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Help with this geometric sequence

Hi guys please look at the table below


So the bacteria doubles every 2 minutes and i want to know an explicit (and recursive if possible) formula to represent this for every nth minute. So far this is what i have come up with

T(0) = 2
T(n) = 2(T_n-1) - 1
T(1) = 2(T_1-1) - 1 The problem is the -1 figure has to change so for:
T(2) = 2(T_2-1) - 2 it changes to -2 and will also be -2 for T(3). The problem is that at T(4) and T(5) the formula will remain:

T(4) = 2(T_4-1) - 4 = 8 But after T(5) it will change to -8 for T(6) and T(7) and then change to -16 for T(8) and T(9).

Is there any way i can formulate this in a proper explicit expression.
Please let me know what you guys think.

Maths is awesome.

Edit: There was a problem in uploading the image so i have done so in the reply below. Please scroll down and have a look
(edited 8 years ago)

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Original post by Jyashi
Hi guys please look at the table belowIt's gibberish.

May I politely suggest you proofread questions before posting them?
Reply 2
Screenshot 2015-11-10 at 17.12.55.pngI couldn't submit the image via editing so did it as a reply
Reply 3
Original post by DFranklin
It's gibberish.

May I politely suggest you proofread questions before posting them?


Hi sorry for that i will remember to do so in the future. Please look below your reply for an image of the table. Thank you
Original post by Jyashi
Hi sorry for that i will remember to do so in the future. Please look below your reply for an image of the table. Thank you
It would have made more sense (and be easier to refer to) if you had transcribed the values into text.

Anyhow, the values you have posted do not form a geometric sequence.
Reply 5
Original post by DFranklin
It would have made more sense (and be easier to refer to) if you had transcribed the values into text.

Anyhow, the values you have posted do not form a geometric sequence.


Well what form of sequence or series do you think it is then?
Reply 6
Original post by Jyashi
Well what form of sequence or series do you think it is then?


you can make a gp for even seconds and odd seconds
Reply 7
Original post by a2874
you can make a gp for even seconds and odd seconds


even seconds have r = 2 (considering t =0 also even )

odd have r= 3
Looking at the table, if you are expliclty saying that the even terms take the form a2n=2×2na_{2n} =2 \times 2^n, and the odd terms a2n+1=3×2na_{2n+1} = 3 \times 2^n, then I think

an=(2+12(1(1)n)(322))2na_n = (2 + \frac{1}{2}(1- (-1)^n)(\frac{3}{\sqrt{2}} - 2))\sqrt{2^n}

will work.
Reply 9
Original post by DFranklin
Looking at the table, if you are expliclty saying that the even terms take the form a2n=2×2na_{2n} =2 \times 2^n, and the odd terms a2n+1=3×2na_{2n+1} = 3 \times 2^n, then I think

an=(2+12(1(1)n)(322))2na_n = (2 + \frac{1}{2}(1- (-1)^n)(\frac{3}{\sqrt{2}} - 2))\sqrt{2^n}

will work.


wouldn't it be easier to consider 2 gp instead of the complicated general term you provided ?
Original post by a2874
wouldn't it be easier to consider 2 gp instead of the complicated general term you provided ?
Yes, but since I gave the formula for both the GP as well and OP asked for an explicit formula...
Reply 11
Original post by DFranklin
Yes, but since I gave the formula for both the GP as well and OP asked for an explicit formula...


If you dont mind me asking, how did you calculate the explicit formula ?
Original post by a2874
If you dont mind me asking, how did you calculate the explicit formula ?
If you start from

an=2×2n/2a_n = 2 \times 2^{n/2} (which is correct for even n), and then look at what happens for odd n, you find you need to replace the first 2 with 32\frac{3}{\sqrt{2}} to get it to work for odd n.

That is, you need to add 322\frac{3}{\sqrt{2}} - 2 to the first 2 when n is odd.

Then I just used that 12(1(1)n)\frac{1}{2}(1-(-1)^n) is 0 for n even and 1 for n odd.

[And if you were wondering - part of the point of doing this was to illustrate that sometimes asking for a single formula is not the most sensible thing to do...]
Reply 13
thank you
Reply 14
Original post by DFranklin
If you start from

an=2×2n/2a_n = 2 \times 2^{n/2} (which is correct for even n), and then look at what happens for odd n, you find you need to replace the first 2 with 32\frac{3}{\sqrt{2}} to get it to work for odd n.

That is, you need to add 322\frac{3}{\sqrt{2}} - 2 to the first 2 when n is odd.

Then I just used that 12(1(1)n)\frac{1}{2}(1-(-1)^n) is 0 for n even and 1 for n odd.

[And if you were wondering - part of the point of doing this was to illustrate that sometimes asking for a single formula is not the most sensible thing to do...]


That may be so. But isn't our scientific community breaking their backs to find one mathematical formula that describes the theory of everything?
(edited 8 years ago)
Reply 15
Original post by DFranklin
Looking at the table, if you are expliclty saying that the even terms take the form a2n=2×2na_{2n} =2 \times 2^n, and the odd terms a2n+1=3×2na_{2n+1} = 3 \times 2^n, then I think

an=(2+12(1(1)n)(322))2na_n = (2 + \frac{1}{2}(1- (-1)^n)(\frac{3}{\sqrt{2}} - 2))\sqrt{2^n}

will work.


Thank you very much for your effort. But feeding this does not seem to generate the right answers. Perhaps this sequence is too complicated to work with.
Original post by Jyashi
Thank you very much for your effort. But feeding this does not seem to generate the right answers.
Just tried it in Excel, it works fine.

Perhaps this sequence is too complicated to work with.
Too complicated for you, possibly.
Reply 17
Original post by DFranklin
Just tried it in Excel, it works fine.

Too complicated for you, possibly.


Relax buddy. I just added that line because I didn't want to seem disrespectful.

And feeding this in a scientific calculator does not give the right results even after triple checking. And I would encourage everyone who is reading to try it out and report back here so mine is not a fluke.
Original post by Jyashi
Relax buddy. I just added that line because I didn't want to seem disrespectful.

And feeding this in a scientific calculator does not give the right results even after triple checking. And I would encourage everyone who is reading to try it out and report back here so mine is not a fluke.
http://www.wolframalpha.com/input/?i=evaluate+%282%2B%281%2F2%29+*+%281+-+%28-1%29%5En%29*%283%2Fsqrt%282%29+-+2%29%29*sqrt%282%5En%29+when+n+%3D0%2C+1%2C+2%2C+3%2C+4%2C+5

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