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Product function (Π) & Casios - Printable Version

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Product function (Π) & Casios - Matt Agajanian - 01-26-2024 07:39 AM

Hi all.

While writing my double factorial program for the 67/97, I came across the product function, Π. This had me wondering. The function is present on the 115 series, but not the 991 calcs.

Maybe it’s just me, but I consider the 991 series to be a more sophisticated lineup than the 115 series. I would even consider the 991EX Classwiz to be an upgrade from any of the 115 models. As such, I would have thought that the product function would be present on the Classwiz.

So, any idea why it’s this way?

Thanks


RE: Product function (Π) & Casios - lrdheat - 01-26-2024 11:18 PM

Easy work around…Use the summation operator for the logs of the function in question, and then take the anti log…


RE: Product function (Π) & Casios - Gerson W. Barbosa - 01-27-2024 03:30 AM

See the 991DEX (DE = Deutschland) version:

https://www.amazon.co.uk/Classico-Scientific-Calculator-Natural-language/dp/B00VB2ISDM


RE: Product function (Π) & Casios - lrdheat - 01-27-2024 03:38 AM

An example to show how it works is the 69! problem. 10^(Summation of log x) from x=1 to 69 produces 1.711224528 *10^98 which is identical to 69!


RE: Product function (Π) & Casios - lrdheat - 01-27-2024 03:44 AM

For answers >10^100, just do summation of, say, log x from 1 to 200. You get 374.8968886 10^.8968886=7.886578674 Final answer is 7.886578674*10^374


RE: Product function (Π) & Casios - lrdheat - 01-27-2024 03:45 AM

Same method applies to any function that you wish to do the product operation on…


RE: Product function (Π) & Casios - Paul Dale - 01-27-2024 04:15 AM

Get yourself a WP 34S which has a product function.
Better still use the summation function because it performs a Kahan sum which will be more accurate.


Pauli


RE: Product function (Π) & Casios - Matt Agajanian - 01-27-2024 04:19 AM

(01-26-2024 11:18 PM)lrdheat Wrote:  Easy work around…Use the summation operator for the logs of the function in question, and then take the anti log…

Brilliant! Thanks!