mstdn.social

Zeitpunkt              Nutzer    Delta   Tröts        TNR     Titel                     Version  maxTL
Fr 26.07.2024 00:00:03   231.304    +104   17.213.069    74,4 Mastodon 🐘                4.3.0...   500
Do 25.07.2024 00:00:14   231.200    +105   17.196.883    74,4 Mastodon 🐘                4.3.0...   500
Mi 24.07.2024 00:01:07   231.095     +96   17.181.363    74,3 Mastodon 🐘                4.3.0...   500
Di 23.07.2024 00:01:06   230.999    +112   17.166.873    74,3 Mastodon 🐘                4.3.0...   500
Mo 22.07.2024 00:01:10   230.887     +66   17.148.519    74,3 Mastodon 🐘                4.3.0...   500
So 21.07.2024 00:01:06   230.821     +76   17.133.203    74,2 Mastodon 🐘                4.3.0...   500
Sa 20.07.2024 00:00:04   230.745     +27   17.120.900    74,2 Mastodon 🐘                4.3.0...   500
Fr 19.07.2024 13:57:31   230.718    +146   17.111.706    74,2 Mastodon 🐘                4.3.0...   500
Do 18.07.2024 00:00:59   230.572     +90   17.086.626    74,1 Mastodon 🐘                4.3.0...   500
Mi 17.07.2024 00:01:10   230.482       0   17.071.683    74,1 Mastodon 🐘                4.3.0...   500

Pedro J. Hdez (@ecosdelfuturo) · 05/2022 · Tröts: 658 · Folger: 1.180

Fr 26.07.2024 13:37

In this particular case, the solution can be checked by equalling the derivative to zero. It works with this particular function but one has to be very careful with the definitions.

In the code below the way I have made it work.

from sympy import Symbol, lambdify from math import e from scipy.optimize import root_scalar f = Symbol('f') Bnu = 2*h*c**(-2)*(f)**3/(e**(h*(f)/(k*T))-1) DBnu = Bnu.diff(f) lam_DBnu = lambdify(f, DBnu) sol = root_scalar(lam_DBnu, bracket=[c/1000e-9, c/100e-9]) 1e9*c/sol.root

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