Solve polynomial interpolation $f(x)=\frac{bc(x - b)(x-c)}{(a - b)(a-c)} + \frac{ac(x - a)(x - c)}{(b - a)(b - c)} + \frac{ab(x-a)(x-b)}{(c-a)(c-b)}$

Question

I think the polynomial

$$f(x)=\frac{bc(x - b)(x-c)}{(a - b)(a-c)} + \frac{ac(x - a)(x - c)}{(b - a)(b - c)} + \frac{ab(x-a)(x-b)}{(c-a)(c-b)}$$

may be easy to find, but I am a bit unexperienced. So, I would be glad to get some help!

$$f(a) = bc,\>\>\>f(b) = ac,\>\>\>f(c) = ab$$

which is what the polynomial should satisfy.

Answer 1

The opening is a standard way but is not easy to simplify. May try in symmetrical manner,

$$af(a)=bf(b)=cf(c)=abc$$

Now,

$$xf(x)-abc=k(x-a)(x-b)(x-c)$$

For $f(x)$ being a polynomial, $k=1$ which is what @Raffaele commented.

Furthermore, you may observe that

• $\dfrac{bc}{(a-b)(a-c)} +\dfrac{ca}{(b-c)(b-a)} +\dfrac{ab}{(c-a)(c-b)} \equiv 1$

• $\dfrac{bc(b+c)}{(a-b)(a-c)} +\dfrac{ca(c+a)}{(b-c)(b-a)} +\dfrac{ab(a+b)}{(c-a)(c-b)} \equiv a+b+c$

• $\dfrac{(bc)^2}{(a-b)(a-c)} +\dfrac{(ca)^2}{(b-c)(b-a)} +\dfrac{(ab)^2}{(c-a)(c-b)} \equiv bc+ca+ab$

Answer 2

Let $f(x)=\alpha x^2 + \beta x + \gamma$. Then \begin{align} & f(a)=\alpha a^2 + \beta a + \gamma = bc \tag1\\ & f(b)=\alpha b^2 + \beta b + \gamma = ca \tag2 \\ & f(c)=\alpha c^2 + \beta c + \gamma = ab \tag3 \\ \end{align} Take (1)-(2) and (1)-(3) to get \begin{align} & \alpha (a+b) + \beta +c= 0\tag4 \\ & \alpha (a+c) + \beta +b = 0\tag5\\ \end{align} Then, take (4)-(5) to get $\alpha =1$. Substitute into (5) to get $\beta = -(a+b+c)$ and, in turn, $\gamma = ab+bc +ca$. Thus $$f(x) = x^2 -(a+b+c)x + ab+bc+ca$$