Wikipedia:Reference desk/Archives/Mathematics/2014 August 12

= August 12 =

What's so special about $$1-\sqrt[n]x$$ ?
OK, the short version of my question would be: What on earth is going on here ?
 * $$\int_0^\infty\exp\Big(-\sqrt[n]x\Big)dx=n!$$


 * $$\int_0^1\Big(1-\sqrt[n]x\Big)^mdx={m+n\choose n}^{-1}$$


 * $$\int_0^1\ln\Big(1-\sqrt[n]x\Big)dx=-H_n$$


 * $$\lim_{n\to\infty}n\Big(1-\sqrt[n]x\Big)=-\ln x$$

As for the long version, I still have trouble formulating it... On one hand, the connection between factorials and binomial coefficients (a.k.a. the $$\Gamma$$ and beta functions) is obvious. On the other hand, the connection between harmonic numbers and the natural logarithm is also well-known. The second and third identities are related by differentiation under the integral sign. But still, I feel like I'm missing something... Like I can't see the forest for the trees... — 79.113.236.1 (talk) 00:40, 12 August 2014 (UTC)


 * You might enjoy reading up on some of the special properties of the roots of unity. I can't really tell what you're getting at, but sometimes identities of exponents and logs make more sense in the context of the complex plane... SemanticMantis (talk) 15:46, 12 August 2014 (UTC)

Mainly for the benefit of other readers: the first one is resolvable by making the substitution $$u=\sqrt[n]x$$, $$du=\frac{1}{n}\frac{\sqrt[n]x}{x} dx = \frac{1}{n}\frac{u}{u^n} dx$$. This changes the integral to $$n \int_0^\infty \frac{u^n}{u} \exp{(-u)} du$$ which is just $$n \Gamma{(n)} = n!$$. The last one can be done using l'Hopital's rule: $$\lim_{n\to\infty} n(1-\sqrt[n]x) = \lim_{n\to\infty} \frac{1-\sqrt[n]x}{1/n} = \lim_{n\to\infty}\frac{-\ln(x)\sqrt[n]{x} \frac{-1}{n^2}}{\frac{-1}{n^2}}=-\ln(x)$$. I myself am a bit curious about the second (the third is done with differentiation under the integral sign with respect to m once the second is known): how does the resulting power series $$\sum_{k=0}^\infty \frac{m!n}{k!(m-k)!(k+n)}$$ turn out to be simply $$\frac{n!m!}{(m+n)!}$$? --Jasper Deng (talk) 07:04, 18 August 2014 (UTC)
 * The first two are simply the beta function and gamma function in disguise. For a more basic proof, based on induction, see here. — 79.113.197.100 (talk) 10:19, 18 August 2014 (UTC)
 * Oh I now understand: if we make the same change of variables as I did above to the second equation, it just becomes $$n\Beta(n,m+1)$$ which has a well-known relation with the gamma function.
 * Probably what is special about this expression is its relation to the binomial theorem, thus the factorial, thus the gamma function, and finally thus the beta function. When one of the terms of the binomial is 1 instead of some other number c, the resulting binomial expansion doesn't have varying powers of c in the terms and thus the coefficients of the power series are just binomial.--Jasper Deng (talk) 16:00, 18 August 2014 (UTC)

Thermal resistance of a wire in partial contact with a disk
Please see attached diagrams in http://i62.tinypic.com/2rdg6yg.jpg

The flow of heat through a solid is governed by the solid's thermal resistivity, R, a property of the solid. For a simple bar shape, the flow of heat from one side to the opposite side as shown in Fig 1 is determined by the particular shape's thermal resistance, R, determined from R and the length, width, and height by the simple formula shown.

For a perfectly heat conducting wire or rod passing through a thermally resistive disk, and in perfect contact with the disk, the thermal resistance from wire to disk perimeter is also given by a standard formula: R = R Ln(Rd/Rw)/2 Pi T -  See fig 2 in the attached file.

I need to find the thermal resistance from a wire to the disk perimeter edge where a pear-shaped hole is cut into the disk, so that the wire makes contact only over a limited angle, as show in Fig 3. The area of the hole can be assumed large compared to the cross section of the wire but small compared with the disk area. Assume heat only flows through solids, and not across any gap, and that there is negligible radiation.

How can I get or derive a formula, either approximate or exact? — Preceding unsigned comment added by 124.178.49.97 (talk) 12:37, 12 August 2014 (UTC)


 * May I ask if this is a real world problem or strictly theoretical ? In the real world, the type of connection will make all the difference.  Any coating or patina on the wire will make a huge difference.  If you do a large number of tests, and record each, you can then develop stats formulas to figure out what the average is, standard deviation, etc. StuRat (talk) 19:16, 14 August 2014 (UTC)
 * It is a real world problem. Any coating, deliberate or incidental (such as your patina), surface roughness, slight misalignment, etc, will increase the thermal resistance.  But there are well known ways of allowing for that, and practical ways of controlling it, described in any good textbook.  Patina etc CAN have a big impact, but it can be made small too.  For example, a drop of oil can dramtically reduce the impact of surface roughnes.  Thermally conductive grease is sold for just that purpose.  Just as in real life, the wire will not be a perfect heat conductor - it will within itself offer some thermal resistance.  But I already know how to deal with that.  — Preceding unsigned comment added by 124.178.49.97 (talk) 01:22, 15 August 2014 (UTC)

Statistical question for a generator.
Lets imagine 2 games:

Game 1: On the first game the guy roll 2d10s, if he roll a 1 on both dices he get 1 point, if he roll a 2 or 3 on a dice and 2 or 3 on the other dice he gets 2 points, on other cares he must reroll until he wins (get 1 or 2 points).

Game 2: Guy roll a d100, if he roll a 1, he get 1 point, if he roll 2, 3, 4 or 5 he gets 2 points. On other results he must reroll until he wins (get 1 or 2 points)

Does both games are statistically equal or not?

201.78.142.165 (talk) 18:21, 12 August 2014 (UTC)


 * This looks like a homework question, so I don't think we can give you the answer, but we can give you a hint: for each game, count the number of ways you can get 1 point, and count the number of ways you can get 2 points, and count the number of ways you have to roll again. (Your teacher wants you to assume that all the dice are fair, and in the first game also wants you to assume that the dice are independent of each other.) RomanSpa (talk) 20:38, 12 August 2014 (UTC)


 * The probability of a re-roll is irrelevant, I think; what counts is the ratio between the two kinds of win. —Tamfang (talk) 23:54, 12 August 2014 (UTC)


 * Tamfang is right, but in this simple case, computing the probabilities of the three results is a very good start:
 * 1 point,
 * 2 points,
 * reroll.
 * Then compare and draw your conclusions. - ¡Ouch! (hurt me / more pain) 07:06, 13 August 2014 (UTC)

Yeah this was not a homework but really for a computer generator program. But I checked wolfram alpha and they had a dice thing. On the 2d0 thing you have 1/100 chance of getting 1 point (1/100 if you include rerolling as a third possible thing) and 4/100 at getting 2 points (again if you include rerolling as a third possible choice). The same thing happens on the second game. So method of convertion XdY rolls into 1dZ rolls works. 201.78.199.49 (talk) 23:42, 13 August 2014 (UTC)