ljc@drux2.UUCP (ClerLJ) (02/20/86)
Can someone out "there" show me an elementary (but, perhaps
involving "tricks") method for integrating the following
function:
int from 0 to inf s sup 3 over {e sup s - 1} ds
or "written" out:
( oo
| 3
| s
| -------------- ds
| s
| e - 1
) 0
By elementary, I mean you can't use the fact that the integral
is related to both the Gamma Function and the Riemann Zeta Func-
tion or use techniques from the theory of complex variables.
Essentially, using only techniques from the standard undergraduate
calculus sequence, find the value of the integral.
However, techniques as used to find the value of:
int from 0 to inf e sup x sup 2 dx
are acceptable.
The value of the integral, if my memory has not failed me, is pi^2/15.
This integral results when considering the radiated power of a black
body, integrated over all wavelengths. Hence, the cross posting to
net.physics.
Note to net.math readers (if it applies to you read it, otherwise
no offense intended!):
I know I'm in for some scathing remarks that this is not mathematics,
(high school algebra: yes, calculus: maybe) in that there is no group
theory, no topology, and no set theory involved. Well, those of us
in applied mathematics are already tired of comments of that sort,
so please don't bother :-).
Maybe what is needed is net.math.appld, and our pure mathematics
brethren wouldn't have to read such (from their perspective)
tom-foolery as this article.
Thanx in advance!
larry cler
ihnp4!drux2!ljcpercus@acf4.UUCP (Allon G. Percus) (02/22/86)
> However, techniques as used to find the value of: > > int from 0 to inf e sup x sup 2 dx > > are acceptable. I'm afraid to say it, but as it happens, inf / 2 [ x I e dx ] / 0 diverges. What you want is: inf / 2 [ - x I e dx ] / 0 To solve, observe that this is just: |----------------------------------- | inf inf | / 2 / 2 | [ - x [ - y | I e dx I e dy | ] ] | / / \| 0 0 Which is: |----------------------------------- | inf inf | / / 2 2 | [ [ - x - y | I I e dx dy | ] ] | / / \| 0 0 Now convert this to polar coordinates: |----------------------------------- | pi/2 inf | / / 2 | [ [ - r | I I e r dr dtheta | ] ] | / / \| 0 0 Which is: |----------------------------------- | pi/2 [ ] inf | / | 2 | | [ | - r | | I | e | dtheta | ] | ----- | | / | - 2 | \| 0 [ ] 0 Or: |----------------------- | pi/2 | / | [ 1 | I - dtheta | ] 2 | / \| 0 Which finally becomes: sqrt(pi) -------- 2 Using a similar technique, you should be able to solve your problem. . ------- |-----| A. G. Percus |II II| (ARPA) percus@acf4 |II II| (NYU) percus.acf4 |II II| (UUCP) ...{allegra!ihnp4!seismo}!cmcl2!acf4!percus |II II| -------
makdisi@yale.ARPA (Makdisi) (02/23/86)
Expires:
Followup-To:
Distribution:
Keywords:
The integral of s^3/[exp(s)-1] from 0 to infinity can be calculated by
elementary means (and a little trickery), except that one needs the result that
the sum of 1/n^4, n = 1 to infinity, is pi^2/90. Here we go:
Write the integrand as s^3*exp(-s)/[1-exp(-s)], and then expand 1/[1-exp(-s)]
into 1 + exp(-s) + exp(-2s) + exp(-ns) + ... . This is justified since s > 0,
so exp(-s) < 1 . The integrand will then be the sum of s^3*exp(-n*s), n = 1 to
infinity (remember the exp(-s) term in the numerator). What remains to be done
is to show that the integral of each term in the integrand is 6/n^4, so that the
sum of the integrals of the terms is 6*pi^2/90, or pi^2/15. In fact, the
indefinite integral of s^3*exp(-n*s) is
3 2
/ s 3s 6s 6 \ -ns
- ( --- + --- + --- + --- ) e + C [note the minus sign!]
\ 2 3 4 /
n n n n
from integrating by parts three times. The definite integral of each term from
0 to infinity is therefore 6/n^4, since as s tends to plus infinity, exp(-ns)
"outdoes" the polynomial, and the product tends to 0.
The only proof I know that the sum of 1/n^4, n = 1 to infinity, is pi^2/90
involves Fourier series -- anyone know a more elementary way of proving this?
P.S. 1/[1-exp(-s)] = 1 + exp(-s) + [exp(-s)]^2 + [exp(-s)]^3 + ...
= 1 + exp(-s) + exp(-2s) + exp(-3s) + ... , just in case that step wasn't
clear.
--
Kamal Khuri-Makdisi
makdisi@yale-cheops.UUCP
LAMAKDISIDKAMAL LAMAKDISIDKAMAL LAMAKDISIDKAMAL
--
Kamal Khuri-Makdisi
makdisi@yale-cheops.ARPA
LAMAKDISIDKAMAL LAMAKDISIDKAMAL LAMAKDISIDKAMALbs@faron.UUCP (Robert D. Silverman) (02/24/86)
> Expires: > Followup-To: > Distribution: > Keywords: > > > The integral of s^3/[exp(s)-1] from 0 to infinity can be calculated by > elementary means (and a little trickery), except that one needs the result that > the sum of 1/n^4, n = 1 to infinity, is pi^2/90. Here we go: > > Write the integrand as s^3*exp(-s)/[1-exp(-s)], and then expand 1/[1-exp(-s)] > into 1 + exp(-s) + exp(-2s) + exp(-ns) + ... . This is justified since s > 0, > so exp(-s) < 1 . The integrand will then be the sum of s^3*exp(-n*s), n = 1 to > infinity (remember the exp(-s) term in the numerator). What remains to be done > is to show that the integral of each term in the integrand is 6/n^4, so that the > sum of the integrals of the terms is 6*pi^2/90, or pi^2/15. In fact, the > indefinite integral of s^3*exp(-n*s) is > 3 2 > / s 3s 6s 6 \ -ns > - ( --- + --- + --- + --- ) e + C [note the minus sign!] > \ 2 3 4 / > n n n n > from integrating by parts three times. The definite integral of each term from > 0 to infinity is therefore 6/n^4, since as s tends to plus infinity, exp(-ns) > "outdoes" the polynomial, and the product tends to 0. > > The only proof I know that the sum of 1/n^4, n = 1 to infinity, is pi^2/90 > involves Fourier series -- anyone know a more elementary way of proving this? > > P.S. 1/[1-exp(-s)] = 1 + exp(-s) + [exp(-s)]^2 + [exp(-s)]^3 + ... > = 1 + exp(-s) + exp(-2s) + exp(-3s) + ... , just in case that step wasn't > clear. > -- > Kamal Khuri-Makdisi > makdisi@yale-cheops.UUCP > > LAMAKDISIDKAMAL LAMAKDISIDKAMAL LAMAKDISIDKAMAL > -- > Kamal Khuri-Makdisi > makdisi@yale-cheops.ARPA > > LAMAKDISIDKAMAL LAMAKDISIDKAMAL LAMAKDISIDKAMAL One minor detail: One needs to establish that the series converges uniformly in order to justify the term by term integration. Bob Silverman
steiner@bgsuvax.UUCP (Ray Steiner) (02/26/86)
> > > Can someone out "there" show me an elementary (but, perhaps > involving "tricks") method for integrating the following > function: > > int from 0 to inf s sup 3 over {e sup s - 1} ds > > or "written" out: > > ( oo > | 3 > | s > | -------------- ds > | s > | e - 1 > ) 0 > > By elementary, I mean you can't use the fact that the integral > is related to both the Gamma Function and the Riemann Zeta Func- > tion or use techniques from the theory of complex variables. > Essentially, using only techniques from the standard undergraduate > calculus sequence, find the value of the integral. > > However, techniques as used to find the value of: > > int from 0 to inf e sup x sup 2 dx > > are acceptable. > > The value of the integral, if my memory has not failed me, is pi^2/15. > > This integral results when considering the radiated power of a black > body, integrated over all wavelengths. Hence, the cross posting to > net.physics. > > Note to net.math readers (if it applies to you read it, otherwise > no offense intended!): > > I know I'm in for some scathing remarks that this is not mathematics, > (high school algebra: yes, calculus: maybe) in that there is no group > theory, no topology, and no set theory involved. Well, those of us > in applied mathematics are already tired of comments of that sort, > so please don't bother :-). > > Maybe what is needed is net.math.appld, and our pure mathematics > brethren wouldn't have to read such (from their perspective) > tom-foolery as this article. > > Thanx in advance! > larry cler > ihnp4!drux2!ljc Apropos of this same article, does anyone out there know if the indefinite integral of e**x*sec(x) is expressible in terms of elementary functions? I have been wrestling with this little teaser for 25 years without finding a solution!! Ray Steiner
moroney@dec-jon.UUCP (02/27/86)
Can anyone evaluate the following differential equation to the form
y=f(x) (i.e. find f(x))
" 2
y y = K y(0)=K y'(0)=K
1 2
K > 0
1
(The " means second derivitive, ' means first derivitive, K, K , K are
fixed constants) 1 2
Thanks in advance.
-Mike Moroney
..decwrl!dec-rhea!dec-jon!moroneytim@ism780c.UUCP (Tim Smith) (02/27/86)
In article <896@yale.ARPA> makdisi@yale-cheops.UUCP (Kamal Khuri-Makdisi) writes: > >The only proof I know that the sum of 1/n^4, n = 1 to infinity, is pi^2/90 >involves Fourier series -- anyone know a more elementary way of proving this? > It depends on what you consider elementary. In "An Introduction to Analytic Number Theory", by Tom Apostol, he evaluates Zeta(2n) for positive integers n. This is done in chapter 12. Your series is Zeta(4). He has to use contour integration. I think it is more elementary than Fourier series. -- Tim Smith sdcrdcf!ism780c!tim || ima!ism780!tim || ihnp4!cithep!tim
jablow@brahms.BERKELEY.EDU (Eric Robert Jablow) (02/28/86)
In article <1396@decwrl.DEC.COM> moroney@jon.DEC (Mike Moroney) writes: > >Can anyone evaluate the following differential equation to the form >y=f(x) (i.e. find f(x)) > > " 2 >y y = K y(0)=K y'(0)=K > 1 2 > >K > 0 > 1 > >(The " means second derivitive, ' means first derivitive, K, K , K are >fixed constants) 1 2 > >Thanks in advance. > >-Mike Moroney > >..decwrl!dec-rhea!dec-jon!moroney Standard trick: in any differential equation of the form f(y, y', y")=0 (no independent variable), let p=y'. Then y"=p'=p*(dp/dy) by the chain rule. Thus you get f(y, p, p(dp/dy))=0. This is a first order ODE in y. Solve it to get g(y)=p=y'. Solve this for y in the obvious fashion. The **best** book on ODEs is Ordinary Differential Equations, by Ince. Dover publishes it, so it is cheap. It is old-fashioned, though. Respectfully, Eric Robert Jablow MSRI ucbvax!brahms!jablow I may be a screwy little wabbit, but at least I'm not going to Alcatraz! --E. Fudd--
weemba@brahms.BERKELEY.EDU (Matthew P. Wiener) (02/28/86)
In article <12087@ucbvax.BERKELEY.EDU> jablow@brahms.UUCP (Eric Robert Jablow) writes: >The **best** book on ODEs is Ordinary Differential Equations, by Ince. >Dover publishes it, so it is cheap. It is old-fashioned, though. The **best** book for someone who just wants to solve a particular ODE, like the person you responded to, is the Schaum's Outline Series on the topic. It too is cheap. I don't like E L Ince's book. I prefer P Hartman's or I G Petrovski's, with the same title. ucbvax!brahms!weemba Matthew P Wiener/UCB Math Dept/Berkeley CA 94720
tim@ism780c.UUCP (Tim Smith) (03/01/86)
makdisi@yale-cheops.UUCP (Kamal Khuri-Makdisi) writes: > > The only proof I know that the sum of 1/n^4, n = 1 to infinity, > is pi^2/90 involves Fourier series -- anyone know a more > elementary way of proving this? > That's pi^4/90. An elementary proof is in "Challenging Mathematical Problems with Elementary Solutions, Volume II", by A.M. Yaglom and I.M. Yaglom. This is problem 145. Here is what they do: ( in the following, things like n,m,j, etc, are integers ) First, we will establish that zeta(2) = pi^2/6. This is just to show how they approach this kind of thing, and besides, we might as well be complete. First, establish the formula sin nx = C(n,1) sin^1 (x) cos^(n-1) (x) - C(n,3) sin^3 (x) cos^(n-3) (x) + C(n,5) sin^5 (x) cos^(n-5) (x) - ... which you can do by induction. This can be rewritten as sin nx = sin^n (x) [ C(n,1) cot^(n-1)(x) - C(n,3) cot^(n-3)(x) + ... ] We let n = 2m+1, and let x = j pi / n, 1 <= j <= m. Then we have sin(x) != 0, and sin(nx) = 0. This gives us C(2m+1,1) cot^(2m)(x) - C(2m+1,3) cot^(2m-2) (x) + ... = 0 or, in other words, the polynomial C(2m+1,1) z^m - C(2m+1,3) z^(m-1) + ... = P(z) has the roots cot^2 ( pi / n ), cot^2 ( 2 pi /n ), ... , cot^2 ( m pi /n ). Now, the sum of the roots of a polynomial A0 y^n + A1 y^(n-1) + ... is -A1/A0. Thus, cot^2 (pi/n) + cot^2 (2 pi/n) + ... + cot^2 (m pi/n) = C(2m+1,3) / C(2m+1,1 ) = m(2m-1)/3 Noting that csc^2 u = cot^2 u + 1, we get that csc^2 (pi/n) + csc^2 (2 pi/n) + ... + csc^2 (m pi/n) = m(2m+2)/3 Now we are ready to go places! From our elementary trig classes we know that 0 < cot w < 1/w < csc w when 0 < w < pi/2 or cot^2 w < 1/w^2 < csc^2 w Using this, and the formulas above for sums of cot^2 and csc^2, we get m(2m-1)/3 < n^2/pi^2 * ( 1/1^2 + 1/2^2 + ... + 1/m^2 ) < m(2m+2)/3 or ( putting in 2m+1 for n), m(2m-1) pi^2 m(2m+2) pi^2 ------------ < 1/1^2 + 1/2^2 + ... + 1/m^2 < ------------ 3 ( 2m+1 )^2 3 ( 2m+1 )^2 As m -> oo, the things on the end both -> pi^2/6. Now for zeta(4)! We need to evaluate the sum cot^4 (pi/n) + cot^4 (2 pi/n) + ... + cot^4 (m pi/n) In other words, the sum of the squares of the roots of of P(z). If we consider a polynomial A0 y^n + A1 y^(n-1) + A2 y^(n-2) + ..., with roots R1,R2,...,Rn, then we have sum Ri = -A1/A0, and sum RiRj ( i<j ) = A2/A0 Note that (sum Ri)^2 = sum( Ri^2 ) + 2 sum RiRj, i<j, or sum( Ri^2 ) =A1^2/A0^2 - 2 A2/A0. Leaving the details to the reader, :-), we get cot^4 (pi/n) + cot^4 (2 pi/n) + ... + cot^4 (m pi/n) = m(2m-1)(4m^2+10m-9)/45 Using csc^4 = cot^4 + 2 cot^2 + 1, you can grind out the corresponding sum for csc^4: 8m(m+1)(m^2+m+3)/45. We then get m(2m-1)(4m^2+10m-9) pi^4 m 8m(m+1)(m^2+m+3) pi^4 ------------------------ < sum 1/k^4 < --------------------- 45 (2m+1)^4 1 45 (2m+1)^4 We now submit to an orgy of mindless manipulation and get pi^4 1 2 3 13 m ---- (1 - ----)(1 - ----)(1 + ---- - --------) < sum 1/k^4 90 2m+1 2m+1 2m+1 (2m+1)^2 1 pi^4 1 11 < ----(1 - --------)(1 + --------) 90 (2m+1)^2 (2m+1)^2 Letting m -> oo, we get zeta(4) = pi^4/90 They point out that we may evaluate sum cot^6, sum cot^8, etc, in the same sort of way, and get zeta(6) = pi^6/945 zeta(8) = pi^8/9450 zeta(10) = pi^10/93555 zeta(12) = 691 pi^12 / 638512875 Some other formulas for pi that they get by playing with trig functions are Vieta's formula: Let R1 = (1/2)^(1/2), and R(n+1) = (1/2 + 1/2 Rn)^(1/2) Then 2/pi = R1 R2 R3 ... Leibniz's formula: pi/4 = 1 - 1/3 + 1/5 - 1/7 + ... and Wallis's formula: pi 2 2 4 4 6 6 8 8 -- = - - - - - - - - ... 2 1 3 3 5 5 7 7 9 This is a great book. Ubizmo, these computers sure suck for typing equations! -- Tim Smith sdcrdcf!ism780c!tim || ima!ism780!tim || ihnp4!cithep!tim
weemba@brahms.BERKELEY.EDU (Matthew P. Wiener) (03/02/86)
In article <3@bgsuvax.UUCP> steiner@bgsuvax.UUCP (Ray Steiner) writes: >> Note to net.math readers (if it applies to you read it, otherwise >> no offense intended!): >> >> I know I'm in for some scathing remarks that this is not mathematics, >> (high school algebra: yes, calculus: maybe) in that there is no group >> theory, no topology, and no set theory involved. Well, those of us >> in applied mathematics are already tired of comments of that sort, >> so please don't bother :-). >> >> Maybe what is needed is net.math.appld, and our pure mathematics >> brethren wouldn't have to read such (from their perspective) >> tom-foolery as this article. I don't mind calculus questions or high school math questions per se. I do mind when they get >10 solutions posted, mostly the same! The same problem plagues net.puzzle, for example. I like both pure and applied mathematics. >Apropos of this same article, does anyone out there know if >the indefinite integral of e**x*sec(x) is expressible in terms >of elementary functions? I have been wrestling with this >little teaser for 25 years without finding a solution!! It isn't. Sorry. ucbvax!brahms!weemba Matthew P Wiener/UCB Math Dept/Berkeley CA 94720
ravikuma@umn-cs.UUCP (03/02/86)
For those interested in combinatorial algorithm design,
here is a deceptively simple problem:
"A and B, two players each independently think of a
positive integer between 1 and 1000. You are allowed
to ask either of them (in any order) questions of the
form " Is your number less or equal to K?" for any K of
your choice, and find out the smaller of the guessed
-------
numbers. Trivially, you can do this using 20 questions,
since 10 questions suffice to find a guess. How much
better can you do? (14 questions suffice, though it is
not easy to prove that it is the lowerbound.)
Note: We are interested, in the worst-case bound.)lambert@boring.uucp (Lambert Meertens) (03/03/86)
> [...] does anyone out there know if the indefinite integral of e**x*sec(x) > is expressible in terms of elementary functions? The indefinite integral can be written as a Fourier-like series: x cos x + sin x cos 3x + 3 sin 3x cos 5x + 5 sin 5x C + 2e (------------- - ----------------- + ----------------- - ... ) . 1 + 1^2 1 + 3^2 1 + 5^2 This can be found by using the formal series ix 3ix 5ix sec x = 2 (e -e + e + ... ) . The integration constant C jumps at the poles of the integrand. I don't see how to rewrite the sum as a finite closed expression, although it does not look beyond hope; in particular, it is reminiscent of the expansions 2m 1 cos x cos 2x cos 3x cosh mx = -- sinh m.pi (---- - ------- + ------- - ------- + ... ) pi 2m^2 m^2+1^2 m^2+2^2 m^2+3^2 2 sin x 2 sin 2x 3 sin 3x sinh mx = -- sinh m.pi (------- + -------- - -------- + ... ) . pi m^2+1^2 m^2+2^2 m^2+3^2 -- Lambert Meertens ...!{seismo,okstate,garfield,decvax,philabs}!lambert@mcvax.UUCP CWI (Centre for Mathematics and Computer Science), Amsterdam
steiner@bgsuvax.UUCP (Ray Steiner) (03/04/86)
Thanks to all of you who answered my earlier query. Does anyone know how to prove that the integral of e**x*sec(x) is not expressible in terms of elementary functions?
larsen@fisher.UUCP (Michael Larsen) (03/11/86)
> > > For those interested in combinatorial algorithm design, > here is a deceptively simple problem: > "A and B, two players each independently think of a > positive integer between 1 and 1000. You are allowed > to ask either of them (in any order) questions of the > form " Is your number less or equal to K?" for any K of > your choice, and find out the smaller of the guessed > ------- > numbers. Trivially, you can do this using 20 questions, > since 10 questions suffice to find a guess. How much > better can you do? (14 questions suffice, though it is > not easy to prove that it is the lowerbound.) > Note: We are interested, in the worst-case bound.) To determine the required number of guesses, define more generally f(x, y) to be the number of questions required to determine the lesser of two integers, one in the interval [1, x], one in the interval [1+x-y, x], where x >= y. It is easy to show that the optimal first question in this situation relates to the larger of the two intervals, [1, x]. This reduces the computation of f(1000, 1000) to a problem requiring ~10^6 memories and ~10^9 operations. The calculation is simplified by defining g(m, n), for m >= f(n, n), to be the largest integer p such that f(p, n) = m. Clearly, (*) f(m+1, n) = f(m, n) + 2^m. Now define h(n) = g(f(n, n), n). If you know h(n), equation (*) gives all other values of g(m, n). Having computed h(n) and f(n, n) for all n such that f(n, n) < M, you compute h(n) for all n with f(n, n) = M as follows: Take the largest k with f(k, k) = M - 1 such that f(M - 1, k) <= n. If no such k exists, then f(n, n) > M. If such a k does exist, then consider p = k + g(M - 1, n - k). If p >= n, then f(n, n) = M, and h(n) = p. Otherwise f(n, n) > M. This gives a linear time, linear memory algorithm for computing f(x, x) for all x < N. Having implemented it, I report the following results: Range of n Value of f(n, n) 1 0 2 2 3 3 4 - 6 4 7 - 10 5 11 - 19 6 20 - 34 7 35 - 62 8 63 - 113 9 114 - 209 10 210 - 387 11 388 - 720 12 721 - 1350 13 In particular, f(1000, 1000) = 13.