حلم طالب الصف الثاني الجامعي

في الرياضيات، حلم طالب الصف الثاني الجامعي sophomore's dream هو زوج من المتطابقات (خصوصاً الأول)

{\displaystyle{\begin{aligned} &\displaystyle\int_{0}^{1}x^{-x}\,dx&&% \displaystyle=\sum_{n=1}^{\infty}n^{-n}\\ &\displaystyle\int_{0}^{1}x^{x}\,dx&&\displaystyle=\sum_{n=1}^{\infty}(-1)^{n+% 1}n^{-n}=-\sum_{n=1}^{\infty}(-n)^{-n}\end{aligned}}}

اكتشفها في 1697 يوهان برنولي.

The numerical values of these constants are approximately 1.291285997... and 0.7834305107..., respectively.

الاسم "حلم طالب الصف الثاني الجامعي" sophomore's dream^[1] يتناقض مع الاسم "حلم الطالب الجامعي الجديد freshman's dream" الذي يُطلق على المتطابقة غير الصحيحة^{[note 1]} ${\textstyle(x+y)^{n}=x^{n}+y^{n}}$ . حلم الصف الثاني الجامعي له مظهر مشابه من كونه أجمل من أن يكون حقيقياً، إلا أنه حقيقي.

البرهان

Graph of the functions y = x^x (red, lower) and y = x^−x (grey, upper) on the interval x ∈ (0, 1].

The proofs of the two identities are completely analogous, so only the proof of the second is presented here. The key ingredients of the proof are:

to write ${\textstyle x^{x}=\exp(x\ln x)}$ (using the notation $ln$ for the natural logarithm and $exp$ for the exponential function);
to expand ${\textstyle\exp(x\ln x)}$ using the power series for $exp$ ; and
to integrate termwise, using integration by substitution.

In details, $x x$ can be expanded as

{\displaystyle x^{x}=\exp(x\log x)=\sum_{n=0}^{\infty}{\frac{x^{n}(\log x)^{n}% }{n!}}.}

ولذلك،

{\displaystyle\int_{0}^{1}x^{x}\,dx=\int_{0}^{1}\sum_{n=0}^{\infty}{\frac{x^{n% }(\log x)^{n}}{n!}}\,dx.}

By uniform convergence of the power series, one may interchange summation and integration to yield

{\displaystyle\int_{0}^{1}x^{x}\,dx=\sum_{n=0}^{\infty}\int_{0}^{1}{\frac{x^{n% }(\log x)^{n}}{n!}}\,dx.}

To evaluate the above integrals, one may change the variable in the integral via the substitution ${\textstyle x=\exp(-{\frac{u}{n+1}}).}$ With this substitution, the bounds of integration are transformed to ${\displaystyle{\displaystyle 0<u<\infty,}}$ giving the identity

{\displaystyle\int_{0}^{1}x^{n}(\log x)^{n}\,dx=(-1)^{n}(n+1)^{-(n+1)}\int_{0}% ^{\infty}u^{n}e^{-u}\,du.}

By Euler's integral identity for the Gamma function, one has

{\displaystyle\int_{0}^{\infty}u^{n}e^{-u}\,du=n!,}

so that

{\displaystyle\int_{0}^{1}{\frac{x^{n}(\log x)^{n}}{n!}}\,dx=(-1)^{n}(n+1)^{-(% n+1)}.}

Summing these (and changing indexing so it starts at $n = 1$ instead of $n = 0$ ) yields the formula.

البرهان التاريخي

The original proof, given in Bernoulli,^[2] and presented in modernized form in Dunham,^[3] differs from the one above in how the termwise integral ${\textstyle\int_{0}^{1}x^{n}(\log x)^{n}\,dx}$ is computed, but is otherwise the same, omitting technical details to justify steps (such as termwise integration). Rather than integrating by substitution, yielding the Gamma function (which was not yet known), Bernoulli used integration by parts to iteratively compute these terms.

The integration by parts proceeds as follows, varying the two exponents independently to obtain a recursion. An indefinite integral is computed initially, omitting the constant of integration ${\displaystyle{\displaystyle+C}}$ both because this was done historically, and because it drops out when computing the definite integral.

Integrating ${\textstyle\int x^{m}(\log x)^{n}\,dx}$ by substituting ${\textstyle u=(\log x)^{n}}$ and ${\textstyle dv=x^{m}\,dx}$ yields:

{\displaystyle{\begin{aligned} \displaystyle\int x^{m}(\log x)^{n}\,dx&% \displaystyle={\frac{x^{m+1}(\log x)^{n}}{m+1}}-{\frac{n}{m+1}}\int x^{m+1}{% \frac{(\log x)^{n-1}}{x}}\,dx\qquad{\text{(for }}m\neq-1{\text{)}}\\ &\displaystyle={\frac{x^{m+1}}{m+1}}(\log x)^{n}-{\frac{n}{m+1}}\int x^{m}(% \log x)^{n-1}\,dx\qquad{\text{(for }}m\neq-1{\text{)}}\end{aligned}}}

(also in the list of integrals of logarithmic functions). This reduces the power on the logarithm in the integrand by 1 (from ${\displaystyle{\displaystyle n}}$ to ${\displaystyle{\displaystyle n-1}}$ ) and thus one can compute the integral inductively, as

{\displaystyle\int x^{m}(\log x)^{n}\,dx={\frac{x^{m+1}}{m+1}}\cdot\sum_{i=0}^% {n}(-1)^{i}{\frac{(n)_{i}}{(m+1)^{i}}}(\log x)^{n-i}}

where ${\textstyle(n)_{i}}$ denotes the falling factorial; there is a finite sum because the induction stops at 0, since $n$ is an integer.

In this case ${\textstyle m=n}$ , and they are integers, so

{\displaystyle\int x^{n}(\log x)^{n}\,dx={\frac{x^{n+1}}{n+1}}\cdot\sum_{i=0}^% {n}(-1)^{i}{\frac{(n)_{i}}{(n+1)^{i}}}(\log x)^{n-i}.}

Integrating from 0 to 1, all the terms vanish except the last term at 1,^{[note 2]} which yields:

{\displaystyle\int_{0}^{1}{\frac{x^{n}(\log x)^{n}}{n!}}\,dx={\frac{1}{n!}}{% \frac{1^{n+1}}{n+1}}(-1)^{n}{\frac{(n)_{n}}{(n+1)^{n}}}=(-1)^{n}(n+1)^{-(n+1)}.}

This is equivalent to computing Euler's integral identity ${\displaystyle{\displaystyle\Gamma(n+1)=n!}}$ for the Gamma function on a different domain (corresponding to changing variables by substitution), as Euler's identity itself can also be computed via an analogous integration by parts.

انظر أيضاً

متسلسلة (رياضيات)

ملاحظات

^ وبينما هي غير صحيحة عموماً، إلى أنها تكون صحيحة حين يعمل المرء على حلقة إبدالية من أعداد أولية مميـِّز $p$ حيث $n$ تكون أس $p$ . النتيجة الصحيحة في سياق تباديلي عام تعطيها مبرهنة ذات الحدين.
^ All the terms vanish at 0 because ${\textstyle\lim_{x\to 0^{+}}x^{m}(\log x)^{n}=0}$ by l'Hôpital's rule (Bernoulli omitted this technicality), and all but the last term vanish at 1 since $log 1 = 0$ .

المراجع

الصيغة

Bernoulli, Johann (1697). Opera omnia. Vol. 3. pp. 376–381.
Borwein, Jonathan; Bailey, David H.; Girgensohn, Roland (2004). Experimentation in Mathematics: Computational Paths to Discovery. Taylor & Francis. pp. 4, 44. ISBN 9781568811369.
Dunham, William (2005). "Chapter 3: The Bernoullis (Johann and ${\displaystyle{\displaystyle x^{x}}}$ )". The Calculus Gallery, Masterpieces from Newton to Lebesgue. Princeton University Press. pp. 46–51. ISBN 9780691095653.
OEIS, (المتتالية A083648 في OEIS) and (المتتالية A073009 في OEIS)
Pólya, George; Szegő, Gábor (1998), "Part I, problem 160", Problems and Theorems in Analysis, Springer, p. 36, ISBN 9783540636403
Eric W. Weisstein, Sophomore's Dream at MathWorld.
Max R. P. Grossmann (2017): Sophomore's dream. 1,000,000 digits of the first constant

الدالة

Literature for x^x and Sophomore's Dream, Tetration Forum, 03/02/2010
The Coupled Exponential, Jay A. Fantini, Gilbert C. Kloepfer, 1998
Sophomore's Dream Function, Jean Jacquelin, 2010, 13 pp.
Lehmer, D. H. (1985). "Numbers associated with Stirling numbers and x^x". Rocky Mountain Journal of Mathematics. 15 (2): 461. doi:10.1216/RMJ-1985-15-2-461.
Gould, H. W. (1996). "A Set of Polynomials Associated with the Higher Derivatives of y = x^x". Rocky Mountain Journal of Mathematics. 26 (2): 615. doi:10.1216/rmjm/1181072076.

الهامش

^ يظهر في Borwein, Bailey & Girgensohn 2004.
^ Bernoulli 1697.
^ Dunham 2005.

[2] وبينما هي غير صحيحة عموماً، إلى أنها تكون صحيحة حين يعمل المرء على حلقة إبدالية من أعداد أولية مميـِّز $p$ حيث $n$ تكون أس $p$ . النتيجة الصحيحة في سياق تباديلي عام تعطيها مبرهنة ذات الحدين.

[5] All the terms vanish at 0 because ${\textstyle\lim_{x\to 0^{+}}x^{m}(\log x)^{n}=0}$ by l'Hôpital's rule (Bernoulli omitted this technicality), and all but the last term vanish at 1 since $log 1 = 0$ .

[1] يظهر في Borwein, Bailey & Girgensohn 2004.

[FOOTNOTEBernoulli1697-3] Bernoulli 1697.

[FOOTNOTEDunham2005-4] Dunham 2005.

[1]

[note 1]

[2]

[3]

[note 2]