So I've got this beautiful little piece of math, $n$ in this formula can be substituted with any non-negative number between 1 and 100.
$$ \frac65 n^3 - 15n^2+100n-140 $$
The expected results are kind of as follows...
- $n = 1 \to 9$
- $n = 2 \to 48$
- $n = 3 \to 39$
- $n = 4 \to 39$
The results I got after plugging in the respective values are such.
- $n = 1 \to -54$
- $n = 2 \to 10$
- $n = 3 \to 57$
- $n = 4 \to 97$
What am I not doing right? How do I figure out where the error is?
Included now are graphs of the expected results according to the person who first found this formula within a game. The formula I am posting about is represented in these graphs by the line with the color purple.
Graph of expected results
Graph of expected results, ratio of n to "level" cubed
If you want to define a cubic polynomial $$ p(n) = an^3+bn^2+cn+d $$ satisfying $p(1)=9,p(2)=49,p(3)=p(4)=39$, there is a unique polynomial like that. You can construct the coefficients by solving the Vandermonde system $$ \begin{pmatrix} 1 & 1 & 1 & 1\\ 1 & 2 & 4 & 8\\ 1 & 3 & 9 & 27 \\ 1 & 4 & 16 & 64 \end{pmatrix} \times \begin{pmatrix} d \\ c \\ b \\ a \end{pmatrix} = \begin{pmatrix} 9 \\ 48 \\ 39 \\ 39 \end{pmatrix} $$ which yields the solution $$ \begin{pmatrix} d \\ c \\ b \\ a \end{pmatrix} = \begin{pmatrix} -135 \\ 215.5 \\ -81 \\ 9.5 \end{pmatrix} $$ and the corresponding polynomial $$ p(n) = \frac{19}{2} n^3-81n^2+\frac{431}{2}n-135 $$ but this seems to have nothing to do with your polynomial.