Finding Taylor Coefficients with Synthetic Division

Horner's method of synthetic division provides an efficient means of performing polynomial division. It is especially simple when dividing by a linear term of the form $x-r$.

Given a polynomial $P(x)$ and base point $r$, we repeatedly divide $P(x)$ and its remainders by $x-r$. This allows us to find a factored form $P(x) = b_n(x-r)^n+b_{n-1}(x-r)^{n-1}+\cdots+b_1(x-r)+b_0$ where $b_k$ is the $k$-th Taylor coefficient of $P(x)$ when centering around $x=r$. In particular, $b_k = {P^{(k)}(r)}/{k!}$.

Why does this work? We can divide $P(x)$ by $x-r$ and get $P(x)=q_1(x)(x-r)+b_0$ where $q_1(x)$ is the quotient and $b_0$ (a constant) is the remainder. Then we can divide $q_1(x)$ by $x-r$ and get $q_1(x)=q_2(x)(x-r)+b_1$. Notice that $P(x)=q_1(x)(x-r)+b_0=(q_2(x)(x-r)+b_1)(x-r)+b_0 = q_2(x)(x-r)^2+b_1(x-r)+b_0$. Continuing in this fashion will eventually yield $P(x) = b_n(x-r)^n+b_{n-1}(x-r)^{n-1}+\cdots+b_1(x-r)+b_0$.

How is the table created?

List the coefficients of $P(x)$ along the top row. These are listed from the leading coefficient down to the trailing coefficient.
Draw a vertical line to the left of the list of coefficients. On the next row write the basepoint $r$ to the left of the vertical line.
We now create rows two at a time. I will refer to these as rows $k$ and $k+1$ (and I will refer to the previously completed row as row $k-1$). We will be working to the right of the vertical line.
In row $k$ draw an arrow pointing down. In row $k+1$ copy $P(x)$'s leading coefficient.
Multiply the last completed entry in row $k+1$ by our basepoint $r$ and place this entry in row $k$ (this is will be up and to the right of the entry we just multiplied).
Sum the latest entry in row $k$ with the entry in row $k-1$ just above it. Place the result below these entries in row $k+1$.
We repeat this process across the entire row.
Draw a line between rows $k$ and $k+1$. In row $k+1$ color all but the last entry red (these are the coefficients of our quotient polynomial). Color the last entry blue (this is our Taylor coefficient).
Repeat this process creating two rows at a time, but only compute up to the last red entry. So each time we create a new pair of rows, they shrink in length by one entry.

The rows of red entries are coefficients of the above mentioned quotients $q_1(x)$, $q_2(x)$, etc. The blue entries at the end of our rows are our Taylor coefficients $P(r)$, $P'(r)$, $P''(r)/2$, etc.

Note: The polynomials can have coefficients in Sage's "symbolic ring" (i.e. SR). So coefficients such as "1+sqrt(2)" and "5+3*i" are allowed. The Sage code tries to clean up inputs (Ex: It changes (x-1)(x-2) to (x-1)*(x-2) and 3x to 3*x), but it cannot anticipate all syntax errors. As warning, check how your input was interpreted. It may also mis-correct certain inputs.

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