how to express a sequence, pattern rule

igormax

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hello! This is a math problem I wrote, and I'm interested to see if I you can help me articulate the answer:

You are moving backwards through time. You experience a series of "time jumps" that increase in both time jumped and in duration. You have a stopwatch that can accurately time each time jump and duration. You jump back 11 minutes (in other words, you go back in time to 11 minutes ago), and then 3 seconds later you jump back 121 minutes, and then 9 seconds later you jump back 1331 minutes, and then in 27 seconds you jump back again. As you can see, the pattern rule for the duration is that each number increases by a factor of 3, and the pattern rule for the time jumps is that each number increases by a factor of 11.

How would you express the totality of this pattern (as opposed to two separate pattern rules) in words?

How would you express it with an algebraic equation?

thank you!
 

JeffM

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You are not going to reduce this to a single formula because you are considering two different things.

One is the time it takes to go back n steps. The other is the how far back in time you go in n steps.

[math]\text {time in seconds to go back n steps} = 3^n.[/math]
[math]\text {total time in minutes gone back in n steps} = \dfrac{11^{n+1} - 1}{11 - 1} - 1.[/math]
Now the second formula may not be correct because it is not entirely clear what you mean. I was assuming that what you meant was first step takes you back 11 minutes, the second step takes you back 11 + 121 = 132 minutes total, the third step takes you back
11 + 121 + 1331 = 1463. If that is not what you are contemplating, please explain further.
 

igormax

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Thanks for the answer! I meant that you go back 121 minutes on the second jump from the point in time at which you jump back, and 1331 on the third, with the same proviso. I did not mean 132 and 1463.
 

Dr.Peterson

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Thanks for the answer! I meant that you go back 121 minutes on the second jump from the point in time at which you jump back, and 1331 on the third, with the same proviso. I did not mean 132 and 1463.
So each time you jump back, you first returned to the present? Because if you jumped back 11 minutes, and then 121 minutes from that time, you would be 132 minutes back. That's what it sounded like you would mean; you said "you jump back [by] 121 minutes", not "you jump back to 121 minutes from the start of the process".

Also, when you are in the past, are you going forward in time the usual way? If so then if you started at time 0, then after the first jump you're at -11:00, then 3 seconds later it would be -10:57, and then you'd jump back to -131:57, and so on.
 

JeffM

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Thanks for the answer! I meant that you go back 121 minutes on the second jump from the point in time at which you jump back, and 1331 on the third, with the same proviso. I did not mean 132 and 1463.
Do you mean 121 seconds farther back, or just 121 minutes back in total?
 

igormax

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So each time you jump back, you first returned to the present? Because if you jumped back 11 minutes, and then 121 minutes from that time, you would be 132 minutes back. That's what it sounded like you would mean; you said "you jump back [by] 121 minutes", not "you jump back to 121 minutes from the start of the process".

Also, when you are in the past, are you going forward in time the usual way? If so then if you started at time 0, then after the first jump you're at -11:00, then 3 seconds later it would be -10:57, and then you'd jump back to -131:57, and so on.
Thanks for the questions!

You would not return to the present at any time. You only move forward in time during the 3 seconds, 9 seconds, and 27 seconds, so the net effect is that you are moving backwards through time. You would jump back to exactly 121 minutes in total prior to the point in time at which you made that jump. You are correct in your second-paragraph assumption.
 

igormax

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Do you mean 121 seconds farther back, or just 121 minutes back in total?
You jump back 11 minutes, and then you move forward in time (at the customary speed) for 3 seconds. Then you jump back 121 minutes. So at that point in the sequence you have travelled backwards in time exactly 131 minutes and 57 seconds from where you were when you started. The pattern continues thusly, moving, on the whole, backwards through time. Hope that makes sense!
 

JeffM

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OK then

4 jumps = (11 + 121 + 1331 + 14641) minutes - (3 + 9 + 27) seconds = 16104 minutes - 39 seconds =

966240 - 39 = 966201 seconds

The formula says, in seconds, after n jumps you have gone back exactly

[math]60 * \left ( \dfrac{11^{(n+1)} - 1}{10} - 1 \right ) - \dfrac{3^n - 1}{2} + 1[/math]
If n = 4, 11^{n+1} = 161501. Subtracting 1 and dividing by 10 gives 16105 . Less 1 is 16104. Multiply that by 60 to get
966240. And 3^n - 1 = 80. Divided by 2 gives 40. Subtract that from 966240 gives 966200. Add 1 gives 966201 seconds. It checks.
 
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