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Help with this geometric sequence watch

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    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
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    (Original post by Jyashi)
    Hi guys please look at the table below
    It's gibberish.

    May I politely suggest you proofread questions before posting them?
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    Name:  Screenshot 2015-11-10 at 17.12.55.png
Views: 70
Size:  4.7 KBI couldn't submit the image via editing so did it as a reply
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    (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
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    (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.
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    (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?
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    (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
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    (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
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    Looking at the table, if you are expliclty saying that the even terms take the form a_{2n} =2 \times  2^n, and the odd terms a_{2n+1} = 3 \times 2^n, then I think

    a_n = (2 + \frac{1}{2}(1- (-1)^n)(\frac{3}{\sqrt{2}} - 2))\sqrt{2^n}

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

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

    a_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 \frac{3}{\sqrt{2}} to get it to work for odd n.

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

    Then I just used that \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...]
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    thank you
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    (Original post by DFranklin)
    If you start from

    a_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 \frac{3}{\sqrt{2}} to get it to work for odd n.

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

    Then I just used that \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?
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    (Original post by DFranklin)
    Looking at the table, if you are expliclty saying that the even terms take the form a_{2n} =2 \times  2^n, and the odd terms a_{2n+1} = 3 \times 2^n, then I think

    a_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.
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    (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.
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    (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.
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    (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...2C+3%2C+4%2C+5
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    Yes I can confirm now that the expression does work properly. Thank you
 
 
 
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