Nth Term Proof Taylor Polynomial

Question

suppose that Msub_n(x) is the nth order maclaurin polynomial for f(x). Show that if k is a constant then Msub_n(kx) is the nth-order maclaurin polynomial for f(kx)

Answer 1

Let $g(x)=f(kx)$. Then by applying the Chain Rule repeatedly, we find that $g'(x)=kf'(kx)$, $g''(x)=k^2f''(kx)$, and in general $g^{(j)}(x)=k^j f^{(j)}(kx)$. Here we are using the notation $p^{(j)}(x)$ for the $j$-th derivative of $p(x)$. (The $0$-th derivative of $p(x)$ is $p(x)$.) We conclude that $$g^{(j)}(0)=k^j f^{(j)}(0),$$ for all $j\le n$.

The $j$-th term of the MacLaurin series expansion of $f(x)$ is $\frac{1}{j!}f^{(j)}(0)x^j$. Thus $$M_n(kx)=\sum_{j=0}^n \frac{1}{j!}f^{(j)}(0)k^jx^j.\tag{1}$$

The $j$-th term of the MacLaurin series expansion of $g(x)$ is $\frac{1}{j!}g^{(j)}(0)x^j$. This is $\frac{1}{j!}k^j f^{(j)}(0)x^j$. It follows that the $n$-th order MacLaurin polynomial of $g(x)$, that is, of $f(kx)$, is $$\sum_{j=0}^n \frac{1}{j!}k^jf^{(j)}(0)x^j.\tag{2}$$ Comparison of (1) and (2) yields the desired result.