# Find the interpolating polynomial of degree $3$ that interpolates $f(x) = x^3$

Find the interpolating polynomial of degree $$3$$ that interpolates $$f(x) = x^3$$ at the nodes $$x_0=0, x_1 = 1, x_2=2, x_3 = 3$$.

Here are my workings below

The basic Lagrange polynomials are:

$$L_0(x) = \frac{(x-1)(x-2)(x-3)}{(0-1)(0-2)(0-3)}$$

$$L_1(x) = \frac{(x-0)(x-2)(x-3)}{(1-0)(1-2)(1-3)}$$

$$L_2(x) = \frac{(x-0)(x-1)(x-3)}{(2-0)(2-1)(2-3)}$$

$$L_3(x) = \frac{(x-0)(x-1)(x-2)}{(3-0)(3-1)(3-2)}$$

Then the interpolating polynomial is:

$$P(x) = L_0(x)+(1)^3L_1(x)+(2)^3L_2(x)+(3)^3L_3(x)$$

Am I allowed to find the interpolating polynomial using these basic lagrange polynomials? and is my $$P(x)$$ correct? I wasn't sure if the first term should be $$L_0(x)$$?

• You would get $$P(x) = 0+ 0.5 (x-3.) (x-2.) (x+0.)-4. (x-3.) (x-1.) (x+0.)+4.5 (x-2.) (x-1.) (x+0.) = x^3$$ – Moo Oct 30 '18 at 16:20
• You're allowed to do Lagrange interpolation, and you're probably allowed to do all sorts of complicated things, but you'd be better off if you just observe that the given $f(x)=x^3$ is already a polynomial of degree 3. – Andreas Blass Oct 30 '18 at 16:33
• sorry I made a mistake, you are correct – mt12345 Oct 30 '18 at 16:58
• @Moo sure, if you could add what error you can expect when using this interpolating polynomial to approximae $f(x) = x^3$ at $x=\frac{1}{2}$ that would be great – mt12345 Oct 30 '18 at 21:07
• It is already there. Regards – Moo Oct 30 '18 at 21:42

The final result is

$$P_3(x) = 0.0 + 0.5 (x-3) (x-2) (x+0)-4 (x-3) (x-1) (x+0)+4.5 (x-2) (x-1) (x+0) = x^3$$

The formula for the error bound is given by:

$$E_n(x) = {f^{(n+1)}(\xi(x)) \over (n+1)!} \times (x-x_0)(x-x_1)...(x-x_n)$$

So, we have

$$E_3(x) = {f^{(4)}(\xi(x)) \over 4!} \times (x-0)(x-1)(x-2)(x-3)$$

The fourth derivative of $$f(x) = x^3$$ is zero, so $$E_3(x) = 0$$.

The reason for this is if $$f(x) =$$ polynomial of degree $$M$$ where $$M \le N$$, then $$f^{(n)}(x) = 0 \implies E_n(x) = 0 ~\forall~ x$$

Therefore $$P_3(x)$$ is an exact representation of $$f(x) = x^3$$.

• one question, did you use my lagrange calculations above to find $p_3(x)$ or did you use another formula? – mt12345 Oct 30 '18 at 21:44
• I used the same approach, but used code I wrote to get my result, but yours is the same (I verified it). There are other approaches too. You can check your result using Wolfram Alpha also. – Moo Oct 30 '18 at 21:46
• okay I didn't know if I could use lagrange to get that polynomial or not – mt12345 Oct 30 '18 at 21:51