NEW: HP 15C Collectors Edition

[Gerson W. Barbosa]

(08-01-2023 05:15 PM)Valentin Albillo Wrote:  .

Quote:Don’t hesitate to get yourself another one [...] one went to my younger brother [...]

As Queen would say: "Big mistake". Younger brothers never deserve such a sacrifice. Nor do older brothers for that matter ...

I did that as a compensation for claiming back my original 15C ten years later. When I bought a 28S I gave the older calculator to him, for his CE studies. That was a mistake, I should have kept it and given him an 11C instead. Anyway, despite some heavy use, it still works nicely and is the one I use on a daily basis because of the very reliable keyboard.

(08-01-2023 05:15 PM)Valentin Albillo Wrote:  

Quote:[...] and one was sold to a friend with no profit,

Another mistake: you should never sell such calc to a friend, either present it as a gift or else keep it.

Yes, I know. I had announced it in a local auction site for a higher price. It turns out he was the first interested buyer and didn’t ask for any rebate, but I offered him a 50% discount just the same. Also, I had already sent him my 33c for free, because I didn’t like it.

(08-01-2023 05:15 PM)Valentin Albillo Wrote:  On another note, you may be interested to know that my new SRC#15, specifically dedicated to the HP-15C & friends (what else ?) will be posted in a few days, though on second thought perhaps I might delay it until September to try and maximize the audience, as most forum members are vacationing in July-August and the participation here seems at a low (or else this thread would be already thrice as long).

I am looking forward to it. Even though lately I haven’t been participating, I like to read the posts.

Back to the approximation, what about making

\(x=5+\frac{7}{5000000000+\left({1000000+\sqrt{1000\times5!-e^5}}\right)\sqrt{5}}\) ?

Now we have 5 5’s for an error of about 5E-25 :-)
But I will go no further as I am running out of digits. Free42 has only 34 SD.

Best regards,

Gerson.

Edited to fix a typo and to include the other part of the approximation.

\(\frac{63}{25}\left({\frac{17+3x\sqrt{x}}{7+3x\sqrt{x}}}\right)\)

Based on the following approximation by Ramanujan:

\(\frac{63}{25}\left({\frac{17+15\sqrt{5}}{7+15\sqrt{5}}}\right)\)