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Math Help - I asked my teacher about this problem...

  1. #1
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    I asked my teacher about this problem...

    Hello everyone

    Forgive me if I am excluding information of some sort. My english mathematical repetoire isn't too impressive and I don't know any discrete math basics.

    Anyway I asked my math teacher how to form the following problem into an simple formula. He said that we're talking discrete math when encountering problems of this sort, and he wouldn't rather sit and do discrete math early in the morning.

    My problem was this.

    We got a base of 20, and a growth of \frac {base} {5}. This number must be a simple number, such as 4 in this case. Any decimals will be ignored (and the number rounded down, example: 5.5 = 5)

    The beginning

     20 + \frac { 20 } { 5 } = 20 + 4 = 24 <- The new base

     24 + \frac { 24 } { 5 } = 24 + 4.8 = 28

    Is there a way to come up with a formula for expressing the result of N times we do this operation?

    I am currently absolutely uninitiated into discrete math, so keep the talk simple and explanatory, please.

    Thanks for all replies
    Last edited by λιεҗąиđŗ; October 10th 2007 at 09:25 AM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by λιεҗąиđ€ŗ View Post
    Hello everyone

    Forgive me if I am excluding information of some sort. My english mathematical repetoire isn't too impressive and I don't know any discrete math basics.

    Anyway I asked my math teacher how to form the following problem into an simple formula. He said that we're talking discrete math when encountering problems of this sort, and he wouldn't rather sit and do discrete math early in the morning.

    My problem was this.

    We got a base of 20, and a growth of \frac {base} {5}. This number must be a simple number, such as 4 in this case. Any decimals will be ignored (and the number rounded down, example: 5.5 = 5)

    The next step is the new base 20 + \frac {20}{5} divided by 5 again.

     \frac { 20 + 4 } { 5 }

    Is there a way to come up with a formula for expressing the result of N times we do this operation?

    I am currently absolutely uninitiated into discrete math, so keep the talk simple and explanatory, please.

    Thanks for all replies
    do we always add \frac {\mbox{Base}}{5} to 20, or do we replace 20 with the previous base?

    If it is always 20 (which probably makes no sense), then a reiteration formula could be:

    b_{n + 1} = 20 + \frac {b_n}5

    if we start at 20, but replace the 20 with the old base when calculating a new base, then a possible reiteration formula is:

    b_{n + 1} =  20 + \frac {b_1 + b_2 + .... + b_n}5

    where b_{n + 1} is the new base, b_n is the previous base

    do you see how i came up with these?
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  3. #3
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    We add the growth to the base, then divide the new base for the new growth.

    The beginning

    20 + \frac { 20 } { 5 } = 20 + 4 = 24, the new base

    24 + \frac { 24 } { 5 } = 24 + 4.8 = 28, we remove all decimals.

    Reviewing how i first put this problem, that way was bad. Use the above, looks much more easy to understand.

    And so it continues. Can we create an simple formula to see what the result would be after N times?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by λιεҗąиđ€ŗ View Post
    We add the growth to the base, then divide the new base for the new growth.

    The beginning

    20 + \frac { 20 } { 5 } = 20 + 4 = 24, the new base

    24 + \frac { 24 } { 5 } = 24 + 4.8 = 28, we remove all decimals.

    Reviewing how i first put this problem, that way was bad. Use the above, looks much more easy to understand.

    And so it continues. Can we create an simple formula to see what the result would be after N times?
    that is what i did.

    let me show you the steps for the last formula.

    b_1 = 20

    \Rightarrow b_2 = b_1 + \frac {b_1}5 = 20 + \frac {b_1}5

    \Rightarrow b_3 = b_2 + \frac {b_2}5 = \left( 20 + \frac {b_1}5\right) + \frac {b_2}5 = 20 + \frac {b_1 + b_2}5

    \Rightarrow b_4 = b_3 + \frac {b_3}5 = \left( 20 + \frac {b_1 + b_2}5\right) + \frac {b_3}5 = 20 + \frac {b_1 + b_2 + b_3}5
    .
    .
    .
    \Rightarrow b_{n + 1} = b_n + \frac {b_n}5 = 20 + \frac {b_1 + b_2 + ... + b_n}5

    and we take the floor function of the base to get rid of decimals
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  5. #5
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    Thank you Jhevon. I have one more question.

    Is there a way to calculate, for example, this operation 10 times, on a not too pc-like calulator?
    Last edited by λιεҗąиđŗ; October 10th 2007 at 10:13 AM.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by λιεҗąиđŗ View Post
    Thank you Jhevon. I have one more question.

    Is there a way to calculate, for example, this operation 10 times, on a not too pc-like calulator?
    not using this formula here. we used an iterative formula, and so it is recursive in nature. in order to calculate new bases, we need to know what comes before. if you wanted to have a straight formula where you, say, plugged in 10 to find the base resulting from doing the operation 10 times, you would need to develope a different, non-recursive, formula. at the moment, i do not see any patterns to do that. i ran the operation 11 times to see if there is some relationship between the bases that i can exploit, but it vain. perhaps the rounding down makes this impossible, or perhaps i just can't "see" it. using this formula, if you want to find the 10th base, you must know the previous 9.
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