Commemorating Victory Day, May 9, 1945

Put $f(x) = r_0 + r_1x + r_2x^2 + \cdots + r_{1945}x^{1945}$ and note that $\int_0^1 f(x) x^k\, dx = \sum_{i = 0}^{1945} r_i \int_0^1 x^{i + k}\, dx = \sum_{i = 0}^{1945} \frac {r_i}{i + k + 1}.$ Expanding the sum on the right and setting it equal to $1$ for $k = 0,1,\dots,1945$ leads to the system $\begin{matrix} r_0 & + & \frac 12 r_1 & + & \cdots & + & \frac 1{1946}r_{1945} & = & 1, \\[0.3em] \frac 12 r_0 & + & \frac 13 r_2 & + & \cdots & + & \frac 1{1947}r_{1945} & = & 1, \\[0.3em] \vdots & & \vdots & & \ddots & & \vdots & & \vdots \\[0.3em] \frac 1{1946} r_0 & + & \frac 1{1947}r_1 & + & \cdots & + & \frac 1{3891}r_{1945} & = & 1, \end{matrix}$ or more compactly, $H \vec r = \vec 1,$ where $H$ is the $1946 \times 1946$ Hilbert matrix, $\vec r$ is the coefficient vector of $f$ , and $\vec 1$ is a column vector of $1$ s. In general, Hilbert matrices are invertible, so $f$ exists and its coefficients are given by $\vec r = H^{-1}\vec 1.$ This implies that $r_i$ is the sum of the entries in the $i$ th row of $H^{-1}$ . Now consider $\int_0^1 [f(x)]^2\, dx = \int_0^1 f(x) \left(\sum_{i = 0}^{1945} r_i x^i \right) dx= \sum_{i = 0}^{1945} r_i \int_0^1 f(x) x^i\, dx = \sum_{i = 0}^{1945} r_i.$ The sum on the right is the sum of all the entries of $H^{-1}$ . There is a useful property of any Hilbert matrix: if $H$ is a $d \times d$ Hilbert matrix, then the sum of all the entries in $H^{-1}$ is $d^2$ . This property implies that $\int_0^1 [f(x)]^2\, dx = 1946^2 = \boxed{3786916}.$ To prove the minimality of this value, suppose that $g$ is square integrable over $[0,1]$ and also has the property that $\int_0^1 g(x) x^k\, dx = 1, \quad k = 0,1,\dots,1945.$ Notice that $\int_0^1 g(x) f(x)\, dx = \sum_{i = 0}^{1945} r_i \int_0^1 g(x) x^i\, dx = \int_0^1 [f(x)]^2\, dx,$ which means that in the Lebesgue space $L^2[0,1]$ $\langle g - f , f \rangle = \int_0^1 \big[ g(x) - f(x) \big] f(x)\, dx = \int_0^1 g(x) f(x)\, dx - \int_0^1 [f(x)]^2\, dx = 0.$ In other words, $g - f$ is orthogonal to $f$ . This result has a nice geometric interpretation. By the Pythagorean theorem for inner product spaces, $\|g\|^2 = \|g - f\|^2 + \|f\|^2 \geq \|f\|^2,$ and thus $\int_0^1 [g(x)]^2\, dx = \|g\|^2 \geq \|f\|^2 = \int_0^1 [f(x)]^2\, dx.$ This establishes the minimality of $f$ with respect to the $L^2[0,1]$ norm. Notice that $f$ is minimal not just among polynomial functions but among all functions that are square integrable over $[0,1]$ .