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+--- Thread: Quiz: calculating a definite integral (/thread-284.html)
RE: Quiz: calculating a definite integral - Bunuel66 - 01-03-2014 09:04 PM
(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
RE: Quiz: calculating a definite integral - Thomas Klemm - 01-03-2014 09:32 PM
(01-03-2014 08:03 PM)W_Max Wrote: My pocket HP30b (yes, not wp34sDid you try to calculate \(\sum_{n=1}^{10}n^{-n}\)?), using simplest rectangle method and step 0.00005 give 1.29128599414 after 3 minutes. Enjoy simple methods
Should take ~0.09s.
Cheers
Thomas
RE: Quiz: calculating a definite integral - W_Max - 01-03-2014 11:08 PM
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.
RE: Quiz: calculating a definite integral - Thomas Klemm - 01-06-2014 10:28 AM
(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
RE: Quiz: calculating a definite integral - Bunuel66 - 01-07-2014 06:12 PM
(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.
RE: Quiz: calculating a definite integral - Thomas Klemm - 01-09-2014 07:45 AM
(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
RE: Quiz: calculating a definite integral - Thomas Klemm - 01-14-2014 02:31 PM
Sophomore's dream
RE: Quiz: calculating a definite integral - walter b - 01-14-2014 02:53 PM
Nice quiz!
d:-)
RE: Quiz: calculating a definite integral - Gerson W. Barbosa - 01-16-2014 08:21 PM
(12-31-2013 01:14 PM)Thomas Klemm Wrote: It takes 2'27" to calculate this integral on a DM-15CC with FIX 9.5 seconds on my outdated iPhone 4s with HP-15C emulator by HP:
It takes 28" to do it with the RPN-15C emulator on my iPhone.
On a real HP-15C it takes probably much longer.
Code:
001- f LBL A
002- CHS
003- y^x
004- g RTN0 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 RTNCheers,
Gerson.