I will start the pseudo code.

Notice that an x of 545 is positive and an x test value of 8756 yields a negative.

I am arguing that with these test values the difference from PNP shows if the desired x is higher or lower than the test value.

There is no reason that calculus won’t give a value when limit (equation) when PNP approaches 0. I know it takes more than that but do you agree the pseudo code will show an indication where x will fall at PNP?

Please join this post. If you don’t believe my code, counter it. There should be a pattern in the pseudo-code.

x = 545 y = 6737 PNP = 4639* y (((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2))) 545 6737 31252943 (566741960869155702888306342808481973/580236226342089968450) 566741960869155702888306342808481973/580236226342089968450 N[566741960869155702888306342808481973/580236226342089968450, 13] 9.767434971132*10^14 Sqrt[9.76743497113228416002242796713336`13.*^14] 3.125289581964*10^7 PNP - 3.125289581963931252118627996719126077423156917`13.\ 301029995663981*^7 47.18036

test second x

x = 8756 y = 6737 PNP = 4639* y (((((x^2*PNP^4 + 2*PNP^2*x^5) + x^8)/ PNP^4) - ((1 - x^2/(2*PNP))))*((PNP^2/x^2))) 8756 6737 31252943 73243982077295884748898890760446202375/74884743323939619512464 N[73243982077295884748898890760446202375/74884743323939619512464, 14] 9.7808951231166*10^14 Sqrt[9.780895123116592692523297300191834`14.*^14] 3.1274422653530*10^7 PNP - 3.127442265353046080516318622627538568244967294`14.\ 301029995663981*^7 -21479.653530