With Horner's algorithm, I can solve f(x$_{0}$) for a polynomial like this: $a_0 + a_1x + a_2x^2 + a_3x^3 + ... + a_nx^n$
By doing this:
b$_n$ = a$_n$
b$_{n-1}$ = a$_{n-1}$ + b$_n$x$_0$
b$_{n-2}$ = a$_{n-2}$ + b$_{n-1}$x$_0$
...
b$_0$ = a$_0$ + b$_1$x$_0$
Where b$_0$ = f(x$_{0}$). The problem is doing Horner's algorithm on something like this: $a_1 + a_2(x + y_1) + a_3(x + y_1)(x + y_2) + a_4(x + y_1)(x + y_2)(x + y_3) + ... + a_{n+1}(x + y_1)(x + y_2)(x + y_3)...(x + y_n)$
So instead of a going from a$_0$ to a$_n$ it goes from a$_1$ to a$_{n+1}$ so for b$_n$, is it supposed to be b$_n$ = a$_{n+1}$? Also let's say we're trying to solve f(1.53) and we have:
a = -1, 3.3, 0, -2.2, 5, -1.6
y = -1, 1, -1, 1, -1
So if n = 5, this is how I did Horner's algorithm:
b$_5$ = -1.6
x$_0$ = (1.53 - 1)(1.53 + 1)(1.53 - 1)(1.53 + 1) = 1.79801281
b$_4$ = 5 + (-1.6)(1.79801281) = 2.123179504
x$_0$ = (1.53 - 1)(1.53 + 1)(1.53 - 1) = 0.710677
b$_3$ = -2.2 + (2.123179504)(0.710677) = -0.691105159
x$_0$ = (1.53 - 1)(1.53 + 1) = 1.3409
b$_2$ = 0 + (-0.691105159)(1.3409) = -0.926702907
x$_0$ = (1.53 - 1) = 0.53
b$_1$ = 3.3 + (-0.926702907)(0.53) = 2.808847459
b$_0$ = -1 + (2.808847459)(1.53) = 3.297536612
The problem is that f(1.53) = 6.65086 which is not what b$_0$ is so what am I doing wrong?
The index needs adjusted, indeed. However, that's not the only change you need to make. Unlike in the case of Horner's method, the second expression is not a polynomial, since the terms do not contain consecutive powers of the same $\,x\,$. To account for that, the recurrence must be modified to:
$$ \begin{align} b_{n} &= a_{n+1} \\ b_{n-1} &= a_n + b_n(x+y_n) \\ b_{n-2} &= a_{n-1} + b_{n-1}(x+y_{n-1}) \\ &\cdots \\ b_{1} &= a_{2} + b_{2}(x+y_2) \\ b_{0} &= a_{1} + b_{1}(x+y_1) \\ \end{align} $$
The above is equivalent to writing the expression as:
$${\displaystyle a_{1}+(x+y_1)\bigg(a_{2}+(x+y_2)\Big(a_{3}+\cdots +(x+y_{n-1})\big(a_{n}+a_{n+1}(x+y_n)\big)\Big)\bigg)}$$