(Spoilers!) Comments and discussion on Valentin's 5th "Then and Now" - Roots

[PeterP]

Ok, short on time and once I saw the factorial I knew that the HP41 was going to be highly limited anyways. But here is how far I have gotten so far:
From the graph shared it is clear that any zero would be between ]0,1[. So to get rid of the logarithm, I made x->10^-y, y in the interval ]0, infinity[

For the denominator D i just noted that zeta(k+1) very quickly goes to 1 (k = 10 or so) within the accuracy of the hp41 or even the 71 I would guess. D is dominated by the factorial (which leaves the 10 digit accuracy of the HP41 past 13! (or k = 12)

Which makes R(y) = 1 + sum [(-y)^k/D] which is an alternating sum, suggesting a pairwise summation (given that we know it converges).

After a little bit of arithmetic (in this forum I'm ashamed to call it even this as most people would have been able to do this - and all of this, and probably have - in their head), I got the following:

R(y) = 1 + sum{n=1,infinity} [y^(2n-1) * (y*(2n-1)*Zeta(2n) - (2n)^2*Zeta(2n+1)) / [(2n-1)* (2n) * (2n)! * Zeta(2n) * Zeta(2n+1)

Where we shall call the new Denominator D'. It is dominated by (2n)!, with the multiplication of the two Zeta functions going even faster to 1. We also see that the HP41 could calculate terms only until about n = 4 for the denominator, unless y is larger and we do very fancy pairwise division. And even then...

so
for n = 1 the term is: y * [y*1*Zeta(2) - (2*1)^2*Zeta(3)] / D'
for n = 2 the term is: y^3 * [y*3*Zeta(4) - (2*2)^2*Zeta(5)] / D'
for n = 3 the term is: y^5 * [y*5*Zeta(6) - (2*3)^2*Zeta(3)] / D'
etc

only for large n relative to Y can the summation pairs get negative:
y(2n-1)Zeta(2n) - (2n)^2Zeta(2n+1)

If for simplicity and a rough aproxmiation we set the two Zeta functions to be the same within the accuracy of the calculator we get
y(2n-1) < (2n)^2

Which, for large n gets close to y < 2n

Given the accuracy of the HP41, I am limited to how far I can take n (because of the factorial in D'), something around 5 or maybe 6 before substantial trickery has to take place.
However, y < 12 does not give negative numbers meaning there is no zero between y = 1 (or x = 10^-1) and y = 12 (x = 10^-12).

I also suspect from looking at the graph and the rapid rise in size of D' that any values of R(y) are very small indeed, maybe even too small for the 10 digit accuracy of the calc. One would have to find a clever way to scale the whole calculation, like many of you came up with in the first problem which dramatically increased the accuracy.

Alas, I have run out of both time and ideas for the moment, so I thought I'd share it. So far, my code consisted of paper and pencil and looking up the values for Zeta (its actually a reMarkable2 with the pen set to pencil)

Would love to hear any corrections (likely many) and suggestions (if this makes any sense at all).