyped '19' as '9', and then, with brain switched off, seen that 9 was out-of-sequence & just moved it.
Great answer! Illustrates my point, i.e., this sequence does not come from a formula that determines each element by virtue of its place in the sequence, nor on the element that precedes it.
Has anyone here read "The man who loved only numbers", about Hungarian mathematician Paul Erdos?
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.