Quiz: calculating a definite integral

01022014, 08:36 AM
(This post was last modified: 01022014 08:42 AM by peacecalc.)
Post: #21




RE: Quiz: calculating a definite integral
Hello,
I think it's : \[ e^{x}^{\ln(x)} \] or \[ e^{x}^{\ln(x)} \]. It could be interesting if this makes a numerical difference. Is anybody good in analysis of numerical mathematics? Greetings peacecalc P.S.: I've seen that \ ( \ ) (with no spaces) doesn't work for inline TeX, with which characters it works? Oh I see it's a problem of my used computer "XP" and so on, so I hope if I use my "win7" machine it will works. 

01022014, 09:42 AM
Post: #22




RE: Quiz: calculating a definite integral  
01022014, 10:12 AM
Post: #23




RE: Quiz: calculating a definite integral
Hello Thomas,
I'm stunning: why the "e" is the problem, otherwise I would say: e is approximatly 3, is that enough? Greetings peacecalc 

01022014, 11:34 AM
(This post was last modified: 01022014 11:36 AM by Gerson W. Barbosa.)
Post: #24




RE: Quiz: calculating a definite integral
(01012014 10:31 PM)Thomas Klemm Wrote:(12312013 10:24 AM)peacecalc Wrote: calculating the integrand: instead of \( x^{x} \) as \( \exp(x\cdot\ln(x)) \)That's a step into the right direction. e^t=1+t+1/2*t^2+1/6*t^3+... =1x*lnx+x^2*(lnx)^2/2... which, upon integration over [0..1], yields 1+1/2^2+1/3^3+1/4^4+... The first ten terms add up to 1.29128599706 Cheers from Camboriú, Brazil Gerson. PS: Sorry for not elaborating enough (slowly typing this on an iphone) Ought to be at the beach instead :) 

01022014, 11:55 AM
Post: #25




RE: Quiz: calculating a definite integral
(01022014 10:12 AM)peacecalc Wrote: e is approximatly 3, is that enough?How do you think we can get the result to 10 or 12 places? I'd prefer: \[ e= \frac{1}{0!}+ \frac{1}{1!}+ \frac{1}{2!}+ \frac{1}{3!}+\cdots \] One method when dealing with difficult problems is breaking them into smaller parts. Hopefully these are easier to handle. In this case we have to integrate \(x^{x}\) which isn't easy. One method to break it apart is using series. Use \(e^{x\log(x)}\) and the power series of \(e^x\). Swap the order of integration and summation. Integrate each of the parts and plug everything together and you have a nice formula which can be used with the calculator. Cheers Thomas 

01022014, 01:14 PM
Post: #26




RE: Quiz: calculating a definite integral
(01022014 11:34 AM)Gerson W. Barbosa Wrote:Bingo! \(\int_{0}^{1}x^{x}=\sum_{n=1}^{\infty}n^{n}\)(01012014 10:31 PM)Thomas Klemm Wrote: That's a step into the right direction. On the WP34S the \(\sum\) key is just beside the \(\int\) key. You can use the same program: 10.001 \(\sum\) 00 Cheers Thomas 

01022014, 06:09 PM
Post: #27




RE: Quiz: calculating a definite integral  
01022014, 11:39 PM
(This post was last modified: 01022014 11:42 PM by Bunuel66.)
Post: #28




RE: Quiz: calculating a definite integral
[img]"http://www.flickr.com/photos/12376860@N07/"[/img]
Sorry, too lazy for using Latex..... If it doesn't work URL: http://www.flickr.com/photos/12376860@N07/ 

01032014, 01:49 AM
Post: #29




RE: Quiz: calculating a definite integral
(01022014 01:14 PM)Thomas Klemm Wrote: On the WP34S the \(\sum\) key is just beside the \(\int\) key. You can use the same program:As much as I like the wp34s, I've preferred the hp 50g for this one: '∑(n=1;10;1/n^n)' Also because it can do that: 'X^2*LN(X)^2/2' INTVX '(1/3*X^3*LN(X)^2+2/9*X^3*LN(X)+2/27*X^3)/2' I'd tried this approach last year, just before leaving, but I quit because I found the overall expression too complicated, even after simplificating the first three terms. Back here again I noticed I was in the right track and decided to give a second look. D'oh! It was only necessary to replace the integration limits for each term individually  only the rational part of the expressions survive the substitution by 0 and 1 which soon makes the pattern evident Thanks for the nice New Year exercise! Cheers, Gerson 

01032014, 05:18 AM
Post: #30




RE: Quiz: calculating a definite integral  
01032014, 06:08 AM
(This post was last modified: 01032014 06:11 AM by walter b.)
Post: #31




RE: Quiz: calculating a definite integral
(01032014 05:18 AM)Thomas Klemm Wrote:(01022014 11:39 PM)Bunuel66 Wrote: iterating: \(\int_{0}^{1}x^n\log(x)^n dx=\frac{(1)^n n!}{n^n}\)This should be: \(\int_{0}^{1}x^n\log(x)^n dx=\frac{(1)^n n!}{(n+1)^{n+1}}\) 'This should be' ?? Christmas is over. Both formulas don't mean the same, there can be only one correct. d:? 

01032014, 07:03 AM
Post: #32




RE: Quiz: calculating a definite integral
(01032014 06:08 AM)walter b Wrote: there can be only one correct.Use integration by parts to get: \(\int_{0}^{1}x^n\log(x)^p dx=\frac{p}{n+1}\int_{0}^{1}x^n\log(x)^{p1}dx\) Iterate starting with \(p=n\) until you end up with: \(\int_{0}^{1}x^n dx=\frac{1}{n+1}\). You can figure out by yourself which is correct. 

01032014, 07:20 AM
Post: #33




RE: Quiz: calculating a definite integral  
01032014, 11:42 AM
Post: #34




RE: Quiz: calculating a definite integral
(01032014 07:03 AM)Thomas Klemm Wrote:(01032014 06:08 AM)walter b Wrote: there can be only one correct.Use integration by parts to get: \(\int_{0}^{1}x^n\log(x)^p dx=\frac{p}{n+1}\int_{0}^{1}x^n\log(x)^{p1}dx\) Sorry, I made a mistake while copying from my paper: too late in the night and with a headhache ;)The right expression is with n+1, it is now corrected. That said 0⁰ is not a problem, it is quite easy to show that it is 1. From a strict mathematical standpoint, I'm a bit puzzled as the Taylor expansion used seems a bit unjustified: the expansion of e^u(x) is not a sum of power of u(x) but a sum of successive derivatives of e^u(x) which doesn't seems to be the same. the first derivative of the function is not defined at 0, then the expansion can't converge, one could have used and expansion around 1 to go that way. I do agree that the sum gives the right result but I don't grasp a rigorous proof. If somebody has some light.... Regards 

01032014, 12:47 PM
Post: #35




RE: Quiz: calculating a definite integral
(01032014 11:42 AM)Bunuel66 Wrote: From a strict mathematical standpoint, I'm a bit puzzled as the Taylor expansion used seems a bit unjustifiedThis is not the Taylorseries expansion: \[x^{x}= 1x log(x)+\frac{1}{2} x^2 log^2(x)\frac{1}{6} x^3 log^3(x)+\frac{1}{24} x^4 log^4(x)\frac{1}{120} x^5 log^5(x)+\frac{1}{720} x^6 log^6(x)+O(x^7) \] The coefficients are not constant but dependent on \(x\). Quote:If somebody has some light....The question is: can we switch integration and summation? As far as I can see the function is well behaved even at \(x=0\) so what can possibly go wrong? It follows from \(\lim_{x\to 0}x\log(x)=0\). HTH Thomas 

01032014, 01:29 PM
Post: #36




RE: Quiz: calculating a definite integral
(01032014 12:47 PM)Thomas Klemm Wrote: [quote='Bunuel66' pid='2007' dateline='1388749354']Great, but then what is the justification for this expansion if it is not based on a Taylor one? It looks a lot like the expansion of exp(u) with u replaced by xlog(x). My view is that it is the Taylor expansion of exp(u) around 0 with u taking the value given by xlog(x), which is numerically correct. But I still miss the justification for getting an equality when integrating on both sides. This equality would have been justified if we had built the Taylor expansion properly. Just as an intuition, if we accept the above serie, as we have an identity for any value of x and as they are both continuous, it is probably feasible to show the identity of the integral by going back to a Riemann sum. This is interesting because it leads to a more general approach like: if f(x) admits a Taylor expansion S(x) then f[u(x)] admits the expansion S[u(x)] and under some conditions to be datailed the equality holds for integrating on both sides. Which is much simpler than the computation of the nth derivative of f[u(x)]. Regards 

01032014, 02:50 PM
Post: #37




RE: Quiz: calculating a definite integral
(01032014 01:29 PM)Bunuel66 Wrote: Great, but then what is the justification for this expansion if it is not based on a Taylor one?We start with \(x^{x}=\exp(x\log(x))\), substitute \(u=x\log(x)\), use the Taylorseries of \(\exp(u)\) and plug \(u\) back into that expression. We just have to make sure that \(u\) is defined for all \(x \in [0, 1]\). Both sides are equal. So integrating both sides yields the same result. Where exactly is the problem? 

01032014, 05:38 PM
(This post was last modified: 01032014 06:21 PM by Bunuel66.)
Post: #38




RE: Quiz: calculating a definite integral
Quote:We start with \(x^{x}=\exp(x\log(x))\), substitute \(u=x\log(x)\), use the Taylorseries of \(\exp(u)\) and plug \(u\) back into that expression. We just have to make sure that \(u\) is defined for all \(x \in [0, 1]\). Well, let's consider the very same approach with the function exp(1/x). Using a similar schema we end up with a serie 11/x+1/2!x²1/3!x³...... who gives the same as exp(1/x) for x<0 and x>0. Then, if we integrate left and right, 1/x will become log(x) which is not defined for x<0 while the integral of exp(1/x) does exist for x<0. This seems to show that having the equality is not enough for keeping it directly after integrating. Regards 

01032014, 08:03 PM
Post: #39




RE: Quiz: calculating a definite integral
My pocket HP30b (yes, not wp34s ), using simplest rectangle method and step 0.00005 give 1.29128599414 after 3 minutes. Enjoy simple methods


01032014, 08:39 PM
Post: #40




RE: Quiz: calculating a definite integral  
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