[VA] SRC#005- April, 1st Mean Minichallenge

[Valentin Albillo]

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Hi, Maximilian:

(04-06-2019 11:41 AM)Maximilian Hohmann Wrote:  No no no! No lack of interest. At least not from my part.

Why, thank you ! It's always reassuring to know for a fact that some people find my productions interesting.

Quote:after a quick read I was not really sure that I understood the problem but didn't want to give away my stupidity by asking silly questions ;-)

There are no such things as silly questions for math-related challenges but in any case, whenever you'd like to ask some question about some challenge of mine and you don't want people to deem your question 'silly', just sent me a PM and I'll promptly answer your question in perfect privacy.

Quote:Will try it out for myself now on the HP-71B and maybe the HP-39GII that I got in the mail today. The latter one should be able to solve the Pi-problem graphically.

Good. If you do, please post your code in this thread, I'd love to see it.

Quote:And by the way, wouldn't it be more elegant for finding the value for e.g. M 5/2, instead of taking the mean between the results M2 and M3, to fit a polynomial through all four results?

I don't understand. The key to this mini-challenge is that all four means mentioned (Harmonic, Geometric, Arithmetic, Quadratic) are but discrete cases of a continuous Generalized Mean which includes those four as particular cases. The Generalized Mean ( M_k for short) is defined as:

        M_k = \( \left( \frac{\sum_{i=1}^n a_i^k}{n} \right) ^ \frac{1}{k} \)

and for k = -1, 0, 1, 2 it gives the Harmonic, Geometric (in the limit), Arithmetic and Quadratic means, respectively (for purely cosmetic purposes I did shift k by 2 to make it 1, 2, 3 and 4 instead).

So you see, as k is a continuous parameter (so not necessarily an integer) from -Inf to +Inf, you can compute M_k directly by using the above formula, no need to "take the mean" of any two results or perform any "polynomial fit", which matter of fact wouldn't succeed as the dependence on k is in the exponents and thus no polynomial fit would ever do.

Thanks again for your interest, much appreciated, and have a nice weekend.

V.
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