another math set problem

[Heewon]

Hi guys! I have another question involving sets. I need to rewrite the set:

A = {..., -4, -1, 2, 5, 8,...}

Into a notation like the example below

B = {..., -4, -2, 0, 2, 4,...}
B = {2x : x in Z }

Thanks

 

[Heewon]

Never mind solved it

 

[Steven G]

Hi guys! I have another question involving sets. I need to rewrite the set:

A = {..., -4, -1, 2, 5, 8,...}

Into a notation like the example below

B = {..., -4, -2, 0, 2, 4,...}
B = {2x : x in Z }

Thanks

Where is your attempt? The common difference between 2 consecutive numbers is 3. So I would think the 3 times table {...-9, -6, -3, 0, 3, 6,...} = {3x | x is in Z}. Can you adjust the numbers in my set so they become the numbers in your set? There are actually an infinite number of ways to do that. You just need to find one. Let us know how you make out.

 

[Heewon]

Yea it seemed that it was just a common pattern :
3x+2 for x in Z
I think I was over thinking the question :lol:

 

[HallsofIvy]

Hi guys! I have another question involving sets. I need to rewrite the set:

A = {..., -4, -1, 2, 5, 8,...}

Into a notation like the example below

B = {..., -4, -2, 0, 2, 4,...}
B = {2x : x in Z }

Thanks

So you want to find some function that gives those numbers? I don't see any very simple function but there always a polynomial of degree no more than n- 1 through any given n points. Here we have 5 points so we can look for \(\displaystyle y= ax^4+ bx^3+ cx^2+ dx+ e\). Taking these to be (-2, -4), (-1, -1), (0, 2), (1, 5), and (2, 8), we have
16a- 8b+ 4c- 2d+ e= -4, a- b+ c- d+ e= -1, e= 2, a+ b+ c+ d+ e= 5, and 16a+ 8b+ 4d+ 2d+ e= 8.

However, there is no guarantee that such a formula will give the values indicated by "...". In fact there is no guarantee that the pattern you give for "B" actually will give the un-listed numbers in B unless you assume that "2x" is the correct formula. It is the simplest formula that will give "-4, -2, 0, 2, 4" but there is no guarantee that the same pattern continues.