Say $$f(x) = x^2 - 10x + 24$$
and we want to find the coefficients of : $$g(x)=f(x+k) = (x+k)^2 - 10(x+k) + 24=\color{blue}{a_2x^2 + a_1x + a_0}$$
Method 1
Using maclaurin series, easy to see that$\color{blue}{a_0} = g(0) = f(k) = k^2-10k+24$
$\color{blue}{a_1} = g'(0) = f'(k) = 2k-10$
$\color{blue}{a_2} = g''(0)/2 = f''(k)/2 = 1$
Method 2
Repeatedly divide $f(x)$ by $\color{red}{x-k}$$f(x) = f_1(x)(\color{red}{x-k})+\color{blue}{a_0}$
$f_1(x) = f_2(x) (\color{red}{x-k})+\color{blue}{a_1}$
$f_2(x) = f_3(x) (\color{red}{x-k})+\color{blue}{a_2}$
I feel both methods are related, but I haven't been able to comprehend exactly how. These methods seem to link "division" to "differentiation". I think the "successive" division in method2 is really throwing me off. I seem to not understand how it actually works... I appreciate if you could help me understand how these two methods are related...
My thoughts so far
- I completely understand how Method1 works, its because maclaurin series for a polynomial gives the same polynomial back.
- But Method2 is still a bit magic to me... I see that Method2 is using $\color{red}{x-k}$ instead of $\color{red}{x+k}$. Is it because the remainder of $\dfrac{f(x)}{x-k}$ is $f(k)$, not $f(-k)$ ? But why is this the reason ? I think I don't get how repeatedly dividing $\color{red}{x-k}$ is spitting out the coefficients.
- Since I don't understand Method2, I thought contrasting it with Method1 might provide some insight..
Method 2. expresses $\,f(x)\,$ as a polynomial in $\,(x-k)\,$, which can indeed be determined via repeated long division:
$$ f(x) = a_0 + a_1 \cdot (x-k) + a_2 (x-k)^2 = a_0 + (x-k) \cdot \underbrace{\big(a_1 + \overbrace{a_2}^{f_2(x)} \cdot (x-k)\big)}_{{f_1(x)}} $$
Then $\require{cancel}\,g(x)=f(\color{blue}{x+k}) = a_0 + a_1 \cdot \big(\color{blue}{x+\bcancel{k}}-\bcancel{k}\big) + a_2 \cdot \big(\color{blue}{x+\bcancel{k}}-\bcancel{k}\big)^2 = a_0+a_1x+ a_2x^2\,$.