Wikipedia:Reference desk/Archives/Mathematics/2016 January 18

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January 18

Increase and decrease

There is an initial number a and every step increases or decreases by $a_{n}b$, and there is n steps.

An example for three steps, where $a_{n}$ is the number at step n

$a_{1}=a\pm ab$

$a_{2}=a_{1}\pm a_{1}b$

$a_{3}=a_{2}\pm a_{2}b$

I figured out the formula for any $a_{n}$.

(1) $a(\pm b+1)^{n}$

But it works only with b>0 for an increase of a>0 and a decrease of a<0, and -1>b<0 for a decrease of a>0 down to 0 (which is impossible to reach, in $a(-b+1)^{n}=0$ the number n must be infinite).

But for a negative (reaching 0 and then going further from 0) decrease of a>0 with b>|1|, it must be thus

(2) $-a(b-1)(b+1)^{n-1}$

where b must be positive, not negative as in (1).

A small alteration of (1) also will work

(3) $-a(b+1)^{n-1}$

An example for a=1 decreasing by 200% (b=2) every step:

$a_{1}=1-1\times 2=-1$

$a_{2}=-1-1\times 2=-3$

$a_{3}=-3-3\times 2=-9$

With formula (1) it won't work:

$1\times (-2+1)^{3}=-1$

but formulas (2) and (3) work fine:

$-1\times (2-1)\times (2+1)^{2}=-9$

$-1\times (2+1)^{2}=-9$

Is there an explanation for this?

P.S. By the way, what branch of mathematics am I doing?--Lüboslóv Yęzýkin (talk) 21:45, 18 January 2016 (UTC)

Recurrence relation or difference equation. Loraof (talk) 23:45, 18 January 2016 (UTC)

I think your calculation for a2 is wrong. It should be $a_{2}=a_{1}-a_{1}(b)=-1-(-1)(2)=1.$ So formula (1) gives the right answer. Loraof (talk) 00:01, 19 January 2016 (UTC)

Yes, theoretically, I tried that and came to the same. But if we imagine the number line, we start from 1, then go to the left by the double of that 1*2 and end at -1. Then we go to the left by the double of what we got that is |1|*2 and we end at -3. Then we go to the left again by that pattern and end at -9, -27, -81, -243... Probably, it is not decreasing in a strict sense but what is then, a "negative stepping back"? With (1) it will always end either at -1 for an odd step or at 1 for an even step. Note also that for the opposite case, with a negative start (e.g. a=-1) and b<-1 (e.g. b=-2) formula (1) will always end at 1.
P.S. Sorry, if my reasoning sounds amateurish, I'm not a mathematician at all and I might have used wrong terminology and understood it all wrong, but this case really intrigues me. Did I hack maths? :) --Lüboslóv Yęzýkin (talk) 01:30, 19 January 2016 (UTC)

Imagine a problem (I'd call it "A Mafia Loan Problem"). A person borrowed $1, the debt grows by the above-mentioned pattern every month. In the 1st month his ballance is -1, he owes $2. In the 2nd the balance is -3 and he owes $4. In the 3rd the balance is -9 and he owes $10. And so on. What would the balance be in 12 months? With (3) it is easy, $-1\times (2+1)^{11}=-177147$. --Lüboslóv Yęzýkin (talk) 01:39, 19 January 2016 (UTC)

So if I'm understanding this right, you want the value to always decrease, whether a is positive or negative (so for a=1, b=2, you get: a0=1, a1= 1 - 2*1= -1, a2= -1 - |2*-1| = -3, etc.). This means your formula is actually $a_{n}=a_{n-1}+|a_{n-1}|b$ where |x| is the absolute value of x (i.e. |-a| = a). I think, given the absolute value expression in here, you're not going to be able to get a simple expression for the result, but perhaps someone who knows more maths than I do can come along and provide it. MChesterMC (talk) 11:09, 19 January 2016 (UTC)

@MChesterMC: Thanks, you seem to be right. Now it is more straightforward: with b>0 it is always goes to the right on the number line, with b<0 goes to the left. We can also avoid the absolute value and write $a_{n}=a_{n-1}+b{\sqrt {a_{n-1}^{2}}}$.

But I have no idea how to generalize a formula for any $a_{n}$. E.g.:

$a_{0}=a$

$a_{1}=a+b{\sqrt {a^{2}}}$

$a_{2}=a_{1}+b{\sqrt {a_{1}^{2}}}=(a+b{\sqrt {a^{2}}})+b{\sqrt {(a+b{\sqrt {a^{2}}})^{2}}}$

$a_{3}=a_{2}+b{\sqrt {a_{2}^{2}}}=((a+b{\sqrt {a^{2}}})+b{\sqrt {(a+b{\sqrt {a^{2}}})^{2}}})+b{\sqrt {((a+b{\sqrt {a^{2}}})+b{\sqrt {(a+b{\sqrt {a^{2}}})^{2}}})^{2}}}$

And so on, more steps the longer and more esoteric the formula becomes. Wolfram Alpha suggests the simplifications: $a_{1}=a(b+1)$, $a_{2}=a(b+1)^{2}$, $a_{3}=a(b+1)^{3}$, but only if a and b are positive, that is the same formula (1) I've suggested at the beginning. But how to find a single formula that will work with any rational number? We returned to the starting question. I suggest we have two different sequences (I have no idea how they are named): one tends to or from zero, the other goes in any direction on the number line, and I've tried to figure out a single formula for the both, but maybe it is impossible?--Lüboslóv Yęzýkin (talk) 09:50, 23 January 2016 (UTC)