My Math Forum Express this as an equation

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 October 5th, 2016, 12:19 AM #1 Newbie   Joined: Oct 2016 From: Israel Posts: 5 Thanks: 0 Express this as an equation (((0.3*0.8+0.3)*0.8+0.3)*0.8) I think it is pretty obvious what I am trying to do here. Add 0.3, take away fifth, rinse repeat. Anyone can tell me how to express this as an equation? Apologies is posted to the wrong subforum, feel free to move. I am NOT attending high school, college, or any math classes, so I have no idea what kind of problem it is that I want to solve.
 October 5th, 2016, 12:43 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,932 Thanks: 1000 I'm not entirely sure what you've got in mind but I'm going to go with you're thinking of a recurrence relation. $a[n+1]=0.8a[n]+0.3,~~~~a[0]=0$ and you'd probably like to know how to find the value for a given $n$ for a general relation of this sort $a[n+1]=c \cdot a[n]+d$ $a[0]=0$ we have $a[n]=d\displaystyle{\sum_{k=0}^{n-1}}c^k$ Now the series is a geometric one and can be simplified $\displaystyle{\sum_{k=0}^{n-1}}c^k=\dfrac{1-c^n}{1-c}$ so we end up with $a[n] = d \dfrac{1-c^n}{1-c}$ and if you plug in your values we have $a[n] = 0.3 \dfrac{1-(0.8 )^n}{1-0.8}=1.5 \left(1 - (0.8 )^n \right)$ To use this to evaluate your original expression you'd note the value you are after is $a[4]-0.3 = 1.5\left(1-( 0.8 )^4 \right) - 0.3 = 1.5(1-0.4096) -0.3 = 0.8856 -0.3 = 0.5856$ yell back if this is all incomprehensible Thanks from Thear
October 5th, 2016, 01:17 AM   #3
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Ok, this is a lot more complex than I expected it to be.

I am trying to figure it out, but still a couple steps away from getting it.
Sorry if I'm going to be wordy, I don't think well in formulas.

I take it that n is how many times I want to do the (add then divide) thing.
What exactly is a? What is it responsible for?
Quote:
 Originally Posted by romsek $\displaystyle{\sum_{k=0}^{n-1}}c^k=\dfrac{1-c^n}{1-c}$
I'm not sure how you did the simplification, or why n has -1. Also, I think you forgot the d in that step?

Last edited by Thear; October 5th, 2016 at 01:27 AM.

October 5th, 2016, 01:29 AM   #4
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Quote:
 Add 0.3, take away fifth, rinse repeat.
I'm not sure that this matches what you are trying to do.. Could you try explaining again ?

October 5th, 2016, 01:42 AM   #5
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Quote:
 Originally Posted by Joppy I'm not sure that this matches what you are trying to do.. Could you try explaining again ?
Take my current number. (any number)
1: Add a sum to it.
2: Divide that sum by a number.
3: Add same sum as step 1 to result.
4: Divide by same number as step 2.
5: Repeat step 3.

I apologize for how terrible I am for explaining things in non-words.
I am currently reading on recurrence equations to figure out what I am doing.

It would be nice to know what specific type of recurrence relation it is that I have, so I could narrow down my search some and not have to wade through a bunch of unrelated stuff.

Last edited by Thear; October 5th, 2016 at 01:59 AM.

October 5th, 2016, 02:37 AM   #6
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Quote:
 Originally Posted by romsek $a[n+1]=0.8a[n]+0.3,~~~~a[0]=0$
Since I am trying to add first, and then divide, wouldn't it be 0.8(a[n] + 0.3) ?
You got the correct answer, so I guess your formula would be correct,
but then what would be the difference if I want to add first and then divide, as opposed to dividing first and then adding?

Still don't get the simplification stage.

 October 5th, 2016, 02:55 AM #7 Global Moderator   Joined: Dec 2006 Posts: 18,956 Thanks: 1602 If 0.3 and 0.8 occur $n$ times in the expression, and S($n$) is the expression's value, expanding the expression gives S($n$) = $0.3*0.8^n + 0.3*0.8^{n-1}\text{ + . . . + }0.3*0.8$. Clearly, 0.8*S($n$) = $0.3*0.8^{n+1}$ + S($n$) $-\, 0.3*0.8$, so S($n$) = $\dfrac{0.3*0.8 - 0.3*0.8^{n+1}}{1 - 0.8} = 0.3*0.8*\dfrac{1 - 0.8^n}{1 - 0.8} = 1.2(1 - 0.8^n)$. Thanks from Thear
October 5th, 2016, 03:29 AM   #8
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Quote:
 Originally Posted by skipjack S($n$) = $0.3*0.8^n + 0.3*0.8^{n-1}\text{ + . . . + }0.3*0.8$.
I tried doing that manually, and I am getting a 0.24 , 0.524 , 0.791 sequence instead of an 0.24 , 0.432 , 0.585 sequence.

Quote:
 Originally Posted by skipjack Clearly, 0.8*S($n$) = $0.3*0.8^{n+1}$ + S($n$) $-\, 0.3*0.8$, so S($n$) = $\dfrac{0.3*0.8 - 0.3*0.8^{n+1}}{1 - 0.8} = 0.3*0.8*\dfrac{1 - 0.8^n}{1 - 0.8} = 1.2(1 - 0.8^n)$.
Thanks mate, and I really appreciate your help. But please do remember that some have not done math in years, so something clear to you is not clear to me at all.

October 9th, 2016, 03:35 PM   #9
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Quote:
 Originally Posted by Thear Thanks mate, and I really appreciate your help. But please do remember that some have not done math in years, so something clear to you is not clear to me at all.
Do you need to understand how it works or do you just need to apply it?

Is skipjack's solution not giving you what you need?

Last edited by skipjack; October 10th, 2016 at 01:01 AM.

October 10th, 2016, 01:23 AM   #10
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Quote:
 Originally Posted by Thear I tried doing that manually, and I am getting a 0.24, 0.524, 0.791 sequence . . .
Can you show your working, so that any mistakes in it can be identified?

Quote:
 Originally Posted by Thear . . . something clear to you is not clear to me at all.
Expand 0.8*S($n$) by using the series I gave for S($n$) and increasing each exponent of 0.8 by 1, then compare the result with the original series prefixed by the extra term I gave and minus its last term.

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