analysistaylor expansion
I feel like I am making this far too difficult for myself!
The question states:
Find $c_0, c_1, c_2, c_3$ from the Taylor Series expansion for ${\cos(z)\over z}$ about $z=1$.
I've tried rewriting the function in terms of $z-1$, expanding and then equating terms with $\sum_{n=0}^\infty c_n(z-1)^n$ but it just gets messier and messier.
Note: Rewritten essentially from the ground up.
Sorry for the somewhat obfuscated previous version. I did not realize at first that you were trying to write the end result as a polynomial in $x$ instead of $(x-x_c)$, and then I was trying to bootstrap that into what I had already written.
I'm going to use $a$ instead of $x_c$, because it is a little simpler to type and harder to confuse with $x$, if you don't mind.
The simplest way to think about what you are doing is that it is the composition of three linear transformations.
First you take a polynomial written in terms of the basis $\beta=[1,x,x^2,\ldots,x^N]$ of the space of polynomials degree at most $N$, and you rewrite it in terms of the basis $\gamma=[1,(x-a),(x-a)^2,\ldots,(x-a)^N]$. That is, a change of coordinates.
Then you take the polynomial written in terms of $\gamma$, and you project down to the subspace generated by $[1,(x-a),\ldots,(x-a)^M]$. That is, you erase the terms of degree $M+1,\ldots,N$. Call this $P^N_M$.
Finally, you take the resulting polynomial, which is written in terms of the basis $\gamma$, and you translated it back into the basis $\beta$; another change of coordinates.
Being a composition of three linear transformations, it can be represented by a matrix which is the product of three matrices, the matrices that represent the three operations.
The matrix for $P^N_M$ is the simplest one: it is a block-diagonal matrix that has the $(M+1)\times (M+1)$ identity in the first block, and the $(N-M)\times(N-M)$ zero matrix as the second block.
The matrix for step 3 is just the inverse of the matrix for step 1. To each real number $r$, we have a matrix whose columns represent the binomial expansion of $(x+r)$: $$B_r = \left(\begin{array}{cccccc} 1 & r & r^2 & r^3 & \cdots & r^N\\\ 0 & 1 & 2r & 3r^2 & \cdots & \binom{N}{1}r^{N-1}\\\ 0 & 0 & 1 & 3r & \cdots & \binom{N}{2}r^{N-2}\\\ 0 & 0 & 0 & 1 & \cdots & \binom{N}{3}r^{N-3}\\\ \vdots & \vdots & \vdots & \ddots & \vdots\\\ 0 & 0 & 0 & 0 & \cdots & \binom{N}{N-1}r\\\ 0 & 0 & 0 & 0 & \cdots & 1 \end{array}\right).$$ When you go from having the basis given by powers of $x$ to the basis given by powers of $x-a$, you do this by multiplying by $B_a$. In order to go from the basis given by the powers of $(x-a)$ to the powers of $x$, the change of variable $u=x-a$ shows that you are going form powers of $u$ to powers of $u+a = u-(-a)$, so you achieve this by multiplying by $B_{(-a)}$. Thus, your operation is given by the matrix $$B_{(-a)}P^N_MB_a.$$ Now, this is a product of three $(N+1)\times (N+1)$ matrices, so you may wonder why I have an $(N+1)\times (N+1)$ matrix and you have an $(M+1)\times (N+1)$ matrix. The reason is that because the last $N-M$ rows of $P_M$ are all zero and all matrices are upper triangular, the last $N-M$ rows of this product will likewise be zero; you are just omitting them.
So, if you have the polynomial $C_0 + C_1x + \cdots +C_Nx^N$, then your end result is the polynomial $K_0 + K_1x+\cdots +K_Nx^N$, where $$\left(\begin{array}{c} K_0\\ K_1\\ \vdots\\ K_N\end{array}\right) = B_{(-a)}P_MB_a\left(\begin{array}{c} C_0\\ C_1\\ \vdots\\ C_N\end{array}\right).$$
For your second example, you have $N=5$ and $M=3$. For $N=5$, you have \begin{align*} B_{a} &= \left(\begin{array}{cccccc} 1 & a & a^2 & a^3 & a^4 & a^5\\ 0 & 1 & 2a & 3a^2 & 4a^3 & 5a^4\\ 0 & 0 & 1 & 3a & 6a^2 & 10a^3\\ 0 & 0 & 0 & 1 & 4a & 10a^2\\ 0 & 0 & 0 & 0 & 1 & 5a\\ 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right)\\ P^5_3 &= \left(\begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right)\\ B_{-a} &=\left(\begin{array}{rrrrrr} 1 & -a & a^2 & -a^3 & a^4 & -a^5\\ 0 & 1 & -2a & 3a^2 & -4a^3 & 5a^4\\ 0 & 0 & 1 & -3a & 6a^2 & -10a^3\\ 0 & 0 & 0 & 1 & -4a & 10a^2\\ 0 & 0 & 0 & 0 & 1 & -5a\\ 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right). \end{align*} If you multiply out $B_{-a}P^5_3B_a$ you get \begin{align*} B_{-a}P^5_3B_a &= \left(\begin{array}{rrrrrr} 1 & -a & a^2 & -a^3 & a^4 & -a^5\\ 0 & 1 & -2a & 3a^2 & -4a^3 & 5a^4\\ 0 & 0 & 1 & -3a & 6a^2 & -10a^3\\ 0 & 0 & 0 & 1 & -4a & 10a^2\\ 0 & 0 & 0 & 0 & 1 & -5a\\ 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right)\left(\begin{array}{cccccc} 1 & a & a^2 & a^3 & a^4 & a^5\\ 0 & 1 & 2a & 3a^2 & 4a^3 & 5a^4\\ 0 & 0 & 1 & 3a & 6a^2 & 10a^3\\ 0 & 0 & 0 & 1 & 4a & 10a^2\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right)\\ &= \left(\begin{array}{rrrrrr} 1 & 0 & 0 & 0 & -a^4 & -4a^5\\ 0 & 1 & 0 & 0 & 4a^3 & 15a^4\\ 0 & 0 & 1 & 0 & -6a^2 & -20a^3\\ 0 & 0 & 0 & 1 & 4a & 10a^2\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right) \end{align*} exactly what you have except for the two extra rows of zeros at the bottom.
Added. If you think about it, you'll see that this process will always produce a block-upper triangular matrix, with an $(M+1)\times(M+1)$ identity in the upper left and an $(N-M)\times(N-M)$ zero matrix in the bottom right; because $B_{-a}B_a = I$, the bottom rows you zero out in $B_a$ do not affect the products in the first $M+1$ columns, and they are all that matters in the last $N-M$ rows. So the only parts that need computation is the block on the upper right. This shows how it relates to your non-square matrices.
Added 2. I should have added this as soon as you said you were programming it...
There is an alternative way of computing this matrix, which may be faster when $M$ is larger than half of $N$. The idea is just to note that the matrix $P^N_M$ can be written as a difference of the identity matrix with the matrix that has the $(N-M)\times(N-M)$ identity on the bottom right and zeros elsewhere; call it $Q^N_M$; that is, $$P^N_M = I_{N+1} - Q^N_M = \left(\begin{array}{ccccc} 1 & 0 & \cdots & 0 & 0\\\ 0 & 1 & \cdots & 0 & 0\\\ \vdots & \vdots & \ddots & \vdots & \vdots \\\ 0 & 0 & \cdots & 1 & 0\\\ 0 & 0 & \cdots & 0 & 1 \end{array}\right) - \left(\begin{array}{cccccc} 0 & 0 & \cdots & 0 & 0\\\ 0 & 0 & \cdots & 0 & 0\\\ \vdots & \vdots & \ddots & \vdots & \vdots\\\ 0 & 0 & \cdots & 1 & 0\\\ 0 & 0 & \cdots & 0 & 1 \end{array}\right).$$ Then you have that $$B_{-a}P^{N}_MB_a = B_{-a}(I-Q^N_M)B_a = (B_{-a}IB_a) - B_{-a}Q^N_MB_a = I-B_{-a}Q^{N}_MB_a.$$ If $M$ is more than half of $N$, then there will be fewer operations to perform in $B_{-a}Q^N_MB_a$ than in $B_{-a}P^N_MB_a$, because $Q^N_MB_a$ has only $N-M$ nonzero rows, whereas $P^N_MB_a$ has $M+1$ nonzero rows. So you can pick which of the two computations to do: either $B_{-a}P^N_MB_a$, or $I-B_{-a}Q^N_MB_a$, and for $N$ and $M$ large, with $M$ close to $N$, using $Q^N_M$ will probably be faster than using $P^N_M$.
If you know the Taylor expansion of the elementary function and the rules to compose one Taylor expansion into another, you can write the Taylor expansion of every function which is expressed by the composition of elementary functions, without computing any derivative.
Best Answer
Hint: for $|z-1|<1$, we have $$ \frac1z=1-(z-1)+(z-1)^2-\cdots $$ In addition, $$ \cos(z)=\cos((z-1)+1)\\=\cos(z-1)\cos(1)-\sin(z-1)\sin(1) $$