Can't figure out how the i could get from that equation

$$\tag{1} p(x)=\sum_{l=0}^n \left(\sum_{j=0}^n a_j\binom{j}{l}\tilde x^{j-l}\right)(x - \tilde x)^{j-l}$$

for that example: "For an polynomial $p: \mathbb{R} \rightarrow \mathbb{R}$ with $p(x)=x^3+2x^2-1$ in the evaluation point $\tilde x = 1$".

The Answer is, that $x^3+2x^2-1$

$=(x-1+1)^3+2(x-1+1)^2-1$ $=(x-1)^3+3(x-1)^2+3(x-1)+1+2(x-1)^2+4(x-1)+1-1$ $\underline{=(x-1)^3+5(x-1)^2+7(x-1)+2}$

is. Please. Can me show someone the read/write steps for equation (1) how to get the same result like that?


Cant see it how to earn the result for that example. big thx

  • $\begingroup$ There seems to be a transcription error in Equation (1). Was it meant to be written the same as in the answer below, where the final exponent is $l$ rather than $j-l$? $\endgroup$ – David K Jun 15 at 18:03

We start with $p(x)=\sum_{j=0}^n a_j x^j$. Then we rewrite all $x$ into $(x-\tilde{x})+\tilde{x}$ and binomial expand all $[(x-\tilde{x})+\tilde{x}]^j$ and collect to get (1) (after you correct the power of $(x-\tilde{x})$ to $l$).

\begin{align*} p(x) &=\sum_{j=0}^n a_j [(x-\tilde{x})+\tilde{x}]^j\\ &=\sum_{j=0}^n\sum_{l=0}^j a_j \binom{j}{l}(x-\tilde{x})^{j-l}\tilde{x}^l\\ &=\sum_{l=0}^n\left(\sum_{j=0}^l a_j \binom{j}{l}\tilde{x}^{j-l}\right)(x-\tilde{x})^{l}\\ &=\sum_{l=0}^n\left(\sum_{j=0}^n a_j \binom{j}{l}\tilde{x}^{j-l}\right)(x-\tilde{x})^{l}\\ \end{align*} the last step follows from $\binom{j}{l}=0$ when $l>j$.

For your example, $n=3$, $(a_0,a_1,a_2,a_3)=(-1,0,2,1)$ and $\tilde x=1$, so \begin{align*} p(x)&=\left(a_0\binom{0}{0}\tilde{x}^{0-0} +a_1\binom{1}{0}\tilde{x}^{1-0} +a_2\binom{2}{0}\tilde{x}^{2-0} +a_3\binom{3}{0}\tilde{x}^{3-0}\right)(x-\tilde{x})^0+\\ &\qquad\left(a_0\binom{0}{1}\tilde{x}^{0-1} +a_1\binom{1}{1}\tilde{x}^{1-1} +a_2\binom{2}{1}\tilde{x}^{2-1} +a_3\binom{3}{1}\tilde{x}^{3-1}\right)(x-\tilde{x})^1+\\ &\qquad\left(a_0\binom{0}{2}\tilde{x}^{0-2} +a_1\binom{1}{2}\tilde{x}^{1-2} +a_2\binom{2}{2}\tilde{x}^{2-2} +a_3\binom{3}{2}\tilde{x}^{3-2}\right)(x-\tilde{x})^2+\\ &\qquad\left(a_0\binom{0}{3}\tilde{x}^{0-3} +a_1\binom{1}{3}\tilde{x}^{1-3} +a_2\binom{2}{3}\tilde{x}^{2-3} +a_3\binom{3}{3}\tilde{x}^{3-3}\right)(x-\tilde{x})^3\\ &=\left(-1\binom{0}{0} +2\binom{2}{0} +\binom{3}{0}\right)(x-1)^0+\left(2\binom{2}{1} +\binom{3}{1}\right)(x-1)^1+\\ &\qquad\left(2\binom{2}{2} +\binom{3}{2}\right)(x-1)^2+\binom{3}{3}(x-1)^3\\ &=2+7(x-1)^1+5(x-1)^2+(x-1)^3. \end{align*}


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