function inception

[gideonzane]

my function right now is:

f(x) = 15 (x² + 14x + 13) - (2y)/5 , 0 ≤ x ≤ 10

what's messing me up is y. for x=1, i want y to be the value of y where x=0. for x=2, i want y to be the value of y where x=1. so, i want y to be y where x=x-1. how do i express this in an equation? is there an operation for this that i don't know about? i've tried doing my own research, but i'm afraid that i don't know enough to really understand exactly what i'm looking for, and my search has been in vain.

 

[DrPhil]

my function right now is:

f(x) = 15 (x² + 14x + 13) - (2y)/5 , 0 ≤ x ≤ 10

what's messing me up is y. for x=1, i want y to be the value of y where x=0. for x=2, i want y to be the value of y where x=1. so, i want y to be y where x=x-1. how do i express this in an equation? is there an operation for this that i don't know about? i've tried doing my own research, but i'm afraid that i don't know enough to really understand exactly what i'm looking for, and my search has been in vain.

Yes, we have a problem of notation. Do you want \(\displaystyle y(x) = f(x-1)\)?

It would help us to know what the original question was, that led you to this equation. Also show any work you have done.

 

[gideonzane]

unfortunately, there was no original question - i making this myself. i'm using this for a d&d campaign. i made a poisoncrafting table where each poison can level up independently. i'm using this function to determine the amount of experience required to go to the next level. as part of the function, i want to include the amount of experience that was needed to get to your current level as a modifier.

y(x)=f(x-1): this is exactly what i want. how would i integrate this into the equation? does it remain a declaration on the side, like the set 0≤x≤10?

 

[DrPhil]

my function right now is:

f(x) = 15 (x² + 14x + 13) - (2y)/5 , 0 ≤ x ≤ 10

unfortunately, there was no original question - i making this myself. i'm using this for a d&d campaign. i made a poisoncrafting table where each poison can level up independently. i'm using this function to determine the amount of experience required to go to the next level. as part of the function, i want to include the amount of experience that was needed to get to your current level as a modifier.

y(x)=f(x-1): this is exactly what i want. how would i integrate this into the equation? does it remain a declaration on the side, like the set 0≤x≤10?

This kind of equation is called "recursive" because each term builds on previous terms. Since you only need values for integer x in the range [0,10], you can make a simple table with \(\displaystyle x\) and \(\displaystyle f(x)\). For each table entry, find the value of f(x-1) from the previous line in the table. [For x=0, there is no previous experience, so treat f(-1) as 0.]

\(\displaystyle f(x) = 15 (x^2 + 14x + 13) - (2/5)f(x-1)\)

It would be possible to write this recursive function as a single function of x [http://www.wikihow.com/Solve-Recurrence-Relations], but I don't think you need that.

 

[gideonzane]

perfect! thanks so much! i'm still messing around with the coefficients, so making tables for each version is a bit of a chore. i'm going to look into writing it out as a single function so i can plug it into a website like wolfram alpha, and toy around with it.

thanks again, man!

cheers