# Thread: Reversing A Formula

1. ## Reversing A Formula [solved]

I'm sure there's an easy way to do this, but its been ages since I did algebra.

I've got an arithmetic series. Each term increases by 10% of the original term. For instance: 40, 44, 48, 52, 56, etc.

The formula I've come up with to calculate the nth term is:

$v_n=v_0+n.v_0/10$

Where $v_n$ is the nth term and $v_0$ is the initial term. This seems to work, although it may not be the most elegant expression of the series.

Now I've decided that, given $n$ and $v_n$, I want to be able to calculate $v_0$. However solving my above equation for $v_0$ has left me going in circles.

I'd just as soon NOT get the answer - a gentle hint in the right direction will be appreciated, helpful, and will hopefully leave my pride intact.

2. Originally Posted by sapph
I'm sure there's an easy way to do this, but its been ages since I did algebra.

I've got an arithmetic series. Each term increases by 10% of the original term. For instance: 40, 44, 48, 52, 56, etc.

The formula I've come up with to calculate the nth term is:

$v_{n}=v_{i}+(.1v_{i}(n-1))$

Where $v_{n}$ is the nth value and $v_{i}$ is the initial value. This seems to work, although it may not be the most elegant expression of the series.

Now I've decided that, given $n$ and $v_{n}$, I want to be able to calculate $v_{i}$. However solving my above equation for $v_{i}$ has left me going in circles.

I'd just as soon NOT get the answer - a gentle hint in the right direction will be appreciated, helpful, and will hopefully leave my pride intact.
Gentle hint: you can take out a common factor.

Also, to ease your notation I would call your starting term $v_0$ instead of $v_i$ - this makes the fact that it is the starting term much clearer. Often you would want to find the " $i^{\text{th}}$" term in your series instead of the " $n^{\text{th}}$" term, so this will also stop that confusion. So in the series you gave you would have $v_0=40, v_1=44, v_2=48, \ldots$. This would mean your formula would be $v_n=v_0+n.v_0/10$.

3. Hm. Maybe I need to let go of my pride. Here are the loops I am going through:

$v_n=v_0+n.v_0/10$ to
$10v_n=10v_0+n.v_0$ to
$10v_n/n=10v_0/n+v_0$ to
$10v_n/n-10v_0/n=v_0$ to
$(10v_n-10v_0)/n=v_0$ to
$10(v_n-v_0)/n=v_0$ Stuck.

$v_n=v_0+n.v_0/10$ to
$v_n-v_0=n.v_0/10$ to
$10(v_n-v_0)=n.v_0$ to
$10(v_n-v_0)/n=v_0$ Stuck.

Because I essentially have an x+xy situation, I am having trouble isolating the x. I tried googling for "x+xy" to see if I could find common solutions. Unfortunately, I forgot that google tends to ignore punctuation in searches, so searching for XXY tends to turn up a lot about genetic disorders.

4. Originally Posted by sapph
Because I essentially have an x+xy situation, I am having trouble isolating the x. I tried googling for "x+xy" to see if I could find common solutions. Unfortunately, I forgot that google tends to ignore punctuation in searches, so searching for XXY tends to turn up a lot about genetic disorders.
If I told you that $x+xy=x(1+y)$, would that help? That is what I was meaning when I said "common factor"...

Also, wolfram alpha is good for helping with this sort of thing. Typing $x+xy$ into this and you will get that it is equivalent to $x(1+y)$.

5. Aha!

$v_n=v_0+v_0*0.1n$ to
$v_n=v_0(1+0.1n)$ to
$v_n/(1+0.1n)=v_0$

Thanks for preserving the aha moment.