جنس$1$ - traduzione in Inglese
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جنس$1$ - traduzione in Inglese

DIVERGENT SERIES
1+1+1+···; 1 + 1 + 1 + 1 + 1 + · · ·; 1 + 1 + 1 + 1 + · · ·; 1 + 1 + 1 + 1 + …; 1 + 1 + 1 + 1 + ...; Zeta(0)
  • alt=A graph showing a line that dips just below the ''y''-axis

جنس      
nationalize
سمك نهري         
جنس من الأسماك
جنس الشبوط الشائع; جنس الكارب العادي; كارب; كارب (سمك); شبوط; سمك نهري; Cyprinus
carp, perch
الأجناس         
  • بالمتكات]] وهي لأعضاء الذكر وتأتي منها [[حبوب اللقاح]].
  • يتحدد نوع النشأ عند ذبابة الفاكهة بوساطة الصبغيان XY مثلما يتحدد عند الإنسان.
  • دورة تكاثر نبات من ذوات البذور angiospermes.
  • مراحل انقسام الخلية ثم اندماج [نصف] خلية من أنثى مع [نصف] خلية من الذكر لتكوين [[خلية]] جديدة ضعفانية تكون نواة لفرد من جيل جديد.
أي من (ذكر أو أنثى) يمكن وضع العديد من الكائنات فيهما ، حسب الوظيفة أو الجهاز التناسلي
جنس بيولوجي; الأجناس; Widower

multi-racial

Definizione

one
the upper limit of intoxication or exhaustion
after the second pint of gin, i was hard one-ing

Wikipedia

1 + 1 + 1 + 1 + ⋯

In mathematics, 1 + 1 + 1 + 1 + ⋯, also written n = 1 n 0 {\displaystyle \sum _{n=1}^{\infty }n^{0}} , n = 1 1 n {\displaystyle \sum _{n=1}^{\infty }1^{n}} , or simply n = 1 1 {\displaystyle \sum _{n=1}^{\infty }1} , is a divergent series, meaning that its sequence of partial sums does not converge to a limit in the real numbers. The sequence 1n can be thought of as a geometric series with the common ratio 1. Unlike other geometric series with rational ratio (except −1), it converges in neither the real numbers nor in the p-adic numbers for some p. In the context of the extended real number line

n = 1 1 = + , {\displaystyle \sum _{n=1}^{\infty }1=+\infty \,,}

since its sequence of partial sums increases monotonically without bound.

Where the sum of n0 occurs in physical applications, it may sometimes be interpreted by zeta function regularization, as the value at s = 0 of the Riemann zeta function:

ζ ( s ) = n = 1 1 n s = 1 1 2 1 s n = 1 ( 1 ) n + 1 n s . {\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}={\frac {1}{1-2^{1-s}}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n^{s}}}\,.}

The two formulas given above are not valid at zero however, but the analytic continuation is.

ζ ( s ) = 2 s π s 1   sin ( π s 2 )   Γ ( 1 s )   ζ ( 1 s ) , {\displaystyle \zeta (s)=2^{s}\pi ^{s-1}\ \sin \left({\frac {\pi s}{2}}\right)\ \Gamma (1-s)\ \zeta (1-s)\!,}

Using this one gets (given that Γ(1) = 1),

ζ ( 0 ) = 1 π lim s 0   sin ( π s 2 )   ζ ( 1 s ) = 1 π lim s 0   ( π s 2 π 3 s 3 48 + . . . )   ( 1 s + . . . ) = 1 2 {\displaystyle \zeta (0)={\frac {1}{\pi }}\lim _{s\rightarrow 0}\ \sin \left({\frac {\pi s}{2}}\right)\ \zeta (1-s)={\frac {1}{\pi }}\lim _{s\rightarrow 0}\ \left({\frac {\pi s}{2}}-{\frac {\pi ^{3}s^{3}}{48}}+...\right)\ \left(-{\frac {1}{s}}+...\right)=-{\frac {1}{2}}}

where the power series expansion for ζ(s) about s = 1 follows because ζ(s) has a simple pole of residue one there. In this sense 1 + 1 + 1 + 1 + ⋯ = ζ(0) = −1/2.

Emilio Elizalde presents a comment from others about the series:

In a short period of less than a year, two distinguished physicists, A. Slavnov and F. Yndurain, gave seminars in Barcelona, about different subjects. It was remarkable that, in both presentations, at some point the speaker addressed the audience with these words: 'As everybody knows, 1 + 1 + 1 + ⋯ = −1/2.' Implying maybe: If you do not know this, it is no use to continue listening.