Quiz: calculating a definite integral
01-02-2014, 08:36 AM (This post was last modified: 01-02-2014 08:42 AM by peacecalc.)
Post: #21
 peacecalc Member Posts: 187 Joined: Dec 2013
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.
01-02-2014, 09:42 AM
Post: #22
 Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: Quiz: calculating a definite integral
(01-02-2014 08:36 AM)peacecalc Wrote:  I think it's : $$(e^{-x})^{\ln(x)}$$ or $$(e^x)^{-\ln(x)}$$.
Then what is $$e$$?
01-02-2014, 10:12 AM
Post: #23
 peacecalc Member Posts: 187 Joined: Dec 2013
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
01-02-2014, 11:34 AM (This post was last modified: 01-02-2014 11:36 AM by Gerson W. Barbosa.)
Post: #24
 Gerson W. Barbosa Senior Member Posts: 1,405 Joined: Dec 2013
RE: Quiz: calculating a definite integral
(01-01-2014 10:31 PM)Thomas Klemm Wrote:
(12-31-2013 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.
What's the next? What's $$\exp(t)$$?

Cheers
Thomas

e^t=1+t+1/2*t^2+1/6*t^3+...
=1-x*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 :-)
01-02-2014, 11:55 AM
Post: #25
 Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: Quiz: calculating a definite integral
(01-02-2014 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
01-02-2014, 01:14 PM
Post: #26
 Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: Quiz: calculating a definite integral
(01-02-2014 11:34 AM)Gerson W. Barbosa Wrote:
(01-01-2014 10:31 PM)Thomas Klemm Wrote:  That's a step into the right direction.
What's the next? What's $$\exp(t)$$?

Cheers
Thomas

e^t=1+t+1/2*t^2+1/6*t^3+...
=1-x*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 :-)
Bingo! $$\int_{0}^{1}x^{-x}=\sum_{n=1}^{\infty}n^{-n}$$
On the WP-34S the $$\sum$$ key is just beside the $$\int$$ key. You can use the same program:

10.001
$$\sum$$ 00

Cheers
Thomas
01-02-2014, 06:09 PM
Post: #27
 Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: Quiz: calculating a definite integral
(01-02-2014 08:11 AM)walter b Wrote:
(01-01-2014 10:31 PM)Thomas Klemm Wrote:  What's $$\exp(t)$$?
$$\exp(t) = e^{t}$$.
(01-02-2014 09:42 AM)Thomas Klemm Wrote:  Then what is $$e$$?
$$e=\exp(1)$$
01-02-2014, 11:39 PM (This post was last modified: 01-02-2014 11:42 PM by Bunuel66.)
Post: #28
 Bunuel66 On Vacation Posts: 29 Joined: Jan 2014
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/
01-03-2014, 01:49 AM
Post: #29
 Gerson W. Barbosa Senior Member Posts: 1,405 Joined: Dec 2013
RE: Quiz: calculating a definite integral
(01-02-2014 01:14 PM)Thomas Klemm Wrote:  On the WP-34S the $$\sum$$ key is just beside the $$\int$$ key. You can use the same program:

10.001
$$\sum$$ 00
As much as I like the wp-34s, 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
01-03-2014, 05:18 AM
Post: #30
 Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: Quiz: calculating a definite integral
(01-02-2014 11:39 PM)Bunuel66 Wrote:  iterating: $$\int_{0}^{1}x^n\log(x)^n dx=\frac{(-1)^n n!}{n^n}$$
This should be: $$\int_{0}^{1}x^n\log(x)^n dx=\frac{(-1)^n n!}{(n+1)^{n+1}}$$

Then you won't run into problems with $$0^0$$.

What did you use to create the formulas then?

Cheers
Thomas
01-03-2014, 06:08 AM (This post was last modified: 01-03-2014 06:11 AM by walter b.)
Post: #31
 walter b On Vacation Posts: 1,957 Joined: Dec 2013
RE: Quiz: calculating a definite integral
(01-03-2014 05:18 AM)Thomas Klemm Wrote:
(01-02-2014 11:39 PM)Bunuel66 Wrote:  iterating: $$\int_{0}^{1}x^n\log(x)^n dx=\frac{(-1)^n n!}{n^n}$$
This should be: $$\int_{0}^{1}x^n\log(x)^n dx=\frac{(-1)^n n!}{(n+1)^{n+1}}$$

Then you won't run into problems with $$0^0$$.

'This should be' ?? Christmas is over. Both formulas don't mean the same, there can be only one correct.

d:-?
01-03-2014, 07:03 AM
Post: #32
 Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: Quiz: calculating a definite integral
(01-03-2014 06:08 AM)walter b Wrote:  there can be only one correct.
Use integration by parts to get: $$\int_{0}^{1}x^n\log(x)^p dx=-\frac{p}{n+1}\int_{0}^{1}x^n\log(x)^{p-1}dx$$
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.
01-03-2014, 07:20 AM
Post: #33
 Ángel Martin Senior Member Posts: 1,251 Joined: Dec 2013
RE: Quiz: calculating a definite integral
(12-31-2013 06:01 PM)Thomas Klemm Wrote:  They are so cool!

Yes they are, but who's got one of those? Balances, on the other hand, are much more frequently available - even a fixture in every household; so what about cutting out the plot and weighing it on a nice density-known printing paper?

;-)
01-03-2014, 11:42 AM
Post: #34
 Bunuel66 On Vacation Posts: 29 Joined: Jan 2014
RE: Quiz: calculating a definite integral
(01-03-2014 07:03 AM)Thomas Klemm Wrote:
(01-03-2014 06:08 AM)walter b Wrote:  there can be only one correct.
Use integration by parts to get: $$\int_{0}^{1}x^n\log(x)^p dx=-\frac{p}{n+1}\int_{0}^{1}x^n\log(x)^{p-1}dx$$
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.

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
01-03-2014, 12:47 PM
Post: #35
 Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: Quiz: calculating a definite integral
(01-03-2014 11:42 AM)Bunuel66 Wrote:  From a strict mathematical standpoint, I'm a bit puzzled as the Taylor expansion used seems a bit unjustified
This is not the Taylor-series expansion:
$x^{-x}= 1-x log(x)+\frac{1}{2} x^2 log^2(x)-\frac{1}{6} x^3 log^3(x)+\frac{1}{24} x^4 log^4(x)-\frac{1}{120} x^5 log^5(x)+\frac{1}{720} x^6 log^6(x)+O(x^7)$
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
01-03-2014, 01:29 PM
Post: #36
 Bunuel66 On Vacation Posts: 29 Joined: Jan 2014
RE: Quiz: calculating a definite integral
(01-03-2014 12:47 PM)Thomas Klemm Wrote:  [quote='Bunuel66' pid='2007' dateline='1388749354']
From a strict mathematical standpoint, I'm a bit puzzled as the Taylor expansion used seems a bit unjustified

This is not the Taylor-series expansion:
$x^{-x}= 1-x log(x)+\frac{1}{2} x^2 log^2(x)-\frac{1}{6} x^3 log^3(x)+\frac{1}{24} x^4 log^4(x)-\frac{1}{120} x^5 log^5(x)+\frac{1}{720} x^6 log^6(x)+O(x^7)$
The coefficients are not constant but dependent on $$x$$.
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
01-03-2014, 02:50 PM
Post: #37
 Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: Quiz: calculating a definite integral
(01-03-2014 01:29 PM)Bunuel66 Wrote:  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.
We start with $$x^{-x}=\exp(-x\log(x))$$, substitute $$u=-x\log(x)$$, use the Taylor-series 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?
01-03-2014, 05:38 PM (This post was last modified: 01-03-2014 06:21 PM by Bunuel66.)
Post: #38
 Bunuel66 On Vacation Posts: 29 Joined: Jan 2014
RE: Quiz: calculating a definite integral
Quote:We start with $$x^{-x}=\exp(-x\log(x))$$, substitute $$u=-x\log(x)$$, use the Taylor-series 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?

Well, let's consider the very same approach with the function exp(-1/x). Using a similar schema we end up with a serie 1-1/x+1/2!x²-1/3!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
01-03-2014, 08:03 PM
Post: #39
 W_Max Junior Member Posts: 21 Joined: Dec 2013
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
01-03-2014, 08:39 PM
Post: #40
 Thomas Klemm Senior Member Posts: 1,447 Joined: Dec 2013
RE: Quiz: calculating a definite integral
(01-03-2014 05:38 PM)Bunuel66 Wrote:  This seems to show that having the equality is not enough for keeping it directly after integrating.
The problem I see is that $$u=-\frac{1}{x}$$ is not defined for $$x=0$$. The Taylor-series of $$\exp(u)$$ is not defined for $$u=-\infty$$.

Cheers
Thomas
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