bre demandé
; on déterminera la valeur approchée
de la différence
par la proportion
![{\displaystyle \operatorname {Log} .(n+1)-\operatorname {Log} .n:1::\operatorname {Log} .(n+x)-\operatorname {Log} .n:}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d06c607f3fca3ebe7762cebcc84b51c69260f0cf)
![{\displaystyle z={\frac {\operatorname {Log} .(n+x)-\operatorname {Log} .n}{\operatorname {Log} .(n+1)-\operatorname {Log} .n}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2363894c1e2581f005d352cce36762e6fef4158e)
On aura donc, en désignant par,
l’erreur commise
![{\displaystyle \varepsilon '={\frac {\operatorname {Log} .(n+x)-\operatorname {Log} .n}{\operatorname {Log} .(n+1)-\operatorname {Log} .n}}={\frac {\operatorname {Log} .\left(1+{\frac {x}{n}}\right)-x\operatorname {Log} .\left(1+{\frac {1}{n}}\right)}{\operatorname {Log} .\left(1+{\frac {1}{n}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d50df6a45ae39d070d48bcc226af87c6fba8f2b)
![{\displaystyle ={\frac {\varepsilon }{\operatorname {Log} .\left(1+{\frac {1}{n}}\right)}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a52ec6e7a0df610db553642a6a402b8bbb6750e3)
Cela donne, au moyen des formules (A) et (B)
![{\displaystyle \varepsilon '={\frac {{\frac {x}{n}}\left({\frac {1-x}{2n}}-{\frac {1-x^{2}}{3n^{2}}}+{\frac {1-x^{3}}{4n^{3}}}-{\frac {1-x^{4}}{5n^{4}}}+\ldots \right)}{{\frac {1}{n}}-{\frac {1}{2n^{2}}}+{\frac {1}{3n^{3}}}-{\frac {1}{4n^{4}}}+\ldots }}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2899bbed45135aaa904e59dbddc668698ad5a2e4)
De là on conclura évidemment
![{\displaystyle \varepsilon '<{\frac {{\frac {x}{n}}.{\frac {1-x}{2n}}}{{\frac {1}{n}}-{\frac {1}{2n^{2}}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10bd4fd6b4edf9f8e51ffbcf7fbb7fbaaf82c201)
ou, en réduisant,
![{\displaystyle \varepsilon '<{\frac {x(1-x)}{2n-1}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9d940173a7da762a826dd600922ff0c2a5dda368)
et,à fortiori,
![{\displaystyle \varepsilon '<{\frac {1}{4(2n-1)}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ccf01fb1037cd5de114a77c4190258c9525aa6e)
Ceci suppose, au surplus, que les logarithmes tabulaires sont par-