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Please translate Hilbert's paper Ein Beitrag zur Theorie des Legendre'schen Polynoms. I cannot find an existing translation for it. Thanks:)

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    $\begingroup$ That's the kind of task you should be prepared to pay for, or to spend more time researching for a translation, not the kind of task to request of MSE volunteers! (Yes, everyone here volunteers their help!) $\endgroup$
    – amWhy
    Jun 24, 2014 at 15:30
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    $\begingroup$ It's only 5 pages, and not a huge amount of text. If it was 10+ and dense, yeah, it's a little much. But this amounts to only 1.5 ish pages of translatable material. $\endgroup$
    – Emily
    Jun 24, 2014 at 15:32
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    $\begingroup$ Go for it, then, @Arkamis! It's all yours! $\endgroup$
    – amWhy
    Jun 24, 2014 at 15:33
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    $\begingroup$ @JackM If you'll notice, I did not vote to close. Nor did I say the OP was prohibited from posting his/her request. I was "put-off" by the imperative-istic title; the OP seems not to have realized that his/her request is significant, and it would have helped for him/her to acknowledge that, and add just wee-bit of context, perhaps. (How would getting a translation help him/her?, e.g.) I for one, sympathize very easily with the OP's on this site (if you haven't noticed), particularly when a request is written as a request, with a bit of motivation added. $\endgroup$
    – amWhy
    Jun 24, 2014 at 16:15
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    $\begingroup$ I agree with @amWhy. The OP is rude, to say the least. $\endgroup$
    – the_fox
    Jun 24, 2014 at 19:49

1 Answer 1

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Acta mathematica 18, printed April 3rd, 1894

A CONTRIBUTION TO THE THEORY OF THE LEGENDRE POLYNOMIAL

BY

DAVID HILBERT

Königsberg, Prussia

This communication at hand is about the question of the smallest value different from $0$ which the integral

$$ I = \int\limits_\alpha^\beta \left[ f(x) \right]^2 \, dx \quad (\beta \gt \alpha) $$

is able to attain, if one chooses for $f(x)$ a polynomial (lit.: "whole rational function") of degree $n - 1$ with integer coefficients and if one understands $\alpha$ and $\beta$ as given constants.

If one sets

$$ f(x) = a_1 x^{n - 1} + a_2 x^{n - 2} + \cdots + a_n $$

the integral turns into a positive quadratic form of the $n$ variables $a_1$, $a_2, \ldots, a_n$:

$$ I = \sum^{i,k} \alpha_{ik} a_i a_k, \quad (i, k = 1, 2, \ldots, n) $$

whose coefficients are given by the formula

$$ \alpha_{ik} = \int\limits_\alpha^\beta x^{2n-i-k} dx = \frac{\beta^{2n-i-k+1}-\alpha^{2n-i-k+1}}{2n-i-k+1} $$

To get an upper bound for the minimum of this quadratic form $I$ it is needed to calculate its discriminant

$$ D_{\alpha \beta} = \left| \begin{matrix} \alpha_{11} & \alpha_{12} & \cdots & \alpha_{1n} \\ \alpha_{21} & \alpha_{22} & \cdots & \alpha_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ \alpha_{n1} & \alpha_{n2} & \cdots & \alpha_{nn} \end{matrix} \right| . $$

(end of page 1)

If we replace in this determinant each element by its integral expression and thereby use for all elements of the first horizontal row the integration variable $x = x_1$, for all elements of the $2^{\tiny\mbox{nd}}, \ldots, n^{\tiny\mbox{th}}$ horizontal row regarding the integration variables $x = x_2, \ldots, x_n$, the discriminant of the quadratic form $I$ is represented as $n$-fold integral of the following form:

$$ D_{\alpha \beta} = \int\limits_\alpha^\beta \ldots \int\limits_\alpha^\beta x_1^{n-1} x_2^{n-2} \ldots x_n^0 \prod^{i,k}(x_i - x_k) dx_1 \ldots dx_n . $$

The interchange of the $n$ integration variables $x_1,\ldots,x_n$ and the addition of the resulting equations gives

$$ D_{\alpha \beta} = \frac{1}{\left|\underline{n}\right.} \int\limits_\alpha^\beta \ldots \int\limits_\alpha^\beta \prod^{i,k} (x_i - x_k)^2 dx_1 \ldots dx_n $$

and if we -- using

$$ x_i = \frac{1}{2}(\beta - \alpha) y_i + \frac{1}{2}(\beta + \alpha) $$

introduce the new integration variables $y_1,\ldots,y_n$, we obtain the formula

$$ D_{\alpha \beta} = \left( \frac{\beta - \alpha}{2} \right)^{n^2} D, $$

where for abbreviation $D = D_{-1, +1}$ has been set.

For example it follows for $\alpha = 0$, $\beta = 1$

$$ D = 2^n \left| \begin{matrix} 1 & 0 & \frac{1}{3} & 0 & \frac{1}{5} & \cdots \\ 0 & \frac{1}{3} & 0 & \frac{1}{5} & 0 & \cdots \\ \frac{1}{3} & 0 & \frac{1}{5} & 0 & \frac{1}{7} & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{matrix} \right| = 2^{n^2} \left| \begin{matrix} 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \cdots & \frac{1}{n} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \cdots & \frac{1}{n + 1} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \cdots & \frac{1}{n + 2} \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac{1}{n} & \frac{1}{n+1} & \frac{1}{n+2} & \frac{1}{n+3} & \cdots & \frac{1}{2n - 1} \end{matrix} \right| . $$

(end of page 2)

To calculate $D$ we develop the polynomial (lit.: "whole rational function") $f(x)$ into a progressing series of Legendre polynomials $X_m$. Because of

$$ x^m = c_m X_m + c_m' X_{m-1} + \cdots + c_m^{(m)} X_0, $$

where

$$ c_m = \frac{\left| \underline{m} \right.}{1 \cdot 3 \cdot 5 \cdots (2m-1)}, $$

we get

$$ \begin{align} f(x) &= a_1 (c_{n-1} X_{n-1} + c_{n-1}' X_{n-2} + \cdots) + a_2 (c_{n-2} X_{n-2} + c_{n-2}' X_{n-3} + \cdots) + \cdots \\ &= c_{n-1} a_1 X_{n-1} + (c_{n-1}' a_1 + c_{n-2} a_2) X_{n-2} + (c_{n-1}'' a_1 + c_{n-2}' a_2 + c_{n-3} a_3 X_{n-3}) + \cdots \end{align} $$

and with the help of the formulas

$$ \int\limits_{-1}^{+1} X_m^2 dx = \frac{2}{2m+1}, \quad \int\limits_{-1}^{+1} X_m X_k dx = 0 \quad (m \ne k) $$

it therefore follows

$$ \left[ I \right]_{\alpha = -1 \atop \beta = +1} = \int\limits_{-1}^{+1} f^2(x) \, dx = \frac{2}{2n-1} b_1^2 + \frac{2}{2n-3} b_2^2 + \frac{2}{2n-5} b_3^2 + \cdots , $$

where

$$ \begin{align} b_1 &= c_{n-1} a_1, \\ b_2 &= c_{n-1}' a_1 + c_{n-2} a_2, \\ b_3 &= c_{n-1}'' a_1 + c_{n-2}' a_2 + c_{n-3} a_3, \\ . \,\,\, & . \,\,\, . \,\,\, . \,\,\, . \,\,\, . \,\,\, . \,\,\, . \,\,\, . \,\,\, . \,\,\, . \,\,\, . \,\,\, . \,\,\, . \,\,\, . \end{align} $$

has been set. Because of this representation as sum of squares of linear forms one wins for the disciminant $D$ the value

$$ \begin{align} D &= \frac{2^n}{1 \cdot 3 \cdot 5 \ldots (2n-1)} (c_0 c_1 \cdots c_{n-1})^2 \\ &= 2^{n^2} \frac{\left\{ 1^{n-1} 2^{n-2} \ldots (n-2)^2 (n-1)^1 \right\}^4}{1^{2n-1} 2^{2n-2} \ldots (2n-2)^2 (2n-1)^1} \end{align} $$

and herein is the right side exactly identical with $\frac{2^{n^2}}{\Delta}$ where $\Delta$ means that value, which I, in my treatise "About the Discriminant of the in the Finite terminating Hypergeometric Series" ("Über die Discriminante der im Endlichen abbrechenden hypergeometrischen Reihe") [1],

[1] Crelle's Journal, Vol. 103, p. 342.

(end of page 3)

got for the discriminant of a $n$-th degree Legendre polynomial under a certain linear transformation. Under consideration of the formula for $D$ stated above, from this follows the result

The discriminant of the quadratic form $\int\limits_0^1 f^2(x)\,dx$ has the value $$ \left| \begin{matrix} 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \cdots & \frac{1}{n} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \cdots & \frac{1}{n + 1} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \cdots & \frac{1}{n + 2} \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac{1}{n} & \frac{1}{n+1} & \frac{1}{n+2} & \frac{1}{n+3} & \cdots & \frac{1}{2n - 1} \end{matrix} \right| = \frac{\left\{ 1^{n-1} 2^{n-2} \ldots (n-2)^2 (n-1)^1 \right\}^4}{1^{2n-1} 2^{2n-2} \ldots (2n-2)^2 (2n-1)^1} $$ and is equal to the reciprocal value of the disciminant of the $n^{\tiny\mbox{th}}$ degree equation $$ \xi^n + \binom{n}{1}^2 \xi^{n-1} + \binom{n}{2}^2 \xi^{n-2} + \ldots + 1 = 0, $$ whose left side, by a linear transformation of the variable $\xi$, can be turned into the Legendre polynomial $X_n$.

We now return to the original posed question. Application of the Stirling forumla, if $N$ is a positive number, delivers the equation

$$ N l 1 + (N-1) l 2 + \ldots + 2 l (N-1) + 1 l N = \frac{1}{2}N^2 l N - \frac{3}{4} N^2 (1 + \varepsilon_N), $$

where $\varepsilon_N$ means a with growing $N$ vanishing number. With the help of this formula one easily finds

$$ l D = (1 + \varepsilon_n') l (2^{-n^2}), $$

(end of page 4)

where $\varepsilon_n'$ vanishes with growing $n$, i.e.

$$ D = \eta_n 2^{-n^2} $$

and further

$$ D_{\alpha \beta} = \eta_n \left( \frac{\beta - \alpha}{4} \right)^{n^2}, $$

where $\eta_n$ approaches the unit with growing $n$.

Now it is, according to a theorem by H. Minkowski [2], always possible, for a definite quadratic form, to choose the $n$ variables such that the value of the quadratic form turns out less than the $n$-fold of the $n^{\tiny \mbox{th}}$ root of its discriminant. If therefore the positive difference $\beta - \alpha$ is assumed less than $4$, it follows that it is always possible to determine a polynomial (lit.: "whole rational function") $f(x)$ with integer coefficients for which the value of the integral $I = \int\limits_\alpha^\beta f^2(x)\,dx$ turns out smaller than $n \eta_n' \left( \frac{|\beta - \alpha|}{4} \right)^n$. But because $\eta_n'=\sqrt[n]{\eta_n}$ approaches the unit too with growing $n$, we get the result

The integral $\int\limits_\alpha^\beta f^2(x) \, dx$ can attain an arbitrary small positive value, if one chooses the polynomial (lit.: "whole integer function") $f(x)$ properly, under the condition that the interval of integration $\alpha$ to $\beta$ is smaller than $4$.

Königsberg in Prussia, March 13th, 1893.

[2] Crelle's Journal, Vol. 107, p. 291.

(end of page 5)

(Special thanks to Donald E. Knuth and the MathJax team)

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    $\begingroup$ A downvote for a work in progress? Come the heck on, people. $\endgroup$
    – Emily
    Jun 24, 2014 at 16:02
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    $\begingroup$ Looks good so far, except that the word-for-word translation "whole rational function" should be "polynomial". (And congratulations on handling the archaic spelling; German has undergone two official spelling "reforms" since this paper was written.) $\endgroup$ Jun 24, 2014 at 16:08
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    $\begingroup$ I like the phrase "we win the formula". We ought to start using it in English! $\endgroup$
    – user940
    Jun 24, 2014 at 17:31
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    $\begingroup$ Wouldn't it be more accurate to say "is able to attain" than "is possible to attain"? $\endgroup$ Jun 24, 2014 at 17:41
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    $\begingroup$ @mvw: That is how they used to write $n!$ (see here). $\endgroup$
    – TonyK
    Aug 1, 2014 at 22:42

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