I have 4 points: $(x_i,y_i),\quad i=1,2,3,4$
and $x_i=x_1+i-1,\quad i=2,3,4$
(so these x-coordinates are evenly spaced)
I want to find a polynomial interpolation of these points, having the following form:
$$y=a_4(x-b)^4+a_2(x-b)^2+c$$
I don't know how to solve the parameters. Could you please give me a help?
Edit
Sorry for the incorrect usage of "coefficient".
I need to solve 4 unknown parameters, especially $b$ and $c$.
Best Answer
For simplicity, I'll adjust your indices down by one. Also, I'll define $z_i := x_i-b$. Finally, I'll write $a_0$ for $c$, for the sake of consistency. Then we have four equations in four unknowns $a_0$, $a_2$, $a_4$, and (the hidden) $b$: $$y_i = a_4z_i^4+a_2z_i^2+a_0 \qquad i=0,1,2,3 \tag{1}$$
Viewing the first three equations (for $i=0,1,2$) as a linear system in the $a_i$, we can solve to get
$$\begin{align} a_0 &= -\frac{ y_0 z_1^2 z_2^2 (z_1^2 - z_2^2) + y_1 z_2^2 z_0^2 (z_2^2 - z_0^2) + y_2 z_0^2 z_1^2 (z_0^2 - z_1^2)}{ (z_0^2 - z_1^2) (z_1^2 - z_2^2) (z_2^2 - z_0^2)} \tag{2}\\[6pt] a_2 &= \phantom{-}\frac{ y_0 (z_1^4 - z_2^4) + y_1 (z_2^4 - z_0^4) + y_2 (z_0^4 - z_1^4)}{ (z_0^2 - z_1^2) (z_1^2 - z_2^2) (z_2^2 - z_0^2)} \tag{3}\\[6pt] a_4 &= -\frac{ y_0 (z_1^2 - z_2^2) + y_1 (z_2^2 - z_0^2) + y_2 (z_0^2 - z_1^2)}{ (z_0^2 - z_1^2) (z_1^2 - z_2^2) (z_2^2 - z_0^2)} \tag{4} \end{align}$$ (Note that the denominators put restrictions on allowable values of some parameters.) Substituting these into the $i=3$ equation, we find after dust settles: $$\begin{align} 0 &=y_0 (z_1^2 - z_2^2) (z_2^2 - z_3^2) (z_3^2 - z_1^2) \tag{5}\\ &-y_1 (z_2^2 - z_3^2) (z_3^2 - z_0^2) (z_0^2 - z_2^2) \\ &+y_2 (z_3^2 - z_0^2) (z_0^2 - z_1^2) (z_1^2 - z_3^2) \\ &-y_3 (z_0^2 - z_1^2) (z_1^2 - z_2^2) (z_2^2 - z_0^2) \end{align}$$
Now, upon substituting back $z_i \to x_i-b$, we see that $$z_i^2 - z_j^2 = (z_i-z_j)(z_i+z_j) = (x_i-x_j)(x_i+x_j-2b) \tag{6}$$ so that, for general $x_i$, each term of $(5)$ has three factors involving $b$ and therefore contributes some multiple of $b^3$ to the overall sum. This makes solving for $b$ possible via the Cubic Formula, but the algebra is messy.
Luckily, for equally-spaced $x_i$ (that is, for $x_i = x_0 + i d$ for some $d$), each term of $(5)$ has a factor of $2x_0+3d-2b$, from the $z_i^2-z_j^2$ factor with $i+j=3$; since the denominators of the $a_i$ also have such a factor, we may assume the value is non-zero. This allows us to divide-through by that factor, as well as $d$s that accumulate from $x_i-x_j$ factors, turning $(5)$ into the following quadratic: $$ \begin{align} 0 &=2 b^2\;(y_0 - 3 y_1 + 3 y_2 - y_3) \\[4pt] &-\phantom{2}b\phantom{^2}\left(\begin{array}{l} \phantom{+3}y_0(4x_0+9 d) - 3y_1(4x_0+7 d) \\[4pt] +3y_2( 4x_0+5 d) - \phantom{3}y_3(4x_0+3 d) \end{array}\;\right) \\[4pt] &+\phantom{3}y_0 (x_0 + 2 d) (2 x_0 + 5 d) - 3 y_1 (2 x_0 + 5 d) (\phantom{2}x_0 + d) \\[4pt] &+ 3 y_2 (x_0 + 2 d) (2 x_0 + \phantom{5}d) - \phantom{3}y_3 (\phantom{2}x_0 + \phantom{9}d) (2 x_0 + d) \end{align} \tag{7}$$
Taking $x_0=0$, this simplifies to $$\begin{align} 0 = 2 b^2&\;(\phantom{10}y_0 - \phantom{1}3 y_1 + 3 y_2 - y_3) \\ -3bd &\;(\phantom{1}3y_0 - \phantom{1}7 y_1+5y_2 - y_3) \\ +d^2&\;(10y_0 - 15 y_1 + 6 y_2 - y_3) \end{align}\tag{8}$$
This quadratic in $b$ is readily solved. Substituting the resulting value(s) into the formulas for the $a_i$ is left as an exercise to the reader.
Example. If we have points $(0,3)$, $(1,1)$, $(2,4)$, $(3,1)$ (so that $d=1$), we find $$(b,a_0,a_2,a_4) = \left(2,4,-\frac{47}{12},\frac{11}{12}\right) \quad\text{or}\quad \left(\frac{19}{22}, \frac{101789}{108416}, \frac{1109}{336}, -\frac{121}{168}\right) \tag{9}$$ The corresponding polynomials have the following graphs: