Reverse Formula

Recable

New member
Joined
Oct 8, 2020
Messages
12
I'm sorry if this isn't in the right section on the forum but I don't actually know where to post this... but anyway - I need help with reversing this formula I made (through trial and error).

(1 / (points + 1)) + (0.1 / (points / 2))
When the points value is below 0, this is the formula I use to decrease the outcomes value. But it decreases by a smaller amount per point. So as an example:
-30
(1 / (1 + 1)) + (0.1 / (1 / 2)) = 0.70x [30% decreased value]
-27
(1 / (2 + 1)) + (0.1 / (2 / 2)) = 0.43x [57% decreased value]
-12
(1 / (3 + 1)) + (0.1 / (3 / 2)) = 0.31x [69% decreased value]
-6
(1 / (4 + 1)) + (0.1 / (4 / 2)) = 0.25x [75% decreased value]
-5
(1 / (5 + 1)) + (0.1 / (5 / 2)) = 0.20x [80% decreased value]

As you can see the next gap is always smaller than the last. Now the problem I'm having is making this reverse, so to make it so the value increases by 30%, 57%, etc.

P.S I know the points would be -1, -2, -3, etc - but I'll be calculating it so when the points are below 0 it uses this formula and then make the points an absolute value to make the above work. Also the values are rounded down to the nearest 2 decimal places.
 
Last edited:

Hope Spirit

New member
Joined
Oct 8, 2020
Messages
9
I don't think this is where this problem goes. I'm pretty sure arithmetic doesn't involve the things you need to solve this problem.
 

Recable

New member
Joined
Oct 8, 2020
Messages
12
I don't think this is where this problem goes. I'm pretty sure arithmetic doesn't involve the things you need to solve this problem.
Yeah I wasn't sure where to put it, I guess in the advanced bit?
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
5,231
I'm sorry if this isn't in the right section on the forum but I don't actually know where to post this... but anyway - I need help with reversing this formula I made (through trial and error).

(1 / (points + 1)) + (0.1 / (points / 2))
When the points value is below 0, this is the formula I use to decrease the outcomes value. But it decreases by a smaller amount per point. So as an example:
-30
(1 / (1 + 1)) + (0.1 / (1 / 2)) = 0.70x [30% decreased value]
-27
(1 / (2 + 1)) + (0.1 / (2 / 2)) = 0.43x [57% decreased value]
-12
(1 / (3 + 1)) + (0.1 / (3 / 2)) = 0.31x [69% decreased value]
-6
(1 / (4 + 1)) + (0.1 / (4 / 2)) = 0.25x [75% decreased value]
-5
(1 / (5 + 1)) + (0.1 / (5 / 2)) = 0.20x [80% decreased value]

As you can see the next gap is always smaller than the last. Now the problem I'm having is making this reverse, so to make it so the value increases by 30%, 57%, etc.

P.S I know the points would be -1, -2, -3, etc - but I'll be calculating it so when the points are below 0 it uses this formula and then make the points an absolute value to make the above work. Also the values are rounded down to the nearest 2 decimal places.
Well this certainly does not belong in arithmetic, but we only ask that people give a good faith effort on that.

However, what you have asked is quite obscure. In your first example, you mention - 30. Is that the number of points? If so the number of points does not appear in your formula.
 

Recable

New member
Joined
Oct 8, 2020
Messages
12
Well this certainly does not belong in arithmetic, but we only ask that people give a good faith effort on that.

However, what you have asked is quite obscure. In your first example, you mention - 30. Is that the number of points? If so the number of points does not appear in your formula.
No that was the current decrease in value, -30, -27 etc as shown on the end of each new statement it says [-30%] etc. But the number of points are 1, 2, 3 etc just increasing by 1.
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
5,231
If I am getting this, your current formula is

\(\displaystyle p \le -1 \implies y = \dfrac{1}{1 - p} + \dfrac{0.1}{\dfrac{-p}{2}} = \dfrac{- p + 0.2 - 0.2p}{(-p)(1 - p)} = - \dfrac{1.2p - 0.2}{p^2 - p}.\)

Let's see if that works.

p = -1, y = -(-1.2 - 0.2)/(1 + 1} = 0.7

p = -2, y = -(-2.4 - 0.2)/(4 + 2) = 2.6/6 = 0.43....

p = -3, y = -(-3.6 - 0.2)/(9 + 3) = 3.8/12 = 0.316

OK I have the computation. Now what do you mean by reversing it?
 

Recable

New member
Joined
Oct 8, 2020
Messages
12
If I am getting this, your current formula is

\(\displaystyle p \le -1 \implies y = \dfrac{1}{1 - p} + \dfrac{0.1}{\dfrac{-p}{2}} = \dfrac{- p + 0.2 - 0.2p}{(-p)(1 - p)} = - \dfrac{1.2p - 0.2}{p^2 - p}.\)

Let's see if that works.

p = -1, y = -(-1.2 - 0.2)/(1 + 1} = 0.7

p = -2, y = -(-2.4 - 0.2)/(4 + 2) = 2.6/6 = 0.43....

p = -3, y = -(-3.6 - 0.2)/(9 + 3) = 3.8/12 = 0.316

OK I have the computation. Now what do you mean by reversing it?
Okay I'm not quite sure what you did but yeah your results are the same as mine (0.7, 0.43, 0.31, etc). But by reverse I mean now I want to make my formula increase instead of decrease, so 1.30, 1.57, 1.69, etc. If that makes sense?

Edit: I mean I'm not sure what you did because I simply don't understand because I'm not that good at maths haha.
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
5,231
Okay I'm not quite sure what you did but yeah your results are the same as mine (0.7, 0.43, 0.31, etc). But by reverse I mean now I want to make my formula increase instead of decrease, so 1.30, 1.57, 1.69, etc. If that makes sense?

Edit: I mean I'm not sure what you did because I simply don't understand because I'm not that good at maths haha.
Still with negative points or with positive points?
 

Recable

New member
Joined
Oct 8, 2020
Messages
12

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
5,231
And do you want y to decrease as p increases (opposite to what happens with negative points)?

And what should happen with 0 points?
 

Cubist

Full Member
Joined
Oct 29, 2019
Messages
645
Also, are the exact values in the OP important? If not then a better, single, function that works over the whole domain of p may be possible. (What's the domain of p values?)
 

Recable

New member
Joined
Oct 8, 2020
Messages
12
And do you want y to decrease as p increases (opposite to what happens with negative points)?

And what should happen with 0 points?
Basically I wanted it so that when points are below 0, it reduces the main value by a certain amount. It decreases less each time per point though. The exact opposite of when points are above 0, so each point increases the main value but also increases it less per point. So 1.30x, 1.57x, etc. Instead of 0.7x, 0.43x etc. Does that make sense?
 

Recable

New member
Joined
Oct 8, 2020
Messages
12
Also, are the exact values in the OP important? If not then a better, single, function that works over the whole domain of p may be possible. (What's the domain of p values?)
What do you mean by the domain of p values?
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
5,231
The domain of p means what values can p be allowed to take?

Must p be an integer? Can it be zero? Are there limits that it cannot surpass?
 

Cubist

Full Member
Joined
Oct 29, 2019
Messages
645
@Recable sorry about asking lots of questions, but there's lots of possible answers.

Here's the graph of a polynomial quotient that might fit the bill? in Desmos you can easily add sliders to "play around":- Desmos (click). By adding the derivative to the graph, you can be more sure that your curve always increases. The derivative ought to get closer the x axis from above as x increases, but it should never dip below.

You could also try looking at the graph of "ln(x)". When you're happy, then obviously substitute variable "p" for "x".
 

Cubist

Full Member
Joined
Oct 29, 2019
Messages
645
...or perhaps you simply want 2-f(-p) where f is your formula? Using JeffM's equation this would be...

\(\displaystyle 2 + \dfrac{-1.2p - 0.2}{p^2 + p}\)

\(\displaystyle =2 - \dfrac{1.2p + 0.2}{p^2 + p}\)

\(\displaystyle =\dfrac{2p^2+2p -(1.2p + 0.2)}{p^2 + p}\)

\(\displaystyle =\dfrac{2p^2+0.8p - 0.2 }{p^2 + p}\)

EDIT: giving 1.3, 1.566..., 1.6833... when p=1,2,3
 

Recable

New member
Joined
Oct 8, 2020
Messages
12
The domain of p means what values can p be allowed to take?

Must p be an integer? Can it be zero? Are there limits that it cannot surpass?
Oh right, yeah it's always a integer and it's never 0. It's only 1, 2, 3, etc and -1, -2, -3, etc.
 
Last edited:

Recable

New member
Joined
Oct 8, 2020
Messages
12
...or perhaps you simply want 2-f(-p) where f is your formula? Using JeffM's equation this would be...

\(\displaystyle 2 + \dfrac{-1.2p - 0.2}{p^2 + p}\)

\(\displaystyle =2 - \dfrac{1.2p + 0.2}{p^2 + p}\)

\(\displaystyle =\dfrac{2p^2+2p -(1.2p + 0.2)}{p^2 + p}\)

\(\displaystyle =\dfrac{2p^2+0.8p - 0.2 }{p^2 + p}\)

EDIT: giving 1.3, 1.566..., 1.6833... when p=1,2,3
Yeah that seems to be correct. Thanks, however after all this I feel really unsatisfied with myself in making the formula but also not fully understanding how to switch it around. Do you have any advice on where I can learn about it more?
 

Cubist

Full Member
Joined
Oct 29, 2019
Messages
645
Yeah that seems to be correct. Thanks, however after all this I feel really unsatisfied with myself in making the formula but also not fully understanding how to switch it around. Do you have any advice on where I can learn about it more?
Try not to feel negative, you'll learn much more if you can stay positive! We all have to start somewhere, don't be afraid of (re?) learning things that might seem basic. There are some resources on this site (click). And when the covid19 restrictions lift then maybe you could look into attending a night class.
 

Recable

New member
Joined
Oct 8, 2020
Messages
12
Try not to feel negative, you'll learn much more if you can stay positive! We all have to start somewhere, don't be afraid of (re?) learning things that might seem basic. There are some resources on this site (click). And when the covid19 restrictions lift then maybe you could look into attending a night class.
Thank you :)
 
Top