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Little explorations with HP calculators (no Prime)
03-21-2017, 10:40 PM (This post was last modified: 03-21-2017 10:42 PM by pier4r.)
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RE: Little explorations with the HP calculators
So I got to another problem and I'm stuck.

Quote:Brilliant.org
Let

\[A= \sum_{n=1}^{2014}{n} ~~~\text{and} ~~~ B= \sum_{n=1}^{2014}{n^3}.\]

Find the value of \(\log_{\sqrt A} {B}\)

Now, I know the closed formulas for the two summations, but I want to solve it with the calculator as much as I can. (one is "n(n+1)/2" the other is the square of the previous value, the more slightly complicated one is the summation of the term "n^2" but this is not the case )

I found out about the summation symbol (see hp50g AUR, almost at the end of the chapter 3. In short 'n' 1 2014 'n' <right shift, sin : for the summation>. Also proper flags have to be set) and I can compute the first sum to 2029105 .

For the second sum, I can compute it too, but I end up with a floating point number (4.11726710102e12). This is further aggravated if I want to push a bit the summation of n^3 until, say 5000 (ending with 1.5631[...]e14), and so on.

I remember that the hp50g is able to print out long numbers, I tried so search on internet a bit and I found no built in functions. I found a post mentioning the library longFloat (how much great work on for those little devices!) but before trying to use it I would like to ask here, where there are a lot of experienced people compared to me.

Am I wrong when I remember that the hp50g can, by default, print out abitrary precision numbers?

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RE: Little explorations with the HP calculators - pier4r - 03-21-2017 10:40 PM



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