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.

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

2. Originally Posted by λιεҗąиđ€ŗ
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.

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?

3. 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?

4. Originally Posted by λιεҗąиđ€ŗ
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.

$b_1 = 20$

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

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

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

and we take the floor function of the base to get rid of decimals

5. 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?

6. Originally Posted by λιεҗąиđ€ŗ
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.