Write an equation for the sequence
- Thread starter Chris S.
- Start date
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,087
I don't understand. This is a sequence of equations, not of numbers, and they all say x is 1. I suspect that something in the formatting changed from what you intended.
Maybe what you meant is that the sequence is 3, 13, 53, ..., and the rest of each line is an equation whose solution is that value of x. But the differences are 10, 40, ..., and it is not clear what pattern is intended.
Please explain further.
EDIT: I think you're saying, not that the difference between terms is 4x+1, but that for any term x, the next term is 4x+1. Is that right?
Yes. Sorry. That is what I meant. I’ll do another one that I already solved but it was easier because the constant is 1.
x=5 (3x+1)/16 =1
x=21 (3x+1)/64 =1
x=85 (3x+1)/256 =1
x=...
This can be expressed as [2^(2x+2)-1]\3
Where x>or=1
And the difference between the x values is 4x+1
But for some reason I can’t express the sequence on the other one because the constants change.
I realized I said difference is 4x+1 so I posted another on deleting that statement. ?
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,087
Please explain what the role of the equations (e.g. (3x+1)/16 =1) is.
Then tell us what [2^(2x+2)-1]\3 means. Is "\" supposed to be division ("/")? Is x the value of the term (5, 21, 85, ...), or the index (1, 2, 3, ...) or what? We'd usually call the index n or k, and refer to the nth term as [MATH]x_n[/MATH].
What does it mean to say "the difference between the x values is 4x+1"? As I said before, you seem to mean that this is a recursive definition, how to find the next term from a given one. That is not a "difference".
And how did you obtain your answer? That will help me see how to adapt your method to the other problem.
Steven G
Elite Member
- Joined
- Dec 30, 2014
- Messages
- 14,383
I think I understand somewhat the OP is saying.
If x=3, then (9x + 5)/32 =1 = (9x+5)/2^5
If x=13, then (9x + 11)/128 =1 = (9x+11)/2^7
If x=53, then (9x + 35)/512 =1= (9x+35)/2^9
Now if x goes from x to 4x+1 then 3 goes to 13 and 13 goes to 53.
How you go from 5 to 11 to 35 might be a mystery. All I can see is 5*2+1 = 11, 11*3 + 2 = 35 and 35*4 + 3 = 143. Let's test this!
4*53 + 1 = 213. So x=213. Now (9*213 + 143)/2^11 does equal 1
Finish up from here.
If x=3, then (9x + 5)/32 =1 = (9x+5)/2^5
If x=13, then (9x + 11)/128 =1 = (9x+11)/2^7
If x=53, then (9x + 35)/512 =1= (9x+35)/2^9
Now if x goes from x to 4x+1 then 3 goes to 13 and 13 goes to 53.
How you go from 5 to 11 to 35 might be a mystery. All I can see is 5*2+1 = 11, 11*3 + 2 = 35 and 35*4 + 3 = 143. Let's test this!
4*53 + 1 = 213. So x=213. Now (9*213 + 143)/2^11 does equal 1
Finish up from here.
There is a reason that we ask to see the original wording. I am wondering if the word "functio" was used and if there were comments about integers.
I suspect that jomo has finally figured out what the background for this problem is
[MATH]x = 3 \implies \dfrac{9x + 2^1 + 3}{2^5} = 1\\ x = 13 \implies \dfrac{9x + 2^3 + 3}{2^7} = 1\\ x = 53 \implies \dfrac{9x + 2^5 + 3}{2^9} = 1.[/MATH]In other words, we have [MATH]f(x,\ n) = \dfrac{9x + 2^{(2n-1)} + 3}{2^{(2n+3)}} = 1 \implies\\ 9x + 2^{(2n-1)} + 3 = 2^{(2n+3)} \implies \\ 9x + 3 = 2^{(2n+3)} - 2^{(2n-1)} \implies \\ 3(3x + 1) = 2^{(2n-1)}(2^4 - 1) = 15 * 2^{(2n-1)} \implies\\ 3x + 1 = 5 * 2^{(2n-1)} \implies \\ x = \dfrac{5 * 2^{(2n-1)} - 1}{3}.[/MATH]Will x necessarily be an integer?
[MATH]x = \dfrac{5 * 2^{(2n-1)} - 1}{3} \implies\\ x = \dfrac{3 * 2^{2n-1)} + 2 * 2^{(2n-1)}- 1}{3} \implies\\ x = 2^{(2n-1)} + \dfrac{2^{(2n)} - 1}{3}.\\ \text {But } \exists \text { integer } k \text { such that } 2^n = 3k \pm 1 \implies \\ 2^{2n} = 9k^2 \pm 6k + 1.\\ \therefore x = 2^{(2n-1)} + \dfrac{9k^2 \pm 6k + 1 1}{3} \implies\\ x = 2^{(2n-1)} + 3k^2 \pm 2k, \text { an integer.}[/MATH]So we can calculate an integer x whenever f(x, n) = 1. For example,
[MATH]f(x,\ 4) = 1 \implies \\ x = \dfrac{5 * 2^{(2*4-1)} - 1}{3} = \dfrac{5 * 2^7 - 1}{3} = 213.[/MATH]Let's check
[MATH]\dfrac{9 * 213 + 2^7 + 3}{2^{11}} = \dfrac{1917 + 128 + 3}{2048} = 1. \ \checkmark[/MATH]
Neat little problem. Pity we had to struggle to determine what it really was.
I suspect that jomo has finally figured out what the background for this problem is
[MATH]x = 3 \implies \dfrac{9x + 2^1 + 3}{2^5} = 1\\ x = 13 \implies \dfrac{9x + 2^3 + 3}{2^7} = 1\\ x = 53 \implies \dfrac{9x + 2^5 + 3}{2^9} = 1.[/MATH]In other words, we have [MATH]f(x,\ n) = \dfrac{9x + 2^{(2n-1)} + 3}{2^{(2n+3)}} = 1 \implies\\ 9x + 2^{(2n-1)} + 3 = 2^{(2n+3)} \implies \\ 9x + 3 = 2^{(2n+3)} - 2^{(2n-1)} \implies \\ 3(3x + 1) = 2^{(2n-1)}(2^4 - 1) = 15 * 2^{(2n-1)} \implies\\ 3x + 1 = 5 * 2^{(2n-1)} \implies \\ x = \dfrac{5 * 2^{(2n-1)} - 1}{3}.[/MATH]Will x necessarily be an integer?
[MATH]x = \dfrac{5 * 2^{(2n-1)} - 1}{3} \implies\\ x = \dfrac{3 * 2^{2n-1)} + 2 * 2^{(2n-1)}- 1}{3} \implies\\ x = 2^{(2n-1)} + \dfrac{2^{(2n)} - 1}{3}.\\ \text {But } \exists \text { integer } k \text { such that } 2^n = 3k \pm 1 \implies \\ 2^{2n} = 9k^2 \pm 6k + 1.\\ \therefore x = 2^{(2n-1)} + \dfrac{9k^2 \pm 6k + 1 1}{3} \implies\\ x = 2^{(2n-1)} + 3k^2 \pm 2k, \text { an integer.}[/MATH]So we can calculate an integer x whenever f(x, n) = 1. For example,
[MATH]f(x,\ 4) = 1 \implies \\ x = \dfrac{5 * 2^{(2*4-1)} - 1}{3} = \dfrac{5 * 2^7 - 1}{3} = 213.[/MATH]Let's check
[MATH]\dfrac{9 * 213 + 2^7 + 3}{2^{11}} = \dfrac{1917 + 128 + 3}{2048} = 1. \ \checkmark[/MATH]
Neat little problem. Pity we had to struggle to determine what it really was.
Thank you Jeff! I’m not reading out of anything just working on something and came across this.
But that equation you came up with is exactly what I was trying to find.
So in my second example is there a better way to write the equation other than what I came up with?
Steven G
Elite Member
- Joined
- Dec 30, 2014
- Messages
- 14,383
Yep!
Well, it looks as though you have got x and n scrambled in your answer to the second example, but if you got your notation cleaned up, I suspect you will be OK.
One more question Jeff then you will have answered everything I was looking for and can proceed.
In your example you factored (2^2n+3) - (2^2n-1) into (2^(2n-1))(2^(4) -1). Will you please explain step by step how you got this. I under stand the rest of it. You have been a tremendous help.
In your example you factored (2^2n+3) - (2^2n-1) into (2^(2n-1))(2^(4) -1). Will you please explain step by step how you got this. I under stand the rest of it. You have been a tremendous help.
As I said, I was not sure of the exact description of the problem. But I guessed that we were looking for answers where both x and n were integers.
So when I got to the answer
[MATH]x = \dfrac{5 * 2^{(2n-1)} - 1}{3}[/MATH],
I did not know whether x would be an integer for every integer value of n. There are probably several ways to try to answer this question, and I started messing around. (This is my typical way to solving math problems: try something.) Now obviously that fraction will equal an integer if and only if the numerator is evenly divisible by 3, and 5 is not divisible by 3. But 5 = 3 + 2, and 3 is certainly divisible by 3. So I gave that a shot.
[MATH]x = \dfrac{2^{(2n-1)}(3 + 2) - 1}{3} = \dfrac{3 * 2^{(2n-1)} + 2 * 2^{(2n-1)} - 1}{3} =\\ 2^{(2n-1)} + \dfrac{2^{2n} - 1}{3}.[/MATH].
Now this is interesting. Having over the years done many problems, I have in my storehouse of math trivia this fact, looking at integers p, q, and r, if
[MATH]p = q - 1 \text { and } q + 1 = r \implies p + 2 = r.\\ Case \ 1: \ \dfrac{q}{3} = \text { integer } s \implies q \text { divisible by } 3.\\ Case \ 2: \ \dfrac{q}{3} = \text {integer } s + \dfrac{1}{3}.\\ \dfrac{p}{3} = \dfrac{q - 1}{3} = \dfrac{q}{3} - \dfrac{1}{3} =\\ s + \dfrac{1}{3} + \dfrac{1}{3} = \text {integer } s + 1 \text { an integer because s an integer} \implies\\ p \text { divisible by } 3.\\ Case \ 3: \ \dfrac{q}{3} = \text {integer } s + \dfrac{2}{3}.\\ \dfrac{r}{3} = \dfrac{q + 1}{3} = \dfrac{q}{3} + \dfrac{1}{3} =\\ s + \dfrac{2}{3} + \dfrac{1}{3} = s + 1 \text { an integer because s an integer} \implies\\ r \text { divisible by } 3. [/MATH]Of three successive integers, exactly one is evenly divisible by 3. Obvious when you think about it, but not taught in third grade arithmetic. Now a power of 2 is not evenly divisible by 3 so that means that for our numerator to be divisible by 3, this power of 2 must be one more than a multiple of 3, right, so as to eliminate that minus 1 in the numerator. Hmm. Sort of a problem because I have no guarantee that an even number is 1 more than a multiple of 3. So I scratch my head for a while, and notice that this is an even power of 2. Does that make a difference? Sure because if I square minus 1 I get plus 1. Let's give that a try. What am I giving a try? Why that if a I square a power of 2, it will be one more than a multiple of 3.
[MATH]\sqrt{2^{2n}} = 2^n.[/MATH]That is an even number not divisible by 3. Therefore either 2^n - 1 is divisible by 3 or 2^n + 1 is divisible by 3. Still with me? How do I express that mathematically?
[MATH]\exists \ k \text { such that } 3k + 1 = 2^n \text { or } 3k - 1 = 2^n\implies\\ 2^n = 3k \pm 1 \implies 2^{2n} = (3k \pm 1)^2 = 9k^2 - 6k + 1.[/MATH]So I substitute that into my numerator (Uh oh. I see I left out a minus sign up above.)
[MATH]x = 2^{(2n-1)} + \dfrac{2^{2n} - 1}{3} =2^{(2n-1)} + \dfrac{9k^2 - 6k + 1 - 1}{3} =\\ 2^{(2n-1)} + \dfrac{9k^2 - 6k}{3} = 2^{(2n-1)} + 3k^2 - 2k, \\ \text { which is an integer because both 2 and k are integers.}[/MATH].
Thus, if f(x, n) = 1 and n is an integer so is x. And we have a formula for x. We are done.
Last edited:
Are you using some other program to write the equations then copying and pasting it on here, or are you actually writing the code out?
I followed your description and I finally understood what you where talking about. I did a few more examples just to make sure and all came out as expected. However, I ran into one that doesn’t work when I factor. I’m guessing it has something to do with the coefficients not being the same, but that doesn’t make much sense since they all are being multiplied together.
5*2^(2n+3) - 2^(2n) - 3
I factored as: 5*2^(2n)*(2^(3) - 1) - 3 =
35*2^(2n) - 3
If I don’t factor it, it comes out as expected but when I try factoring it, it stops working. Why will this not work?
5*2^(2n+3) - 2^(2n) - 3
I factored as: 5*2^(2n)*(2^(3) - 1) - 3 =
35*2^(2n) - 3
If I don’t factor it, it comes out as expected but when I try factoring it, it stops working. Why will this not work?