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| In mathematics, '''sophomore's dream''' is a name occasionally used for the [[identity (mathematics)|identities]] (especially the first)
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| :<math>\begin{align}
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| \int_0^1 x^{-x}\,dx &= \sum_{n=1}^\infty n^{-n}&&(\scriptstyle{= 1.29128599706266354040728259059560054149861936827\dots)} \\
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| \int_0^1 x^x \,dx &= \sum_{n=1}^\infty (-1)^{n+1}n^{-n} = - \sum_{n=1}^\infty (-n)^{-n} &&(\scriptstyle{= 0.78343051071213440705926438652697546940768199014\dots})
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| \end{align}</math>
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| discovered in 1697 by [[Johann Bernoulli]].
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| The name "sophomore's dream", which appears in {{Harv|Borwein |Bailey |Girgensohn |2004}}, is in contrast to the name "[[freshman's dream]]" which is given to the incorrect<ref group="note">Incorrect unless one is working over a [[Field (mathematics)|field]] or [[Unital algebra|unital]] [[commutative ring]] of prime [[characteristic (algebra)|characteristic]] ''n'' or a factor of ''n''. The correct result is given by the [[binomial theorem]].</ref> equation {{nowrap|1=(''x'' + ''y'')<sup>''n''</sup> = ''x''<sup>''n''</sup> + ''y''<sup>''n''</sup>}}. The [[sophomore]]'s dream has a similar too-good-to-be-true feel, but is in fact true.
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| == Proof ==
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| [[Image:Sophdream.png|thumb|right|200px|Graph of the functions ''y'' = ''x''<sup>''x''</sup> and ''y'' = ''x''<sup>−''x''</sup> on the interval ''x'' ∈ (0, 1].]]
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| We prove the second identity; the first is completely analogous.
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| The key ingredients of the proof are:
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| * Write ''x''<sup>''x''</sup> = exp(''x'' log ''x'').
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| * Expand exp(''x'' log ''x'') using the [[power series]] for exp.
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| * Integrate termwise.
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| * [[Integration by substitution|Integrate by substitution]].
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| Expand ''x''<sup>''x''</sup> as
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| : <math>x^x = \exp(x \log x) = \sum_{n=0}^\infty \frac{x^n(\log x)^n}{n!}.</math>
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| Therefore we have : <math>\int_0^1 x^xdx = \int_0^1 \sum_{n=0}^\infty \frac{x^n(\log x)^n}{n!} \, dx. </math>
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| In order to exchange the order of summation and integration on the right-hand side, we may use [[monotone convergence theorem]] in the following way: The summation can be written as
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| :<math> \sum_{n=0}^{\infty} \frac{x^n(\log x)^n}{n!}=\sum_{k=0}^{\infty}\left( \frac{x^{2k}(\log x)^{2k}}{(2k)!}+\frac{x^{2k+1}(\log x)^{2k+1}}{(2k+1)!}\right).</math> | |
| The second sum has all positive terms for <math>\scriptstyle k\geq 2</math> hence we may exchange the order of the integral and the sum to obtain:
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| :<math>\int_0^1 \sum_{n=0}^\infty \frac{x^n(\log x)^n}{n!} \, dx= \sum_{k=0}^{\infty}\int_0^1\left( \frac{x^{2k}(\log x)^{2k}}{(2k)!}+\frac{x^{2k+1}(\log x)^{2k+1}}{(2k+1)!}\right)\,dx.</math>
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| Since each integral has only two summands now we can change the order of sum and the integral again and deduce the following:
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| : <math>\int_0^1 x^xdx = \sum_{n=0}^\infty \int_0^1 \frac{x^n(\log x)^n}{n!} \, dx. </math>
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| To evaluate the above integrals we perform the [[Integration by substitution|change of variable in the integral]] <math>\scriptstyle x=\exp\, {-\frac{u}{n+1}}</math>, with <math>\scriptstyle 0 < u < \infty </math>, so the integral
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| :<math>\int_0^1 x^n(\log\, x)^n dx</math>
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| writes
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| :<math>(-1)^n (n+1)^{-(n+1)} \int_0^\infty u^n e^{-u} du\, . </math>
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| By the well-known [[Gamma function|Euler's integral identity]] for the [[Gamma_function#Definition|Gamma function]]
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| :<math>\int_0^\infty u^n e^{-u} du=n!</math>
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| so that:
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| :<math>\int_0^1 \frac{x^n (\log x)^n}{n!}\; dx
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| = (-1)^n (n+1)^{-(n+1)}.</math>
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| Summing these (and changing indexing so it starts at ''n'' = 1
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| instead of ''n'' = 0) yields the formula.
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| === Historical proof ===
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| The original proof, given in {{Harvtxt|Bernoulli|1697}}, and presented in modernized form in {{Harvtxt|Dunham|2005}}, differs from the one above in how the termwise integral <math>\int_0^1 x^n(\log\, x)^n\,dx</math> 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.
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| 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]] <math>+ C</math> both because this was done historically, and because it drops out when computing the definite integral. One may integrate <math>\scriptstyle \int x^m (\ln x)^n\; dx</math> by taking ''u'' = (ln ''x'')<sup>''n''</sup> and ''dv'' = ''x''<sup>''m''</sup> ''dx'', which yields:
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| : <math>\begin{align} | |
| \int x^m (\ln x)^n\; dx
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| & = \frac{x^{m+1}(\ln x)^n}{m+1} - \frac{n}{m+1}\int x^{m+1} \frac{(\ln x)^{n-1}}{x} dx \qquad\mbox{(for }m\neq -1\mbox{)}\\
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| & = \frac{x^{m+1}}{m+1}(\ln x)^n - \frac{n}{m+1}\int x^m (\ln x)^{n-1} dx \qquad\mbox{(for }m\neq -1\mbox{)}
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| \end{align}
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| </math>
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| (also in the [[list of integrals of logarithmic functions]]). This reduces the power on the logarithm in the integrand by 1 (from <math>n</math> to <math>n-1</math>) and thus one can compute the integral [[mathematical induction|inductively]], as
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| : <math> | |
| \int x^m (\ln x)^n\; dx
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| = \frac{x^{m+1}}{m+1}
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| \cdot \sum_{i=0}^n (-1)^i \frac{(n)_i}{(m+1)^i} (\ln x)^{n-i}</math>
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| where (''n'')<sub> ''i''</sub> denotes the [[falling factorial]]; there is a finite sum because the induction stops at 0, since ''n'' is an integer.
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| In this case ''m'' = ''n'', and they are integers, so
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| :<math>\int x^n (\ln x)^n\; dx
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| = \frac{x^{n+1}}{n+1}
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| \cdot \sum_{i=0}^n (-1)^i \frac{(n)_i}{(n+1)^i} (\ln x)^{n-i}.</math>
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| Integrating from 0 to 1, all the terms vanish except the last term at 1,<ref group="note">All the terms vanish at 0 because <math>\scriptstyle\lim_{x \to 0^+} x^m (\ln x)^n \, = \, 0</math> by [[l'Hôpital's rule]] (Bernoulli omitted this technicality), and all but the last term vanish at 1 since ln(1) = 0.</ref> which yields:
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| :<math>\int_0^1 \frac{x^n (\ln x)^n}{n!}\; dx
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| = \frac{1}{n!}\frac{1^{n+1}}{n+1}
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| (-1)^n \frac{(n)_n}{(n+1)^n} = (-1)^n (n+1)^{-(n+1)}.</math>
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| From a modern point of view, this is (up to a scale factor) equivalent to computing Euler's integral identity for the Gamma function, <math>\Gamma(n+1) = n!\,</math> 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.
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| ==See Also==
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| * [[Series (mathematics)]]
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| ==Notes==
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| {{reflist|group=note}}
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| ==References==
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| ===Formula===
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| {{refbegin}}
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| * Johann Bernoulli, 1697, collected in Johannis Bernoulli, ''Opera omnia,'' vol. 3, pp. 376–381
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| * {{citation|authorlink1=Jonathan Borwein |first1=Jonathan |last1=Borwein |authorlink2=David H. Bailey |first2=David H. |last2=Bailey |first3=Roland |last3=Girgensohn |title=Experimentation in Mathematics: Computational Paths to Discovery |isbn= 978-1-56881-136-9 |year=2004 |pages= 4, 44}}
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| * {{citation|first=William |last=Dunham |title=The Calculus Gallery, Masterpieces from Newton to Lebesgue |publisher=Princeton University Press |location= Princeton, NJ |year=2005 |isbn= 978-0-691-09565-3 |chapter= 3: The Bernoullis (Johann and <math>x^x</math>) |pages = 46–51}}
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| * [[OEIS]], {{OEIS|id=A083648}} and {{OEIS|id=A073009}}
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| * {{citation|authorlink1=George Pólya|first1=George |last1= Pólya |authorlink2= Gábor Szegő |first2= Gábor |last2= Szegő |title= Problems and Theorems in Analysis |chapter=part I, problem 160 |year= 1998 |isbn= 978-3-54063640-3| page= [http://books.google.com/books?id=b9l2NqGEFzgC&pg=PA36 36]}}<!-- They just state the result, without a name or proof -->
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| * {{MathWorld|urlname=SophomoresDream |title=Sophomore's Dream}}
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| {{refend}}
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| * [http://mrpg.pw/sophomores_dream Sophomore's dream.] 50,000 digits of the first constant
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| ===Function===
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| {{refbegin}}
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| * [http://math.eretrandre.org/tetrationforum/showthread.php?tid=426 Literature for x^x and Sophomore's Dream], Tetration Forum, 03/02/2010
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| * ''[http://eretrandre.org/rb/files/JayFantini1998_203.pdf The Coupled Exponential],''<!-- Actual title unclear --> Jay A. Fantini, Gilbert C. Kloepfer, 1998
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| * [http://math.eretrandre.org/tetrationforum/attachment.php?aid=788 Sophomore's Dream Function], Jean Jacquelin, 2010, 13 pp.
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| * {{cite doi|10.1216/RMJ-1985-15-2-461}}
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| * {{cite doi|10.1216/rmjm/1181072076}}
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| {{refend}}
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| [[Category:Integrals]]
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Hi there, I am Alyson Pomerleau and I think it seems fairly great when you say it. Credit authorising is how she tends to make a residing. One of the things she loves most is canoeing and she's been doing it for fairly a whilst. For a while I've been in Alaska but I will have to free online tarot card readings transfer in a year or two.
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