Sum of infinitely repeating equation

[finks]

Hello. Sorry if I am posting this in the wrong section, as it is my first post and it is sort of a weird one question. This isn't for any sort of assignment, it was just something I randomly thought of and had no clue how to write it out correctly to solve.

So the main equation is...

x = (a * b) / c

b and c are always fraction (decimal representation of fractions)

That part is easy enough for me to figure out, but what I'm trying to do is take the solution for x from equation, and place it in a for the next round of the same equation. I basically want this to run infinitely (obviously this would keep going forever, but maybe have it stop once x becomes a few decimal places below 1) and always adding x from each round to the sum.

Since that might be a bit confusing, I'll fill out an example equation for 3 rounds...

b always = .1
c always = .7

a starts at 10,000

x = (10,000*.1)*.7
x = 700 (x becomes a in next cycle)

x = (700*.1)*.7
x = 49 (x becomes a in next cycle)

x = (49*.1)*.7
x = 3.43

so far the sum of x for 3 rounds is 700 + 49 + 3.43 = 752.43

I'm guessing this is some sort of multi part equation, but I assume there is a way to write this and have an online calculator, or Excel or something similar to calculate this I would think?

Again I apologize for the layman explanation of what i'm trying to figure out. Any help would be greatly appreciated!

 

[ksdhart2]

Well, first off, I have to question if your equation is actually x = (a * b) / c, as you show at the beginning, or x = (a * b) * c, as you show in your example. I'll assume that the first one is correct, but the same logic will apply even if it's actually the second one. The basic idea is that you've got an iterative equation, where the same process is done but you use the value from the previous output as the new input. You can see that the second iteration is the formula, but as you say "x becomes a." So, what if you just replaced x with its definition?

\(\displaystyle x_0=\dfrac{ab}{c}\)

\(\displaystyle x_1=\dfrac{x_0 b}{c} \implies x_1=\dfrac{\left( \dfrac{ab}{c} \right) b}{c}\)

What do you get if you simplify that? Going one step further, you can replace the x in the third cycle with its definition:

\(\displaystyle x_2=\dfrac{x_1 b}{c} \implies x_2=\dfrac{\left( \dfrac{\left( \dfrac{ab}{c} \right) b}{c} \right) b}{c}\)

After you simplify that kind of nasty equation, what do you get? Are you seeing a pattern? What does that make the general formula for \(\displaystyle x_n\)?

 

[finks]

Well, first off, I have to question if your equation is actually x = (a * b) / c, as you show at the beginning, or x = (a * b) * c, as you show in your example. I'll assume that the first one is correct, but the same logic will apply even if it's actually the second one. The basic idea is that you've got an iterative equation, where the same process is done but you use the value from the previous output as the new input. You can see that the second iteration is the formula, but as you say "x becomes a." So, what if you just replaced x with its definition?

\(\displaystyle x_0=\dfrac{ab}{c}\)

\(\displaystyle x_1=\dfrac{x_0 b}{c} \implies x_1=\dfrac{\left( \dfrac{ab}{c} \right) b}{c}\)

What do you get if you simplify that? Going one step further, you can replace the x in the third cycle with its definition:

\(\displaystyle x_2=\dfrac{x_1 b}{c} \implies x_2=\dfrac{\left( \dfrac{\left( \dfrac{ab}{c} \right) b}{c} \right) b}{c}\)

After you simplify that kind of nasty equation, what do you get? Are you seeing a pattern? What does that make the general formula for \(\displaystyle x_n\)?

I messed up and the equation I was trying to figure out. It is actually the one from the example, not the one I first posted. So...

(a * b) * c

I kind of get what you are saying, but just to an extent. It almost is like an exponential loop, and when b and c are less than 1, it would make it an exponentially decaying loop.


if we are using the (ab)c equation, then it is something like...

x_n = (x_n-1)b*c


the problem to me though is, that only works if you define x₀ and your limit of n is 1, so that version I came up with doesn't really help much as it doesn't really keep acting like a feedback loop.

Would I need to be using a summation (sigma)? Basically stating to start at x₀ and end at x₁₀ or however far down I find necessary for the decimal value to seem negligible? I'm unsure how I would write that equation if so.

As you can tell, I don't have the best math skills, it has been at least 10 years since I've really studied any form of math... thanks so far for all the help!

 

[ksdhart2]

I kind of get what you are saying, but just to an extent. It almost is like an exponential loop, and when b and c are less than 1, it would make it an exponentially decaying loop.

Yes, that's exactly what it is. Because the (corrected) formula is just x_n = x_n-1 * b * c, we can simplify that to say that x_n = x_n-1 * d, where d = b * c. You've correctly identified that if d < 1, the sequence will decay to 0, as n approaches infinity. Each time, we're just multiplying again by d, and multiplying by a number less than 1 makes the result smaller. The other cases are: d = 1, where the sequence will remain it's original value forever; d = 0, where it immediately decays to 0 in one iteration; and d > 1 where the sequence grows without bounds.

If you want to stop after a fixed number of iterations, say n = 10, then you'd find the closed form solution of x_n based on x₀. In this case, it would be x_n = x₀ * dⁿ. Then you can plug in the desired value. If, on the other hand, you want to stop when the sequence decays to a sufficiently small value (let's say 0.0001), you can just set up and solve an inequality. Namely:

x₀ * dⁿ < 0.0001

Edit: Fixed superscript notation

 

[finks]

Yes, that's exactly what it is. Because the (corrected) formula is just x_n = x_n-1 * b * c, we can simplify that to say that x_n = x_n-1 * d, where d = b * c. You've correctly identified that if d < 1, the sequence will decay to 0, as n approaches infinity. Each time, we're just multiplying again by d, and multiplying by a number less than 1 makes the result smaller. The other cases are: d = 1, where the sequence will remain it's original value forever; d = 0, where it immediately decays to 0 in one iteration; and d > 1 where the sequence grows without bounds.

If you want to stop after a fixed number of iterations, say n = 10, then you'd find the closed form solution of x_n based on x₀. In this case, it would be x_n = x₀ * dⁿ. Then you can plug in the desired value. If, on the other hand, you want to stop when the sequence decays to a sufficiently small value (let's say 0.0001), you can just set up and solve an inequality. Namely:

x₀ * dⁿ < 0.0001

Edit: Fixed superscript notation

Thanks so much for the help! I'm surprised I wasn't too far off, but the funny part is, I set up the original equation incorrectly, so this all doesn't help me solve my problem, which is fine as it was just a fun exercise for me.

The silver lining though is that this is the first time I've used/learned more than just my basic everyday arithmetic since highschool 11 years ago (went to an art school for college, not a lot of advanced math involved!).


The problem I wanted to solve was essentially a finance/business math problem, but I tried to phrase it without the context in the hopes of learning a larger math lesson.

Essentially, a friend of mine and I donated a product we made to charity. Usually we sell it in his store with a 70%/30% split, where I receive 70% and he receives 30%.

Since he cannot split the value of a donation, he was going to find the $ amount by which it would reduce his income, and then do our 70%/30% of that amount.

This all seems easy enough, right? Take the amount we are splitting, lets say 1,000 dollars, and then do our 70/30 split, meaning I would get $700, and he would get $300.

For the equation I thought I came up with, a=1,000, b=0.7

The problem I wanted to solve would be that once he gives me the $700 dollars, I would then have to pay income tax on that amount, which is c, lets say c is 20%, so c=0.2.

Now all of a sudden, I actually have 20% less of the $700.

Since He doesn't have to pay income tax on his $300, now all of a sudden I'm only receiving $560 instead, as $140 are going to taxes.

I basically was trying to figure out how to keep 70/30 splitting and splitting and splitting the taxable amount on my end, to find a new percentage split with a bit more on my end, a bit less on his end, to cover some of the income tax loss on my end.

I know that is all a bit tangled and convoluted, but I realized the formula you helped me with isn't actually the correct formula to figure this out...


But! In real life, he will just give be giving me the 700 dollars as a personal check since it is post-tax money anyway, so this equation isn't actually needed, I just thought it would be fun to figure it out as if it were the situation.

 

[finks]

You didn't understand what I gave you?

i = -1 + b*c = -1 + .1*.7= -.93

After 3rd transaction:
10000 * (1 + i)^3 = 3.43

As you can see, a simple formula...

Hi Denis, I understand what you posted, I just mentioned at the beginning of my next post that my original intention was incorrect. I was trying to come up with a formula for a slightly different situation that I asked for by accident.