à l'ordre p avec reste intégral

English translation: p-order Taylor's theorem with integral remainder

GLOSSARY ENTRY (DERIVED FROM QUESTION BELOW)
French term or phrase:Formule de Taylor à l'ordre p avec reste intégral
English translation:p-order Taylor's theorem with integral remainder
Entered by: Tony M

15:53 Mar 13, 2018
French to English translations [PRO]
Science - Mathematics & Statistics / Licence Sciences, Technologie & Santé Option Mathématiques
French term or phrase: à l'ordre p avec reste intégral
Objectif: Proursuivre l'étude des functions de la variable réelle et des suites réelles
Prérequis: Analys 1
Programme:
- Fonction d'une variable réelle à values réelles: Limite, continuity, continuity uniforms, théorème de Heine.
- Suites de nombre réels: convergent, divergent, suites extradites, suites de Cauchy, théorème de Bolzano-Weierstrass.
- Dérivation des functions d'une variable réelle à valeurs réelles: opérations, inégalité et égalité des accroissements finis. Extremums local des functions drivable à valeurs réelles. Théorème de Rolle. Fonction de Classe Cp. Formule de Leibniz. Formule de Taylor /////à l'ordre p avec reste integral///// pour une fonction de Classe Cp+1; formulae de Taylor-Lagrange. Etude locale des functions: développement limits et asymptomatiques, operations sur les développement limits. Théorème de Taylor-Young.
- Etude de suites récurrentes

From this "Licence" programme description.
Thanks for your help.
Ghyslaine LE NAGARD
New Caledonia
Local time: 15:39
p-order Taylor's theorem with integral remainder
Explanation:
I am not a maths / statistics specialist, whence the lowish cofnidence level — I have found plenty of references for both parts of the expression separately, but have not dug deeper to find them collocated, so my word-order suggestion is only a guess: usually, 'X-order' anything is a pre-qualifier, though I don't know if that holds good here.


Taylor's Theorem - Integral Remainder - Penn Math

https://www.math.upenn.edu/~kazdan/361F15/Notes/Taylor-integ...

Remark In this version, the error term involves an integral. Because of this, we assume that fk+1 is continuous, whereas previously we only assumed this derivative exists. However, we get the valuable bonus that this integral version of Taylor's theorem does not involve the essentially unknown constant c. This is vital in ...

The Integral Form of the Remainder in Taylor's Theorem MATH 141H

www.math.umd.edu/~jmr/141/remainder.pdf

The Integral Form of the Remainder in Taylor's Theorem. MATH 141H. Jonathan Rosenberg. April 24, 2006.
Let f be a smooth function near x = 0. For x close to 0, we can write f(x) in terms of f(0) by using the Fundamental Theorem of Calculus: f(x) = f(0) + ∫ x. 0 f (t) dt. Now integrate by parts, setting u = f (t), du = f (t) dt, v = t ...

Théorème de Taylor — Wikipédia

https://fr.wikipedia.org/wiki/Théorème_de_Taylor

En mathématiques, plus précisément en analyse, le théorème de Taylor (ou formule de Taylor), du nom du mathématicien anglais Brook Taylor qui l'établit en 1715, montre qu'une fonction plusieurs fois dérivable au voisinage d'un point peut être approchée par une fonction polynôme dont les coefficients dépendent ...
‎Formule de Taylor-Young · ‎Formule de Taylor ... · ‎Formule de Taylor avec ...

Taylor's theorem - Wikipedia

https://en.wikipedia.org/wiki/Taylor's_theorem

Explicit formulas for the remainder - for some number ξ between a and x. This version covers the Lagrange and Cauchy forms of the remainder as special cases, and is proved below using Cauchy's mean value theorem. The statement for the integral form of the remainder is more advanced than the ...
Selected response from:

Tony M
France
Local time: 06:39
Grading comment
Thanks
4 KudoZ points were awarded for this answer



Summary of answers provided
3 +1p-order Taylor's theorem with integral remainder
Tony M


Discussion entries: 1





  

Answers


21 mins   confidence: Answerer confidence 3/5Answerer confidence 3/5 peer agreement (net): +1
formule de Taylor à l'ordre p avec reste integral
p-order Taylor's theorem with integral remainder


Explanation:
I am not a maths / statistics specialist, whence the lowish cofnidence level — I have found plenty of references for both parts of the expression separately, but have not dug deeper to find them collocated, so my word-order suggestion is only a guess: usually, 'X-order' anything is a pre-qualifier, though I don't know if that holds good here.


Taylor's Theorem - Integral Remainder - Penn Math

https://www.math.upenn.edu/~kazdan/361F15/Notes/Taylor-integ...

Remark In this version, the error term involves an integral. Because of this, we assume that fk+1 is continuous, whereas previously we only assumed this derivative exists. However, we get the valuable bonus that this integral version of Taylor's theorem does not involve the essentially unknown constant c. This is vital in ...

The Integral Form of the Remainder in Taylor's Theorem MATH 141H

www.math.umd.edu/~jmr/141/remainder.pdf

The Integral Form of the Remainder in Taylor's Theorem. MATH 141H. Jonathan Rosenberg. April 24, 2006.
Let f be a smooth function near x = 0. For x close to 0, we can write f(x) in terms of f(0) by using the Fundamental Theorem of Calculus: f(x) = f(0) + ∫ x. 0 f (t) dt. Now integrate by parts, setting u = f (t), du = f (t) dt, v = t ...

Théorème de Taylor — Wikipédia

https://fr.wikipedia.org/wiki/Théorème_de_Taylor

En mathématiques, plus précisément en analyse, le théorème de Taylor (ou formule de Taylor), du nom du mathématicien anglais Brook Taylor qui l'établit en 1715, montre qu'une fonction plusieurs fois dérivable au voisinage d'un point peut être approchée par une fonction polynôme dont les coefficients dépendent ...
‎Formule de Taylor-Young · ‎Formule de Taylor ... · ‎Formule de Taylor avec ...

Taylor's theorem - Wikipedia

https://en.wikipedia.org/wiki/Taylor's_theorem

Explicit formulas for the remainder - for some number ξ between a and x. This version covers the Lagrange and Cauchy forms of the remainder as special cases, and is proved below using Cauchy's mean value theorem. The statement for the integral form of the remainder is more advanced than the ...

Tony M
France
Local time: 06:39
Native speaker of: Native in EnglishEnglish
PRO pts in category: 35
Grading comment
Thanks

Peer comments on this answer (and responses from the answerer)
agree  Francois Boye
2 hrs
  -> Merci, François !
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