General square form of Taylor expansion polynomial

Question

Suppose we have a Taylor expansion for a function $f$ with respect to t up to $M$-th order.

$$ \begin{equation} T_M = \sum^M_{k=0}\frac{1}{k!}f^k(x)\Delta t^k = f(x) + f'(x)\Delta t + \frac{1}{2}f''(x)\Delta t^2 + \cdots \end{equation} $$

What would be the general form of $T_M^2$ with respect to $\Delta t$?

$$ \begin{equation} T_M\times T_M = (\sum^M_{k=0}\frac{1}{k!}f^k(x)\Delta t^k)^2 = (???) + (???)\Delta t + (???)\Delta t^2 + \cdots \end{equation} $$

Perhaps Multinomial theorem can help?

Answer 1

Since we are squaring a sum we can utilize $$ \left( \sum_{k=1}^{n} a_k \right)^2 = \sum_{k=1}^{n} \sum_{j=1}^{n} a_ka_j $$ Applying this formula onto our problem yields

$$ \left(\sum_{k=0}^{M} \frac{1}{k!}f^k(x)\Delta t^k\right)^2 = \sum_{k=0}^{M}\sum_{j=0}^{M} \frac{1}{k!}\frac{1}{j!}f^k(x)f^j(x)\Delta t^{k+j} $$ Taking a closer look at the coefficients of $\Delta t^{k+j}$, we see that there are $k+j+1$ different combinations of $k$ and $j$. For example ($k+j=3$ cf. Jose Brox' answer), we have the combinations for: $$ (k,j) \in \{(0,1),(1,2),(2,1),(3,0)\}$$

Thus we can rewrite the above sum $$ \sum_{k=0}^{M}\sum_{j=0}^{M} \frac{1}{k!}\frac{1}{j!}f^k(x)f^j(x)\Delta t^{k+j} = \sum_{k=0}^{M}\left[ \sum_{j=0}^{k}\frac{1}{j!}\frac{1}{(k-j)!}f^j(x)f^{k-j}(x)\right]\Delta t^k$$ where the coefficient corresponding to the $k$-th exponent is $$\sum_{j=0}^{k}\frac{1}{j!(k-j)!}f^j(x)f^{k-j}(x) $$

Answer 2

Like with usual polynomial multiplication, the term of degree $k$ comes from adding up all products of two monomials whose degrees sum to $k$. For example, for degree $3$ you have $(0,3)+(1,2)+(2,1)+(3,0)$.