Quiz: calculating a definite integral

[Bunuel66]

(01-03-2014 08:39 PM)Thomas Klemm Wrote:  

(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

Don't get the point, \(\exp(-\infty)\)=0. The serie is converging whatever the sign of x (more and more slowly as you're closing to 0-....). Then we have two expressions who provides similar values whatever the sign of x, and after integration we have a new set of expressions with one which is no more defined on one side (x<0). And as you mention, this is not exactly a Taylor serie in the sense that the sum is not using the derivatives of u(x). The problem is maybe a little bit more subtle (at least for me) than it seems ;-(...

Regards

[Thomas Klemm]

(01-03-2014 08:03 PM)W_Max Wrote:  My pocket HP30b (yes, not wp34s ), using simplest rectangle method and step 0.00005 give 1.29128599414 after 3 minutes. Enjoy simple methods

Did you try to calculate \(\sum_{n=1}^{10}n^{-n}\)?
Should take ~0.09s.

Cheers
Thomas

[W_Max]

Not exactly, but close to. I wrote simple RPN program.

0 STO4
LBL00 RCL3 INPUT +/- Y^X
STO+4 RCL1 STO+3 RCL2 RCL3 ?> GT00
RCL4 RCL1 * Stop

0.00005 STO1, 1 STO2 0 STO3

It takes about 2+ min to complete (20000 cycles or 166cycles/sec! ). As HP30b is relatively fast machine - such a 'brute force' method gives acceptable result too.

[Thomas Klemm]

(01-03-2014 09:04 PM)Bunuel66 Wrote:  Don't get the point, \(\exp(-\infty)\)=0.

The domain of \(\exp(x)\) is \(\mathbb{R}\), but \(-\infty \notin \mathbb{R}\). Thus you can not just plug \(-\infty\) into the Taylor-series of this function and expect everything works. You can calculate \(\lim_{x\to\infty}\exp(x)\) but that's not the same as \(\exp(-\infty)\). This expression is just not defined.

HTH
Thomas

[Bunuel66]

(01-06-2014 10:28 AM)Thomas Klemm Wrote:  

(01-03-2014 09:04 PM)Bunuel66 Wrote:  Don't get the point, \(\exp(-\infty)\)=0.

The domain of \(\exp(x)\) is \(\mathbb{R}\), but \(-\infty \notin \mathbb{R}\). Thus you can not just plug \(-\infty\) into the Taylor-series of this function and expect everything works. You can calculate \(\lim_{x\to\infty}\exp(x)\) but that's not the same as \(\exp(-\infty)\). This expression is just not defined.

HTH
Thomas

Could have been rewriten as a limit to be more rigorous...;-) That said the serie gives the same value than the function also for x<0. Doesn't seems to be the point. And as you mention this is not a Taylor serie strictly speaking.

Regards.

[Thomas Klemm]

(01-07-2014 06:12 PM)Bunuel66 Wrote:  Could have been rewriten as a limit to be more rigorous...;-)

Maybe these posts are helpful:

• Change of order of double limit of function sequence
• When can you switch the order of limits?
• When can we exchange order of two limits?
  

Cheers
Thomas

[Thomas Klemm]

Sophomore's dream

[walter b]

Nice quiz!

d:-)

[Gerson W. Barbosa]

(12-31-2013 01:14 PM)Thomas Klemm Wrote:  It takes 2'27" to calculate this integral on a DM-15CC with FIX 9.
It takes 28" to do it with the RPN-15C emulator on my iPhone.
On a real HP-15C it takes probably much longer.

5 seconds on my outdated iPhone 4s with HP-15C emulator by HP:

Code:

001- f LBL A
002-   CHS
003-   y^x
004- g RTN

0 ENTER 1 f Integrate --> 1.291285997 (blinking, but that's another story)

Estimated +4 hours on a real HP-15C.

As a comparison the following return the same result (no blinking, of course!) in 13.7 and 13.4 seconds, respectively, on my 30-year old HP-15C:

Code:


001- f LBL A            001- f LBL A
002-   0                002- f MATRIX 1
003-   STO 0            003-   8
004-   9                004-   STO I
005-   STO I            005- f LBL 0
006- f LBL 0            006-   RCL 1
007-   RCL I            007-   RCL+ I
008-   ENTER            008-   ENTER
009-   CHS              009-   CHS
010-   y^x              010-   y^x
011-   STO+ 0           011-   STO+ 0
012- f DSE I            012- f DSE I
013-   GTO 0            013-   GTO 0
014-   RCL 0            014-   RCL 0
015- g RTN              015- g RTN

Cheers,

Gerson.