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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?

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  • $\begingroup$ $\lambda_1$ (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). $\endgroup$ – Arthur Oct 26 '18 at 18:42
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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.

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  • $\begingroup$ can you check if i did everything alright? $\endgroup$ – C. Cristi Oct 26 '18 at 19:29
  • $\begingroup$ It looks fine, except that $C_k^1$ should be $C_1^k$ (or $\binom k1$). $\endgroup$ – José Carlos Santos Oct 26 '18 at 20:52
  • $\begingroup$ and the sum changes, right? no longer from $1$ to $n$, but from ..? $\endgroup$ – C. Cristi Oct 27 '18 at 13:16
  • $\begingroup$ $\lambda_k=\sum_{j=k}^n(\ldots)$. $\endgroup$ – José Carlos Santos Oct 27 '18 at 16:08
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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 $a$.

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