Taylor Polynomial of Product

Question

Problem Show that $$P_{n,a,fg}=[P_{n,a,f}*P_{n,a,g} ]_n$$ where $P_{n,a,f}$ is the Taylor polynomial of degree n around a of f and $[P]_n$ denotes truncation of $P$ to degree $n$, the sum of all terms of $P$ of degree $\leq n$. Give a proof using obvious facts about products involving terms of the form $R_n$, where $R_n$ is the remainder of the taylor polynomial of degree $n$, i.e $f(x)=P_{n,a,f}+R_{n,a,f}$.

What I tried: I tried to use the derivative of products but I don't know what to do from there, so it didn't get me anywhere.

Answer 1

Assume $a=0$. For any fixed $r\geq0$ Leibniz' formula says that $$(fg)^{(r)}=\sum_{k=0}^r{r\choose k}f^{(r-k)}\,g^{(k)}\ .$$ This implies $${(fg)^{(r)}(0)\over r!}x^r =\sum_{k=0}^r\ {f^{(r-k)}(0)\over (r-k)!}x^{r-k}\ \ {g^{(k)}(0)\over k!}x^k\ .\tag{1}$$ Here we have on the LHS the term of degree $r$ in any Taylor expansion $P_{n,\> fg}$ with $n\geq r$, and on the RHS the collected terms of exact degree $r$ when we multiply out the two Taylor polynomials $P_{n,\,f}$ and $P_{n,\, g}$ of any order $n\geq r$.

Now let an $n\geq0$ be given. Then we may sum $(1)$ over $r$ from $0$ to $n$, and obtain $$P_{n,\,fg}(x)=\bigl[P_{n,\,f}(x)\cdot P_{n,\,g}(x)\bigr]_n\ .$$

Answer 2

Hints:

On one hand, the convolutions are sum of terms of the form $f^{(j)}(a)f^{(k-j)}(a)\dfrac {t^jt^{k-j}}{j!(k-j)!}$.

On the other hand, the $k^{th}$ derivative of a product is given by a sum of terms $\displaystyle\binom kjf^{(j)}(a)f^{(k-j)}(a)$.

For example, to the third order,

$$f(a)+f'(a)t+\frac{f''(a)}{2}t^2+\frac{f''(a)}{3!}t^3+\cdots$$ $$g(a)+g'(a)t+\frac{g''(a)}{2}t^2+\frac{g''(a)}{3!}t^3+\cdots$$

gives the following coefficients, by increasing power of $t$:

$$f(a)g(a)$$ $$f'(a)g(a)+f(a)g'(a)$$ $$\frac{f''(a)}2g(a)+f'(a)g'(a)+f(a)\frac{g''(a)}2$$ $$\frac{f'''(a)}{3!}g(a)+\frac{f''(a)}2\frac{g'(a)}2+\frac{f'(a)}2\frac{g''(a)}2+f(a)\frac{g'''(a)}{3!}$$ $$\cdots$$

Compare to the corresponding coefficients oftained by direct evaluation:

$$f(a)g(a)$$ $$f'(a)g(a)+f(a)g'(a)$$ $$\frac{f''(a)g(a)+2f'(a)g'(a)+f(a)g''(a)}2$$ $$\frac{f'''(a)g(a)+3f''(a)g'(a)+3f'(a)g''(a)+f(a)g'''(a)}{3!}$$ $$\cdots$$