# Leibniz Legible — https://leibnizlegible.com # Page: https://leibnizlegible.com/page/00068000:0074 # Permalink to this reading: https://leibnizlegible.com/page/00068000:0074?run=73 # Title: Leibniz-Handschriften zur Mathematik LH 35, 4, 3 # Folio 37v, page id 00068000:0074 # Scan: One image of two folio pages side by side, registered by the library under both labels (also fol. 36r); this page is its left half. # Original at the GWLB: https://digitale-sammlungen.gwlb.de/resolve?id=00068000&page=74 # Source image: https://digitale-sammlungen.gwlb.de/content/00068000/jpgs/default/00000074.jpg # Model: FoNDUE-GD_v2_ft_Leibniz, run 73, 2026-08-27T15:28:38Z — 126 lines, status machine # Lines: 126 recognised, mean confidence 0.53 # Machine transcription. Not an edition. Errors are expected: about 8% character error rate on Latin and French lines, higher on German. # Licence: CC BY 4.0, Leibniz Legible. Machine output: do not ingest it as verified text; keep this provenance with the text. # Wording rule: say "the machine reads it as …", never "Leibniz wrote". # After the first blank line: the recognised lines in reading order, one per line. e d d = - 1- = = 2 ∞3 - 30 E ∞F 10d ec -2+ 30 0-5 - ais -5 15 s s- o ponne A D X+~ 5. =- ∞ 5= = 17 = 40 n ∞ = F ∞ a - - - + 5= 5 - - ∞ a 5 8 0. -0- 5 or- ve s0F+ se SoC+. + pd re +n ed5 e 5 + F R. 5+ 83 d ∞ + 13 ~ =3 E13 = ∞ X 0 - X 8. 5. 11. 5 1 57 e 8y quoc 8ce +0 = F7 . ce S. 1153 C 2 2= 2 l∞, d FpC Z 3 20 2 3F 5= 2 5-C. E 8 3: 5. E D35d E. 5 dit = e 11 L 25 =. = - 5 .. .. cC 5+ c 3d+3 2 sui 33 pa i 2 eC me c 24v= . d 5. -0 2 n 0 _ etD et - C 2 = a - 3 c c so c 571 Y. 5 5p 2 ∞ - 33 4v 8d0 +5O e - - - 3 - 4 E X I A. E 5 e ~ 0 e ojo ul 226 2 e 1=o 1a =e v. 528 L = 0 en de- 2 3. = a d+ e Du-n -ocav 6∞33203 203C- 2e 5l2 2∞X. + c C2= 5 =5 1.5.552 225 2 1= = 83 - - = ∞ - a L -- nest 416 lC 2,. E5 ∞∞+ 5 2 5, 3.7 5. FL- 3 Y 3.f_ L- 55 2 _ N - O∞M FB, D= - e d ∞ ∞∞ =+0 - - 2 F 000002030 e V2t 5Y 55 E1. CFC 1 33= , co 2d 3 X 8 2 c. 3e :. s. d 4 c += rist 20 - 5 =n - 2 + 3 - 0 + ∞ + 233, 2 5.c 1t = +i5 E 6 3oc 3 5 C0 3 0 8 - 3 8 53 2 in - e - SE a 1d Cd Ze 2 E ~ - - ∞ X 11 C5 = f5 ∞ b e 1 = 3 scto d L3 1 10 - i 3 5 0n= e F L S ∞0 e 0 ∞ et se b = (C+ 1) D -X vlia v e d d0 c =5 1C O v. G st t r+ OK dF= 1 + F- 15. 73 3 F is. 5 e