An extrapolation problem: finding the constant term of a polynomial

Question

$f(x)$ is a degree $6$ polynomial and $f(n)=\frac{1}{n}$ when $n=1, 2, 3, 4, 5, 6, 7$.
How to find the constant term of $f(x)$?

Answer 1

By the well-known properties of the forward difference operator,
if $f$ is a polynomial with degree $\leq 6$ we have $$ \sum_{k=0}^{7}\binom{7}{k}(-1)^k\,f(r+k) = 0 $$ for any $r\in\mathbb{R}$. In our case, by picking $r=0$ and exploiting the binomial theorem we get $$ f(0)=\sum_{k=1}^{7}\binom{7}{k}\frac{(-1)^{k+1}}{k} =\int_{0}^{1}\frac{1-(1-x)^7}{x}\,dx=\int_{0}^{1}\frac{x^7-1}{x-1}\,dx=H_7=\color{red}{\frac{363}{140}}.$$ The approach suggested by Lord Shark the Unknown in the comments leads to $$ f(0)=\left.\frac{d}{dx}(x-1)\cdots(x-7)\right|_{x=0}\stackrel{\text{LogDerivative}}{=} H_7$$ the same outcome, which is not surprising.

Answer 2

Leading questions: What degree is $xf(x)-1$? Since we know seven of its roots, what form must it have? What constant term could $xf(x)-1$ possibly have (i.e. what is its value at $x=0$)? Therefore, what polynomial must it be? Now what is $f(x)$?

Answer 3

Write $$f(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+a_{5}x^{5}+a_{6}x^{6}$$

Now form a system of equations:

$$1=f(1)=a_{0}+a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}$$

$$\frac{1}{2}=f(2) =a_{0}+a_{1}2+a_{2}2^{2}+a_{3}2^{3}+a_{4}2^{4}+a_{5}2^{5}+a_{6}2^{6}$$

$$\frac{1}{3}=f(3) =a_{0}+a_{1}3+a_{2}3^{2}+a_{3}3^{3}+a_{4}3^{4}+a_{5}3^{5}+a_{6}3^{6}$$

$$\frac{1}{4}=f(4) =a_{0}+a_{1}4+a_{2}4^{2}+a_{3}4^{3}+a_{4}4^{4}+a_{5}4^{5}+a_{6}4^{6}$$

$$\frac{1}{5}=f(5) =a_{0}+a_{1}5+a_{2}5^{2}+a_{3}5^{3}+a_{4}5^{4}+a_{5}5^{5}+a_{6}5^{6}$$

$$\frac{1}{6}=f(6) =a_{0}+a_{1}6+a_{2}6^{2}+a_{3}6^{3}+a_{4}6^{4}+a_{5}6^{5}+a_{6}6^{6}$$

$$\frac{1}{7}=f(7) =a_{0}+a_{1}7+a_{2}7^{2}+a_{3}7^{3}+a_{4}7^{4}+a_{5}7^{5}+a_{6}7^{6}$$

Now you have a linear system of equations and you know that the Vandermonde matrix is invertible in this case, so the solution is unique. Then use Cramer rule to find $a_0,\ldots,a_6$.