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[VA] SRC #012b - Then and Now: Root
11-18-2022, 01:44 AM
Post: #24
RE: [VA] SRC #012b - Then and Now: Root
 
Hi, Albert Chan,

Thanks for your recent additional comments, I appreciate it. However, I have a thing or two to comment back, read on ... (all highlights are mine)

Albert Chan Wrote:
Valentin Albillo Wrote:Line 2 finds the minimum absolute value for polynomials of degrees 10, 20, 30, ... until two consecutive minimum values are within 0.01, which essentially gives us the result correct to 2 digits (~0.81)

Slight error in logic.

Not at all, see below.

Albert Chan Wrote:If consecutive minimum abs both ~0.81, we cannot deduce trend apply to higher degree. (we can assume trend continues, but have to later test validity of assumption)

And I did test, it's just that I didn't want to make an already long post any longer by including unneeded data that most people won't be interested in, as they understand the scope of my articles and challenges and trust my results (which they can verify by themselves, if in doubt), but as you seem to like said data, here you are, the checks I did (which I saved but didn't post):
    Degree  Min. Magnitude
    -----------------------
      1     .66666 6666667
      2     .63245 5532034
      4     .65224 6975033
      8     .73550 1548954
     16     .76890 4440166
     32     .80252 6072477
     64     .80650 0035750
     96     .80651 362173
    128     .80651 3599285

    140     .80651 3599258
    145     .80651 3599260
    150     .80651 3599261
    160     .80651 3599261
    165     .80651 3599261
    170     .80651 3599261
    175     .80651 3599261
    180     .80651 3599261
    185     .80651 3599261
    190     .80651 3599261 
    195     .80651 3599261
    200     .80651 3599261
and I was more than satisfied that the trend continued alright and converged to the correct solution, namely .806513599261.

Albert Chan Wrote:
Valentin Albillo Wrote:Line 3 now uses this 2-digit value to compute an estimation to the minimum necessary degree to get a fully accurate result, which it then obtains and displays, correct to 12 digits.

Close, but not quite.

If this "minimum necessary degree" polynomial also gives abs ≤ 0.81, we are done.
However, if it gives bigger abs, we have to repeat again, with even higher degree polynomial.

But it didn't give "bigger abs", it gave  .806513599261, which is less than 0.81, so your comment doesn't apply at all and my statement is fully correct, not merely "close".

Now a word on the scope for my articles and challenges. In a past thread in which you took part (you posted 5 times no less) and so you surely read my posts there, I said:
    "My articles are intended as just that, articles to be published in a physical fan-made magazine [...] or else on the WWW (MoHPC) for all kinds of fans, most of them not scholars, so my articles do not have the structure nor goals of a formal peer-reviewed paper."

In other words, my goal is first and foremost to provide entertainment to the forum members and HP calc fans in general, enticing them to think about the challenge and how to use their vintage HP calc to solve it, and if additionally they learn something new and interesting (and even useful) from my productions then so much the better. That is my goal.

Yours is obviously different, seemingly centered on lecturing, posting symbolic proofs and theoretical ramblings and lots of data obtained in Xcas sessions, lua, Mathematica, the works. Good for you and for the people who like (lots of) posts like that. Not my cup of tea here.

And please leave aside topics having little or nothing to do with my present Problem 2 (the references to integration, Borwein integrals and Kahan), save that for your own threads or where it's appropriate. Thanks.

V.

  
All My Articles & other Materials here:  Valentin Albillo's HP Collection
 
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