If I am given a function $P = ax^3 + bx^2 + cx + d$ and following:
$P(0) = P_1$,
$P(1) = P_2$,
$P^\prime(0) = v_0$,
$P^\prime(1) = v_1$.
How would I solve for $a$, $b$, $c$ and $d$?
Edit:
I have tried following:
$P(0) = P_1$ => d = P1
$P^\prime(0) = v_1$ => c = v1
However, I am not sure how would I solve for $a$ and $b$?
$P(x)=ax^3+bx^2+cx+d$, $P'(x)=3ax^2+2bx+c$.
From condition 1, that $P(0)=P_1$, we learn that $d=P_1$.
From condition 2, that $P(1)=P_2$, we learn that $a+b+c+d=P_2=a+b+c+P_1$.
From condition 3, that $P′(0)=v_0$, we learn that $c=v_0$.
From condition 4, that $P′(1)=v_1$, we learn that $3a+2b+c=v_1=3a+2b+v_0$.
Thus, we can express
$a=P_2-(b+c+d)$.
$a=P_2-(b+c+P_1)$.
$a=\frac{v_1-v_0-2b}{3}$.
$a=\frac{v_1-c-2b}{3}$.
$b=P_2-(a+c+d)$.
$b=P_2-(a+c+P_1)$.
$b=\frac{v_1-v_0-3a}{2}$.
$b=\frac{v_1-c-3a}{2}$.
$c=v_0$.
$d=P_1$.