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crisdean
Joined: 04 Feb 2010
Location: Seoul Special City
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 Posted: Fri Mar 30, 2012 7:03 am Post subject: A call to my fellow Math/Comp. Sci./Logic nerds |
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So I was really bored today, and came up with 3 "Math" problems for my students just to see if they could come up with an answer
So here are 3 sequences, can you list the next 3 elements for each sequence, give an algebraic rule (if possible), and tell me if the sequences are convergent, divergent, or neither.
first:
0
10
1110
3110
second:
1
4
11
20
31
third:
2
3
10
15
26 |
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silkhighway
Joined: 24 Oct 2010
Location: Canada
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| Posted: Fri Mar 30, 2012 11:35 am Post subject: Re: A call to my fellow Math/Comp. Sci./Logic nerds |
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| crisdean wrote: |
So I was really bored today, and came up with 3 "Math" problems for my students just to see if they could come up with an answer
So here are 3 sequences, can you list the next 3 elements for each sequence, give an algebraic rule (if possible), and tell me if the sequences are convergent, divergent, or neither.
first:
0
10
1110
3110
second:
1
4
11
20
31
third:
2
3
10
15
26 |
first:
....42 42 42
Formula: (-22595x^6+534141x^5-493485x^4+22532535x^3
-52883480x^2+59823684x-25048800) / 360
second:
...42 42 42
third:
..42 42 42
Lagrange Polynomials (Or why those "logical pattern" tests on standardized tests are so stupid).  |
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crisdean
Joined: 04 Feb 2010
Location: Seoul Special City
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| Posted: Fri Mar 30, 2012 5:12 pm Post subject: Re: A call to my fellow Math/Comp. Sci./Logic nerds |
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| silkhighway wrote: |
| Lagrange Polynomials (Or why those "logical pattern" tests on standardized tests are so stupid). |
While Lagrange Polynomials are great for interpolating data, they are decidedly less useful for extrapolating data. In particular, if the datapoints are implied to be Integers, as they would be in each of my sequences (perhaps I should have made that explicit.) Were one to use Lagrange Polynomials to extrapolate future data point it is extremely unlikely you would get Integer solutions.
You remind me of those people at university that complained about physics problems being stupid because we ignored things like air-resistance. Pattern recognition is one of the many factors of intelligence, and it can be an important skill. |
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transmogrifier
Joined: 02 Jan 2012
Location: Seoul, South Korea
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| Posted: Fri Mar 30, 2012 6:20 pm Post subject: |
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First:
132110
1113122110
311311222110
No clue on the others |
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crisdean
Joined: 04 Feb 2010
Location: Seoul Special City
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| Posted: Fri Mar 30, 2012 6:50 pm Post subject: |
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| transmogrifier wrote: |
First:
132110
1113122110
311311222110
No clue on the others |
Good! If you were one of my students I'd give you a chupa-chup.
That's one of the two answers I had in mind for that sequence; the other has a similar construction. |
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Konglishman
Joined: 14 Sep 2007
Location: Nanjing
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| Posted: Sat Mar 31, 2012 2:05 am Post subject: Re: A call to my fellow Math/Comp. Sci./Logic nerds |
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| Quote: |
second:
1
4
11
20
31 |
After subtracting the squares 1, 4, 9, 16, and 25, I arrived at 0, 0, 2, 4, and 6. I then figured that replacing the 1st 0 with -2 would make a linear pattern for the difference. So, this new "difference" would be described by 2n-4. Adding the squares back in, I got n^2 + 2n - 4 or (n + 1)^2 - 5 which describes the sequence -1, 4, 11, 20, and 31. Thus, we can get the correct sequence by taking the absolute value.
Therefore, our sequence is described by the formula,
A_n = abs[(n + 1)^2 - 5] where n refers to the nth term. Consequently, the next three terms are 44, 59, and 76.
And of course, this sequence will diverge to infinity. |
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Konglishman
Joined: 14 Sep 2007
Location: Nanjing
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| Posted: Sat Mar 31, 2012 2:28 am Post subject: Re: A call to my fellow Math/Comp. Sci./Logic nerds |
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| Quote: |
third:
2
3
10
15
26 |
After again subtracting the squares, we get the sequence 1, -1, 1, -1, and 1. Therefore, the original sequence can be described by the formula,
A_n = n^2 + (-1)^(n + 1).
So, the next three terms will be 35, 50, and 63. Once again, this sequence will diverge to infinity. |
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crisdean
Joined: 04 Feb 2010
Location: Seoul Special City
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| Posted: Sat Mar 31, 2012 6:56 am Post subject: Re: A call to my fellow Math/Comp. Sci./Logic nerds |
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| Konglishman wrote: |
| Quote: |
second:
1
4
11
20
31 |
After subtracting the squares 1, 4, 9, 16, and 25, I arrived at 0, 0, 2, 4, and 6. I then figured that replacing the 1st 0 with -2 would make a linear pattern for the difference. So, this new "difference" would be described by 2n-4. Adding the squares back in, I got n^2 + 2n - 4 or (n + 1)^2 - 5 which describes the sequence -1, 4, 11, 20, and 31. Thus, we can get the correct sequence by taking the absolute value.
Therefore, our sequence is described by the formula,
A_n = abs[(n + 1)^2 - 5] where n refers to the nth term. Consequently, the next three terms are 44, 59, and 76.
And of course, this sequence will diverge to infinity. |
Wasn't the answer I had intended, but it definitely works, so full points for you.
| Konglishman wrote: |
| Quote: |
third:
2
3
10
15
26 |
After again subtracting the squares, we get the sequence 1, -1, 1, -1, and 1. Therefore, the original sequence can be described by the formula,
A_n = n^2 + (-1)^(n + 1).
So, the next three terms will be 35, 50, and 63. Once again, this sequence will diverge to infinity. |
this is exactly what I had in mind, I'm pretty sure there is only one valid answer for this sequence, but I could be wrong. I wrote the arithmetic expression different, but it's equivalent A_n = |n - i^n|^2 ::edit:: this is incorrect, as noted below.
Last edited by crisdean on Sat Mar 31, 2012 8:51 am; edited 1 time in total |
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ontheway
Joined: 24 Aug 2005
Location: Somewhere under the rainbow...
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| Posted: Sat Mar 31, 2012 7:28 am Post subject: Re: A call to my fellow Math/Comp. Sci./Logic nerds |
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| crisdean wrote: |
| Konglishman wrote: |
| Quote: |
second:
1
4
11
20
31 |
After subtracting the squares 1, 4, 9, 16, and 25, I arrived at 0, 0, 2, 4, and 6. I then figured that replacing the 1st 0 with -2 would make a linear pattern for the difference. So, this new "difference" would be described by 2n-4. Adding the squares back in, I got n^2 + 2n - 4 or (n + 1)^2 - 5 which describes the sequence -1, 4, 11, 20, and 31. Thus, we can get the correct sequence by taking the absolute value.
Therefore, our sequence is described by the formula,
A_n = abs[(n + 1)^2 - 5] where n refers to the nth term. Consequently, the next three terms are 44, 59, and 76.
And of course, this sequence will diverge to infinity. |
Wasn't the answer I had intended, but it definitely works, so full points for you.
| Konglishman wrote: |
| Quote: |
third:
2
3
10
15
26 |
After again subtracting the squares, we get the sequence 1, -1, 1, -1, and 1. Therefore, the original sequence can be described by the formula,
A_n = n^2 + (-1)^(n + 1).
So, the next three terms will be 35, 50, and 63. Once again, this sequence will diverge to infinity. |
this is exactly what I had in mind, I'm pretty sure there is only one valid answer for this sequence, but I could be wrong. I wrote the arithmetic expression different, but it's equivalent A_n = |n - i^n|^2 |
I got the same results as Konglishman.
Your alternate equation doesn't seem to work, however. Perhaps you miswrote it here.
Since i = square root of -1
n=1 ...... x = |2-2i|
n=2 ...... x = 9
n=3 ...... x = |8 + 6i|
etc. |
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crisdean
Joined: 04 Feb 2010
Location: Seoul Special City
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| Posted: Sat Mar 31, 2012 7:45 am Post subject: Re: A call to my fellow Math/Comp. Sci./Logic nerds |
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| ontheway wrote: |
| crisdean wrote: |
| A_n = |n - i^n|^2 |
I got the same results as Konglishman.
Your alternate equation doesn't seem to work, however. Perhaps you miswrote it here.
Since i = square root of -1
n=1 x = |2-2i|
n=2 x = 9
n = 3 x = |8 + 6i|
etc. |
I think perhaps you're misunderstanding some part of the expression:
^ is not multiplication it's exponentiation
and for complex numbers |a+ib| = [(a + ib)(a - ib)]^1/2 (this is the absolute value function or the magnitude r when expressing a complex number in polar form)
thus |a + ib|^2 = (a + ib)(a - ib) = a^2 - b^2
A_n = |n - i^n|^2
= (n - i^n)(n + i^n) ::edit:: I realise now that I have an error here, and this is only true when n is an odd integer, for some reason I had convinced myself it was true for all integers. Your right "ontheway" bonus points to you for pointing out teachers mistake.
= n^2 - i^2n
= n^2 - (-1)^n
= n^2 + (-1)^n+1
Last edited by crisdean on Sat Mar 31, 2012 8:45 am; edited 2 times in total |
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ontheway
Joined: 24 Aug 2005
Location: Somewhere under the rainbow...
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| Posted: Sat Mar 31, 2012 8:10 am Post subject: Re: A call to my fellow Math/Comp. Sci./Logic nerds |
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| crisdean wrote: |
| ontheway wrote: |
| crisdean wrote: |
| A_n = |n - i^n|^2 |
I got the same results as Konglishman.
Your alternate equation doesn't seem to work, however. Perhaps you miswrote it here.
Since i = square root of -1
n=1 x = |2-2i|
n=2 x = 9
n = 3 x = |8 + 6i|
etc. |
I think perhaps you're misunderstanding some part of the expression:
^ is not multiplication it's exponentiation
and for complex numbers |a+ib| = [(a + ib)(a - ib)]^1/2 (this is the absolute value function or the magnitude r when expressing a complex number in polar form)
thus |a + ib|^2 = (a + ib)(a - ib) = a^2 - b^2
A_n = |n - i^n|^2
= (n - i^n)(n + i^n)
= n^2 - i^2n
= n^2 - (-1)^n
= n^2 + (-1)^n+1 |
I didn't misunderstand that you are squaring.
You have now implied above that:
|a + ib|^2 = |a - ib|^2
This will only be true when b = 0.
Try your computation again. Don't use "i" but substitute something else. Try w. Solve. Then replace w with (-1)^1/2. Then use i. |
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crisdean
Joined: 04 Feb 2010
Location: Seoul Special City
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| Posted: Sat Mar 31, 2012 8:22 am Post subject: Re: A call to my fellow Math/Comp. Sci./Logic nerds |
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| ontheway wrote: |
You have now inferred above that:
|a + ib|^2 = |a - ib|^2
This will only be true when b = 0.
Try your computation again. Don't use "i" but substitute something else. Try w. Solve. Then replace w with (-1)^1/2. Then use i. |
I've inferred it because it's true by definition, I'm guessing you didn't study complex number theory, check out the definition of the complex conjugate. |
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ontheway
Joined: 24 Aug 2005
Location: Somewhere under the rainbow...
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| Posted: Sat Mar 31, 2012 8:42 am Post subject: |
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Let's solve this:
x = |n - i^n|^2
where n = 2
x = |2 - i^2|^2
x = |2 - (-1)|^2
x = |2 + 1|^2
x = |3|^2
x = 9 |
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crisdean
Joined: 04 Feb 2010
Location: Seoul Special City
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| Posted: Sat Mar 31, 2012 8:47 am Post subject: |
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| ontheway wrote: |
Let's solve this:
x = |n - i^n|^2
where n = 2
x = |2 - i^2|^2
x = |2 - (-1)|^2
x = |2 + 1|^2
x = |3|^2
x = 9 |
I have an edit in my previous, I admit I made a false assertion. |
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ontheway
Joined: 24 Aug 2005
Location: Somewhere under the rainbow...
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| Posted: Sat Mar 31, 2012 8:53 am Post subject: |
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| crisdean wrote: |
| ontheway wrote: |
Let's solve this:
x = |n - i^n|^2
where n = 2
x = |2 - i^2|^2
x = |2 - (-1)|^2
x = |2 + 1|^2
x = |3|^2
x = 9 |
I have an edit in my previous, I admit I made a false assertion. |
Thanks for the link to complex conjugates.
Something new and fun. |
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