# question about solution of $1,(x-5),(x-5)^2,...,(x-5)^m$ as a basis for $\mathcal{P}_m(\mathbf{R})$

axler 3.F.8 here is another answer for but I have problem for understanding highlighted part:

first why $$\varphi_j(0)$$?? second isn't any problem for $$j=0$$?

The $$j$$th derivative of $$a_0+a_1(x-5)+\ldots +a_m(x-5)^m$$ is $$j!\cdot a_j+b_{j+1}(x-5)+b_m(x-5)^{(m-j)}$$ (for some constants $$b_{j+1},\ldots ,b_m$$ that we don't need to know, but that one can easily calculate), so when you evaluate it a $$x=5$$, you get $$j!\cdot a_j$$, and then $$\varphi_j(p)=a_j$$. On the other hand, exactly as in the question you have cited, here we are supposing that $$p(x)=a_0+a_1(x-5)+\ldots +a_m(x-5)^m$$ is the null vector, so $$\varphi_j(p)=\varphi_j(0)=0$$.
When $$j=0$$, conventionally one have to take $$p^{(0)}$$ as $$p$$, see for example here (find "zeroth derivative" in the first paragraph) or here (paragraph 1.2, "Taylor series").
Note that actually there is a typo, the coefficients $$a_0, \ldots ,a_m$$ are in $$\mathbb{R}$$ not in $$\mathbb{F}$$, but actually you can consider also polynomials over any infinite field (viewed as an extension of $$\mathbb{Q}$$).
We have an equality of polynomials $$a_0 + a_1(x-5) + a_2(x-5)^2 + \cdots + a_m(x-5)^m = 0$$ Applying $$\varphi_j$$ to both sides, we keep the equality and get $$\varphi_j(a_0 + a_1(x-5) + a_2(x-5)^2 + \cdots + a_m(x-5)^m) = \varphi_j(0)$$ The main difficulty here is actually recognizing that the left-hand side is equal to $$a_j$$, not that the left-hand side and right-hand side are equal.
As for what happens when $$j = 0$$, there is no issue. Differentiating $$0$$ times is the same as just not doing anything, so $$p^{(0)}$$ is just $$p$$. And $$0!$$ is $$1$$. So $$\varphi_0(p)$$ turns out to just be $$p(5)$$. Which for our particular (left-hand side) polynomial turns out to be just $$a_0$$.
• for first question I got that ${a_j} =$\varphi({a_j})$but why also equal to$\varphi_j(0)$Dec 7, 2020 at 9:42 • @negar As I say in my answer, the two polynomials$a_0 + a_1(x-5) + a_2(x-5)^2 + \cdots + a_m(x-5)^m$and$0$are actually the same polynomial (that's what the first equality establishes), so applying$\varphi_j\$ to both of them must yield the same result. Dec 7, 2020 at 9:44