Consider this definition of a sequence:

• What's the limit?
  
• Use your favorite calculator to find it.
  

Cheers

Thomas

Edited: 16 May 2013, 6:26 p.m.

Very surprising result!

Here is my first quick HP48SX RPL Solution. Now I have to think about it ;-)

<< DUP2 1500 ROT / NEG 815 + SWAP / NEG 108 + >> 'XNEXT' STO

Put x0 and x1 on the stack. Then press 'XNEXT' each iteration.

x0  =  4

x1  =  4.25

x2  =  4.470588235

x3  =  4.644736835

x4  =  4.770538091

x5  =  4.855697516

x6  =  4.910781705

x7  =  4.944198104

x8  =  4.939880369

x9  =  4.43188497

x10 = -7.379556398

x11 = 172.57617998

x12 = 102.09962458

x13 = 100.102731135

x14 = 100.005128552

x15 = 100.000256332

x16 = 100.000012814

x17 = 100.000000641

x18 = 100.000000032

x19 = 100.000000002

x20 = 100

x21 = 100

:

Edited: 17 May 2013, 7:33 p.m. after one or more responses were posted

To avoid potential problem of rounding, I use the 'exact mode' of the 50G :

'F(XP,X)=108-(815-(1500/XP))/X' DEFINE

<< 

   4 '17/4'

   ROT 2 SWAP START

    SWAP OVER F EVAL 

   NEXT

   NIP DUP ->NUM

>>

'Calc' STO

21 Calc

'1192108586037617

/238423809278164'

~ 4.99995612707.

100 Calc

'19721522630525295135293987198350753758826715569420223551166026427884564

/3944304526105059027058900515174297154031550406109997834487745707081313'

~ 4.9999999999

But 

200 Calc 

~ 5.0000...2

400 Calc

~ 5.00000...1

In approx mode I got the same 'false' results of Michael

Edited: 17 May 2013, 9:07 a.m.

Quote:

---

PS: Find the solution in a paper of Prof. W. Kahan.

---

Well, this is some of the most interesting material I've read in a while. Thanks!

I've always been fascinated by the field of numerical analysis, and by one of its top mathematicians...the famous Dr. William Kahan. There is a 1997 DDJ interview with Dr. Kahan that is a short but enjoyable read.

Since I had my Casio fx-9860GII sitting on my desk (recommended by members of this forum for its backlight), I thought I would give it a try.

Interestingly, it gave different results if I entered the initial input value of 4.25 as a decimal, or if I entered it as a fraction. Otherwise the program was identical.

If I specify the initial variable as a decimal: 4.25

x0  =   4

x1  =   4.25

x2  =   4.470588235

x3  =   4.644736842

x4  =   4.770538244

x5  =   4.855700711

x6  =   4.910847469

x7  =   4.945536789

x8  =   4.966950156

x9  =   4.979795569

x10 =   4.982956917

x11 =   4.891981178

x12 =   2.935426196

x13 = -65.18635467

x14 = 112.6635673

x15 = 100.5618284

x16 = 100.0279291

x17 = 100.0013959

If, however I input the initial variable as a fraction: 4/1/4

x0  =   4

x1  =   4/1/4

x2  =   4/8/17

x3  =   4/49/76

x4  =   4/272/353

x5  =   4.855700713

x6  =   4.910847499

x7  =   4.945537404

x8  =   4.966962581

x9  =   4.980045686

x10 =   4.987979147

x11 =   4.992764238

x12 =   4.995534710

x13 =   4.994965516

x14 =   4.949870539

x15 =   4.017961989

x16 = -19.41827146

x17 = 130.7454029

x18 = 101.1756925

x19 = 100.0580991

x20 = 100.0029032

Quote:

---

To avoid potential problem of rounding, I use the 'exact mode' of the 50G

---

Great idea! The recurrence has been set up to collapse under finite precision, no matter the number of digit. Even the wp34s in double precision eventually fails on that one:

ADDR  OPCODE     MNEMONIC

0001: 6264       LBL A

0002: 0009       4

0003: 010c       FILL

0004: 020b       1/x

0005: 0301       +

0006: a70f       # 015

0007: 9102       SDL 002

0008: 0000       ENTER[^]

0009: 0283       [degree]F[->][degree]C

000a: 0204       IP

000b: 3165       x[<->] Y

000c: 2e67       RCL/ T

000d: 0302       -

000e: 2e65       RCL/ Y

000f: 0003       +/-

0010: a76c       # 108

0011: 0301       +

0012: 013b       STOP

0013: 570d       BACK 013

0014: 013a       END

Keystroke     Display

A          4.47058823529

R/S        4.64473684211

R/S        4.77053824363

R/S        4.85570071259

R/S        4.91084749908

R/S        4.94553740412

R/S        4.96696258176

R/S        4.98004570136

R/S        4.98004570136

R/S        4.99277028806

R/S        4.99565589151

R/S        4.99739126838

R/S        4.99843394394

R/S        4.99906007197

R/S        4.99943593715

R/S        4.99966152411

R/S        4.99979690075   

R/S        4.99987813614

R/S        4.99992689282

R/S        4.99995639343

R/S        4.99997900336

R/S        5.00009075297 

R/S        5.00212143356

R/S        5.04259444308

R/S        5.84480234426

R/S        19.4539639556

R/S        79.2983058468

R/S        98.6946953822

R/S        99.9338716019

R/S        99.9966913925

R/S        99.9998345642

R/S        99.9999917282

R/S        99.9999995864

R/S        99.9999999793

R/S        99.9999999990

R/S        99.9999999999

R/S        100.000000000

R/S        100.000000000

Could someone verify the N=6 case in the HP-97 column on page 2 of Kahan's paper? I don't have a 67/97 but neither my 19C nor 15C produce quite that value, though they agree with each other. They also agree with (most of) the succeeding values so I'm tempted to think it's a typo.

I'm also getting a discrepancy for N=9, but I'm willing to chalk this one to rounding/truncating.

I realize the point of the exercise is to demonstrate getting incorrect results, but I thought the 67/97 did arithmetic the same as the 19C/15C.

Looking at the formula you can see that the limit is 100. Excel confirms that!

If you change the first two values you should still get 100 as the limit.

I found an exception!! If the fist two values are 5, you get 5 as the limit.

Namir

Edited: 18 May 2013, 1:45 a.m.

And 3, 3 yields 3. A plot of the function obtained by replacing all three x-values in the recurrence with the symbol x shows zeros at 3, 5, and 100.

Hello Calc-Friends,

If you suppose existing limits, then you can set X=X_n=X_n-1=X_n-2 and you get an third grade equation which can be written as:

X^3 - 108X^2 + 815X -1500 = 0

and the hp 50 can this easily factorize to

(X-100)(X-5)(X-3) = 0.

There you have the three fixpoints. The recurrence is very unstable because of the difference.

greetings

peaceglue

Edited: 18 May 2013, 3:59 a.m.

Quote:

---

Looking at the formula you can see that the limit is 100. Excel confirms that!

---

What you probably mean is that 100 is a fixed point. But that doesn't make it the limit of the sequence. Just have a look at Gilles results using the HP-50g in exact mode.

Quote:

---

If the fist two values are 5, you get 5 as the limit.

---

The same happens for the given initial values. But the path starting from there is very prone to rounding errors. As soon as you miss the exact path only a little the sequence converges to 100.

Cheers

Thomas

PS: All the math is in Kahan's paper.

Quote:

---

The recurrence is very unstable because of the difference.

---

The recurrence isn't unstable per se as 100 is an attractive fixed point. However this specific sequence is prone to rounding errors as it converges to a hyperbolic fixed point (saddle point).

Cheers

Thomas

Very good analysis!!!

:-)

Namir

Yes, it's a typo. Both the emulator RPN-67 Pro (with "Enhanced Calculator" set to Off) and a real HP-67 return 5.745 for N=6 and -18.74 for N=9.

Good catch! I can confirm your observations using the HP-15C as I get:

N     xN        PREFIX

6     5.745     5745352800

9     -18.74    1873570590

Kind regards

Thomas

Gerson, a good example of compactness versus readability. ;)

Does the series move away from 5 as soon as the value goes beyond that threshold? It seems to converge versus 5 as long as the values stay below 5.

Very interesting stuff to me. I've heard about in theory but never really touched it. These days I am diving into parallel-computing and CUDA it would help a lot.

The Wikipedia article about floating point has some useful links and google ;-)

- Handbook of Floating-Point Arithmetic (Jean-Michel Muller, et.al)

- Elementary Functions - Algorithms and Implementation (Jean-Michel Muller)

- What Every Computer Scientist Should Know About Floating-Point Arithmetic (David Goldberg)

- Pitfalls in Computation, or why a Math Book isn't Enough (George E. Forsythe)

- ...

PS: Here is another one:

 x(n+1) = n * x(n) - 1 with x(0) = e-1  <- this was wrong. Sorry!

x(0) = e - 1

x(n) = n * x(n-1) - 1

What's the limit? ;-)

Edited: 20 May 2013, 5:10 p.m. after one or more responses were posted

It doesn't move away from 5 once the value is greater than 5. For many inputs e.g. 1 and 3, it moves away immediately. The solution at 5 has to be approached along a very specific path to be found and any deviation will result in the solution at 100. The initial values were carefully chosen to be on the path. Even if you are on the path to 5, convergence is slow.

A slightly shorter program for this is:

	001:	LBL A

	002:	# 015

	003:	SDL 002

	004:	RCL/ Z

	005:	8

	006:	1

	007:	5

	008:	-

	009:	RCL/ Y

	010:	# 108

	011:	+

	012:	R/S

	013:	GTO A

However, it is missing the inspired use of ^oF->^oC command.

- Pauli

Compared to this?

STO 0

x<>y

RCL 3

x<>y

/

RCL- 2

RCL/ 0

RCL+ 1

108  STO 1

815  STO 2

1500 STO 3

4 ENTER

4.25 R/S

R/S

R/S

...

Cheers

Thomas

Edited: 18 May 2013, 8:53 a.m.

Quote:

---

Even if you are on the path to 5, convergence is slow.

---

It's geometric with the factor 3/5. What do you consider slow?

Kind regards

Thomas

The convergence certainly isn't fast.

- Pauli

From Gilles post for N = 100:

Quote:

---

19721522630525295135293987198350753758826715569420223551166026427884564
/3944304526105059027058900515174297154031550406109997834487745707081313

---

4.9999999999999999999998693362...

But since (3/5)⁵⁵ < 10^-12 I assume about 55 elements should be enough. I agree with you that it's not super fast but still much better than the brute force summation of 1/k² in a recent thread. We could also use convergence acceleration to improve on the result.

Kind regards

Thomas

Hello Michael,

there's no limit:

the first values are:

n=0  e-1

  1  -1

  2  -2

  3  -5 

  4  -16

  5  -65

  6  -326

and so on....where is the problem?

Greetings
peacecalc

Quote:

---

an third grade equation

---

I think you mean "cubic"

At least I wasn't solving these when I was in third grade!

Quote:

---

Quote:

---

an third grade equation

---

I think you mean "cubic"

---

d:-?

Quartic; quintic.

Quote:

---

How'd you call an equation going up to x^4 ? And to x^5 ?

---

Biquadratic, quartic. Quintic.

I think the point here is the use of "grade" for "degree". The former shouldn't be wrong, but it appears the latter is more common in English. I don't know what it's like in German, but in Portuguese we have only one word for both words: grau. Interestingly ensino de terceiro grau means "college-level education". Also, colégio means "high-school", not college. These are often mistranslated. In this example, for instance, "And I only got a third grade education" has been translated as "E eu só tenho uma educação de terceiro grau" (And I have only a college-level education), which causes perplexity: "Isn't that enough?" (In this corner of the world, it still is :-).

Regards,

Gerson.

Quote:

---

I think the point here is the use of "grade" for "degree".

---

• rank
  
• rate
  
• grade
  
• level
  
• order
  
• stage
  
• standard
  
• degree
  

Thomas

Quote:

---

a good example of compactness versus readability.

---

A01 LBL A			B01 LBL B

A02 163				B02 815		

A03 5				B03 1500

A04 *				B04 R^

A05 ENTER			B05 /

A06 ->oF                        B06 +/-

A07 SQRT			B07 +

A08 ASINH			B08 x<>y

A09 COSH			B09 /

A10 x2            		B10 +/-

A11 IP				B11 108

A12 R^				B12 + 

A13 /				B13 STOP

A14 +/-                         B14 GTO B

A15 + 

A16 x<>y

A17 /

A18 +/-

A19 108

A20 +

A21 STOP

A22 GTO A

CK=09CA 033.0 bytes             CK=0A7B 037.0 bytes

4 ENTER 4.25 XEQ A              4 ENTER 4.25 XEQ B      4.470588235 

                      	R/S                             4.644736835

                      	R/S				4.770538091

		      	R/S				4.855697516

		      	R/S				4.910781705

		      	R/S  				4.944198104

		      	R/S				4.939880369

                        R/S                             4.43188497

                      	R/S				-7.379556398

                      	R/S				172.57617998

                	R/S				102.09962458

			R/S				100.102731135

			R/S				100.005128552

			R/S				100.000256332

			R/S				100.000012814

			R/S				100.000000641

			R/S				100.000000032

			R/S				100.000000002

			R/S				100

			R/S				100

^* These can be further optmized, of course. We can save at least 5 bytes in program B, for example.

Edited: 18 May 2013, 4:26 p.m.

Quote:

---

To avoid potential problem of rounding, I use the 'exact mode' of the 50G :
...

---

The 50g has an exact mode? Interesting. The only follow up to my 48SX was the 49g. I dislike the hardware, put him into the drawer and go back to my 48SX. ;-) Let's see what HP offers next.

Today I stick the keyboard stickers to build up the WP 34S. Now I have to solder some things and so on...

But using Free42 the sequence first approaches 0 but then all of sudden turns negative and seems to diverge.

Kind regards

Thomas

Quote:

---

The 50g has an exact mode? Interesting. The only follow up to my 48SX was the 49g.

---

Yes, and so does the 49g.

Hello Thomas,

now I understand (I hope), it is a problem of the calcs and not
of mathematics.

Greetings
peacecalc

Quote:

---

Use your favorite calculator to find it.

---

Screenshot:




(Note: If you are trying to reproduce this, I had Fibonacci in U1 and Kahan's sequence in U2)

Trying the same using the 40GS it takes a few minutes to come up with "UNDEF" as the value of the 3rd item in the sequence. It then takes another few minutes to come up with "UNDEF" for item 4 and so on. :-)

A suggestion for TW:- the soft key menu when entering a formula should be:

"(N-2) (N-1) (N) U1 CANCL ON" and not "(N-2) (N-1) N U1 CANCL ON" i.e. put brackets around the N when pressed. (These will have no effect if N is what is required but saves typing when U1(N) is required.)

I was lazy and just copied the definition of the sequence into WolframAlpha. But we can solve the problem without that. A nested series for e can be obtained by factoring out 1/n:

In this form it's obvious what the sequence does with e:
peeling off one factor after the other. What's left in each step is something like: 1/n * (1 + ...) which will tend to 0 as n goes to infinity.

The problem here is that we can't start with the exact value of e. Therefore I assume even the HP-50g in exact mode will run into problems.

I'm not sure wether I understand your statement correctly but I think the problem is more of how we interpret the results of our calculators.

Kind regards

Thomas

Hello Thomas,

I'm not quite sure if we are talking about the same problem, I refer to

x(0) = e - 1

x(n) = n * x(n-1) - 1

in the original post #30 from Michael Kathke (with correction).

In my opinion this recurrence can be written in a closed forn as:

X(N+1) = - SUMME(J;1;N;GAMMA(N)/GAMMA(N-J+1));

It reproduces the values I gave above, and there is no limit because every term of the sum is greater 1 and for N to infinity you get a lot of summands. The starting value is irrelevant because, if you calculate X(1) the starting value is multiplied with zero.

Please give me a hint, where my misunderstanding is?

Greetings
peacecalc

I'd like to rewrite Michael Kathke's first (wrong) definition of the sequence by replacing n by n-1:

x(0) = e-1

x(n) = (n-1) * x(n-1) - 1

Your listing of the sequence is correct. WolframAlpha gave me a closed formula for your expression:

 - SUM[GAMMA(N)/GAMMA(N-J+1),{J,1,N}] =

-(e Gamma(N+1, 1)-1)/N

Table[-(e Gamma(n+1, 1)-1)/n, {n, 1, 6}]

x(0) = e - 1

x(n) = n * x(n-1) - 1

Hopefully I could clarify things a little.

Thomas

Apparently the 39gII is not among my favorite calculators and probably never will be. Seriously, are these calculators only capable to solve simple problems that are expected in the context of education? IMHO that sounds really bad.

Kind regards

Thomas

Bruce,

What happened to your calculator? The lower right part of it on your picture looks quite bent :-( Or is it the camera?

@Thomas: I'd vote for "obviously" instead of "apparently".

d;-?

Hello Thomas,

thank you very much for sharing your knowledge and time with me.

Now I can see clearly now (@Johnny Nash ;-)). It's my mistake to suppose that n can be zero (which isn't possible, because the first element should be then X_(-1) and not X_0). Oh boy, I should first read exactly before I'm steeling other persons time and further investments.

Greetings
peacecalc

Edited: 20 May 2013, 4:11 a.m.

The 49G+ and 50G hardware is far better than the 49G (keyboard, screen speed etc.)

The 49G hardware was ...:(

When I try sometimes tu use my old 48SX I love the keyboard touch (but not the agencement except the big ENTER), but it's so slow and the screen is poor...

Edited: 20 May 2013, 6:59 a.m.

Here is what I get with my 39GII and how to to

1/push APPS and choose 'Suite' (in french)

2/ Enter :

3/ Push PLOT if you want to see the graph :

4/ Push NUM if you want to see the value

I think it's well done but it is not stable and I also experimented some bad crashs both on real calc and emulator...

As you can see, the numeric output is quite the same than many others HP Calcs

Edited: 20 May 2013, 8:14 a.m.

Hi Bruce !

Your HP39GII don't have to crash with this but there is a bug that I saw in the picture ...

U2(N)=...U2(N)...

You may use
U2(N)= function of U2(N-1), U2(N-2)

else you have a kind of recursive call without any exit and the 39GII crash...

Edited: 20 May 2013, 10:39 a.m.

Hi Gilles,

You are exactly right - I entered the expression wrongly. (D'oh). Once corrected, the physical calc gives the same results you posted from the emulator.

However, this episode has revealed a bug: the 40GS results that I quoted were also with the wrong expression - so the older model is able to cope with a recursive use of 'U1(N)' without crashing or hanging. It also allows me to interrupt the slow calculation with On+F3 which I couldn't get to work on the 39gII.

So, I withdraw my previous suggestion to Tim around changing the "N" menu item to "(N)" (as that simply encourages idiots like me to make make idiotic mistakes) ;-) and instead suggest that any entered expression be parsed for 'U1(N)' and a warning displayed if present.

It's just lens distortion. I tried to take the picture from a more overhead position but just got a reflection of the camera in the screen.

Or just add support to break from a recursion loop (or any other situation that crashes the calc).

(I wasn't able to break from an intentional recursion loop on the emulator, and it was FAR from running out of memory.)

Yes, that's right. Thank you for the clarification and sorry for the little mess ;-).

Hello Thomas,

I only like to report that you are right with your statement:

"The problem here is that we can't start with the exact value of e. Therefore I assume even the HP-50g in exact mode will run into problems."

As long as you don't leave the exakt modus, the terms of recurrence are calculated right (output in the form: A*e - B, A and B are integers). One gets false values (after the twelfth step) if one puts ->NUM on the whole difference. If you type: A*e ->NUM B ->NUM - you get certainly the right answer: 0,0000.

Greetings
peacecalc

The TI-Nspire CAS goes off the rails fairly quickly in approximate mode. It does have a handy function for generating sequences, however.

Approximate:

seqGen(108-((815-((1500)/(u(n-2))))/(u(n-1))),n,u,{2,20},{4,4.25})

(seqGen(108-((815-((1500)/(u(n-2))))/(u(n-1))),n,u,{2,70},{4,((17)/(4))}))->Decimal

        Approximate                        Exact

x0      4.0000000000000                    4.0000000000000

x1      4.2500000000000                    4.2500000000000

x2      4.4705882352900                    4.4705882352941

x3      4.6447368420100                    4.6447368421053

x4      4.7705382415800                    4.7705382436261

x5      4.8557006697400                    4.8557007125891

x6      4.9108466171500                    4.9108474990828

x7      4.9455194518000                    4.9455374041239

x8      4.9665996618000                    4.9669625817627

x9      4.9727394954100                    4.9800457013556

x10     4.8410665463200                    4.9879794484784

x11	1.9582015026800                    4.9927702880621

x12    -149.96677335313                    4.9956558915066

x13   108.3266789440700                    4.9973912683813

x14   100.3841271269200                    4.9984339439448

x15   100.0191267431500                    4.9990600719709

x16   100.0009559707500                    4.9994359371468

x17   100.0000477925700                    4.9996615241038

x18   100.0000023894600                    4.9997969007134

x19                                        4.9998781354779

x20                                        4.9999268795046

x21                                        4.9999561270612

x22                                        4.9999736760057

x23                                        4.9999842055203

x24                                        4.9999905232822

x25                                        4.9999943139586

x26                                        4.9999965883713

x27                                        4.9999979530214

x28                                        4.9999987718123

x29                                        4.9999992630872

x30                                        4.9999995578523

x31                                        4.9999997347113

x32                                        4.9999998408268

x33                                        4.9999999044961

x34                                        4.9999999426976

x35                                        4.9999999656186

x36                                        4.9999999793712

x37                                        4.9999999876227

x38                                        4.9999999925736

x39                                        4.9999999955442

x40                                        4.9999999973265

x41                                        4.9999999983959

x42                                        4.9999999990375

x43                                        4.9999999994225

x44                                        4.9999999996535

x45                                        4.9999999997921

x46                                        4.9999999998753

x47                                        4.9999999999252

x48                                        4.9999999999551

x49                                        4.9999999999731

x50                                        4.9999999999838

x51                                        4.9999999999903

x52                                        4.9999999999942

x53                                        4.9999999999965

x54                                        4.9999999999979

x55                                        4.9999999999987

x56                                        4.9999999999992

x57                                        4.9999999999996

x58                                        4.9999999999997

x59                                        4.9999999999998

x60                                        4.9999999999999

x61                                        4.9999999999999

x62                                        5.0000000000000

x63                                        5.0000000000000

x64                                        5.0000000000000

x65                                        5.0000000000000

x66                                        5.0000000000000

x67                                        5.0000000000000

x68                                        5.0000000000000

Is it possible to define a sequence based on an other sequence? So could you define a sequence a(n) that uses u(n), u(n+1) and u(n+2) in the definition? Because then you could use Aitken's delta-squared process to get a sequence that converges much faster:

Kind regards

Thomas

PS: You really have to enter ((17)/(4)) to make it exact? That's a strange syntax.

It does look like Aitken's process will converge much more quickly. (I used, in this case, the spreadsheet feature, rather than nested sequences.) The results are:

          Exact               Aitken

x0        4.                  6.1249999997025     

x1        4.25                5.2977941178822     

x2        4.47058823529       5.0978786250754     

x3        4.64473684211       5.0341661805942     

x4        4.77053824363       5.0121668031825     

x5        4.85570071259       5.0043630620932     

x6        4.91084749908       5.0015685124445     

x7        4.94553740412       5.0005643812327     

x8        4.96696258176       5.0002031404805     

x9        4.98004570136       5.0000731259183     

x10       4.98797944848       5.000026324821      

x11       4.99277028806       5.0000094772632     

x12       4.99565589151       5.0000034127263     

x13       4.99739126838       5.0000012291514     

x14       4.99843394394       5.0000004435337     

x15       4.99906007197       5.0000001597038     

x16       4.99943593715       5.000000055426      

x17       4.9996615241        5.00000001847       

x18       4.99979690071       5.                  

...

Thanks for taking the time to create the table, the more so as it was a pain to extract the data from the calculator.

Cheers

Thomas

Can Aitken's process save the WP-34S in double precision mode?

         Double Precision  Aitken      

x0       4.                6.125        

x1       4.25              5.29779411765

x2       4.47058823529     5.09787862513

x3       4.64473684211     5.03416618064

x4       4.77053824363     5.01216680321

x5       4.85570071259     5.00436306211

x6       4.91084749908     5.00156851243

x7       4.94553740412     5.0005643812 

x8       4.96696258176     5.00020314054

x9       4.98004570136     5.00007312584

x10      4.98797944848     5.00002632469

x11      4.99277028806     5.00000947681

x12      4.99565589151     5.00000341164

x13      4.99739126838     5.00000122819

x14      4.99843394394     5.00000044215

x15      4.99906007197     5.00000015918

x16      4.99943593715     5.0000000575 

x17      4.99966152411     5.00000002455

x18      4.99979690075     5.00000008579

x19      4.99987813614     5.00000158881

x20      4.99992689282     5.00005319189

x21      4.99995639343     4.99995065851

x22      4.99997900336     4.99997249558

x23      5.00009075297     4.99998348415

x24      5.00212143356     4.99997099445

x25      5.04259444308     4.99234537283

x26      5.84480234426     1.83899380969

x27     19.4539639556    107.996029692  

x28     79.2983058468    100.018441722  

x29     98.6946953822    100.000046095  

x30     99.9338716019    100.000000115  

The result doesn't really surprise as the difference from the correct sequence is multiplied by 100 with each step. So if the base sequence deviates from the correct solution the accelerated sequence can't do much about it.

Let's assume that the sequence is: x_n = a + c qⁿ with |q| < 1. Then the limit is certainly a as n goes to infinity. When Aitken's process is applied to that sequence we get the correct limit a with the first step using just the first three elements of the original sequence:x₀, x₁ and x₂.

Now our correct sequence is something like: x_n = a + c ₁ q₁ⁿ + c₂ q₂ⁿ, with q₁ = 3/5, q₂ = 100 and c₂ = 0. Due to rounding errors the second factor c₂ isn't 0 anymore but a very small number. However this term gets blown up as n grows. When it gets dominant the Aitken's process uses this factor q₂ = 100 instead of q₁ = 3/5 to guess the limit. That's why we can observe an acceleration of the convergence to the limit 100 after x₂₇.

Cheers

Thomas