# Find coordinates of polynom $Q$ with respect to basis $B$

Let $$P_k(x)=\frac {(x-a)^k}{k!},\forall a\in\mathbb{R}$$ and $$k$$ from $$0$$ to $$n$$. Find coordinates of an arbitrary polynom $$Q\in \mathbb{R}_{\leq n}[X].$$

$$P_k(x)$$ form the basis $$B=\{1,x-a,\frac 1{2!}(x-a)^2\,...,\frac {(x-a)^n}{n!}\}$$.

Let $$Q(x) = a_0+a_1x+a_2x^2+...+a_nx^n.$$

so then $$Q(x)=\lambda_1\times1+\lambda_2\times(x-a)+...+\lambda_n\times\frac{(x-a)^n}{n!}\implies a_0+a_1x+a_2x^2+...+a_nx^n = \lambda_1\times1+\lambda_2\times(x-a)+...+\lambda_n\times\frac{(x-a)^n}{n!}.$$

so we have

$$\lambda_1=a_0$$

$$\lambda_2=a1$$

$$...$$

$$\lambda_n=n!\times a_n$$

and $$\lambda_1,....,\lambda_n$$ are the coordinates of $$Q$$ with respect to $$B$$, have I done something wrong? something doesn't look right for me.

UPDATE: that's for $$a=0$$.

so I calculated using $$Q((x-a)+a)$$ and found out that:

$$\lambda_0=\sum_{k=1}^n a_k\times a^k$$

$$\lambda_1=\sum_{k=1}^n C_k^1\times a_k \times a^{k-1}$$

$$...$$

$$\lambda_n=\sum_{k=1}^n a_k.$$

is that correct?

• $\lambda_1$ (which should be $\lambda_0$) isn't the only constant term if you expand the product (for instance, you get $\frac1{n!}a^n\lambda_n$ from the highest degree term). – Arthur Oct 26 '18 at 18:42

From the equality$$\lambda_0+\lambda_1(x-a)+\lambda_2\frac{(x-a)^2}{2!}+\cdots+\lambda_n\frac{(x-a)^n}{n!}=a_0+a_1x+a_2x^2+\cdots+a_nx^n,$$you can deduce that $$\lambda_0=a_0$$, $$\lambda_1=a_1$$ and so on if $$a=0$$. In the general case, write\begin{align}Q(x)&=Q\bigl((x-a)+a\bigr)\\&=a_0+a_1\bigl((x-a)+a\bigr)+a_2\bigl((x-a)+a\bigr)^2+\cdots+a_n\bigl((x-a)+a\bigr)^n,\end{align}and expand each binomial $$\bigl((x-a)+a\bigr)^k$$ in order to get the $$\lambda_k$$'s.
• It looks fine, except that $C_k^1$ should be $C_1^k$ (or $\binom k1$). – José Carlos Santos Oct 26 '18 at 20:52
• and the sum changes, right? no longer from $1$ to $n$, but from ..? – C. Cristi Oct 27 '18 at 13:16
• $\lambda_k=\sum_{j=k}^n(\ldots)$. – José Carlos Santos Oct 27 '18 at 16:08
You want to write $$Q(x)=c_0P_0(x)+c_1P_1(x)+\dots+c_nP_n(x)$$ You can observe that $$P_0'(x)=0$$ and, for $$k>0$$, $$P_k'(x)=P_{k-1}(x)$$; then \begin{align} Q'(x)&=c_1P_0(x)+c_2P_1(x)+\dots+c_nP_{n-1}(x) \\ Q''(x)&=c_2P_0(x)+\dots+c_nP_{n-2}(x) \\ &\;\vdots\\ Q^{(n)}(x)&=c_nP_0(x) \end{align} so we immediately get that $$c_0=Q(a),\quad c_1=Q'(a),\quad c_2=Q''(a),\quad \dots,\quad c_n=Q^{(n)}(a)$$ It's simply the Taylor expansion at $$a$$.