# 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:

$\displaystyle v_n=v_0+n.v_0/10$

Where $\displaystyle v_n$ is the nth term and $\displaystyle v_0$ 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 $\displaystyle n$ and $\displaystyle v_n$, I want to be able to calculate $\displaystyle v_0$. However solving my above equation for $\displaystyle v_0$ 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:

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

Where $\displaystyle v_{n}$ is the nth value and $\displaystyle v_{i}$ 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 $\displaystyle n$ and $\displaystyle v_{n}$, I want to be able to calculate $\displaystyle v_{i}$. However solving my above equation for $\displaystyle v_{i}$ 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 $\displaystyle v_0$ instead of $\displaystyle v_i$ - this makes the fact that it is the starting term much clearer. Often you would want to find the "$\displaystyle i^{\text{th}}$" term in your series instead of the "$\displaystyle n^{\text{th}}$" term, so this will also stop that confusion. So in the series you gave you would have $\displaystyle v_0=40, v_1=44, v_2=48, \ldots$. This would mean your formula would be $\displaystyle v_n=v_0+n.v_0/10$.

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

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

$\displaystyle v_n=v_0+n.v_0/10$ to
$\displaystyle v_n-v_0=n.v_0/10$ to
$\displaystyle 10(v_n-v_0)=n.v_0$ to
$\displaystyle 10(v_n-v_0)/n=v_0$ 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 $\displaystyle x+xy=x(1+y)$, 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 $\displaystyle x+xy$ into this and you will get that it is equivalent to $\displaystyle x(1+y)$.

5. Aha!

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

Thanks for preserving the aha moment.