Solving H_n = 100

The previous post includes code for solving the equation

H_n = m

i.e. finding the value ofn for which thenth harmonic number is the closest tom. It works well for small values ofm. It works for largem in the sense that the solution is very close tom, but it's not necessarily the best solution.

For example, setm = 100. The code returns

n = 15092688622113830917200248731913020965388288

and indeed for that value ofn,

H_n -100 3 * 10^-15

and that's as much as we could hope for with IEEE 754 floats.

The approximation

n = exp(m -)

is very good for large values ofm. Using Mathematica we can find the exact value ofn.

f[n_] := Log[n] + EulerGamma + 1/(2 n) - 1/(12 n^2)n = Floor[Exp[100 - EulerGamma]];N[f[n], 50]100.00000000000000000000000000000000000000000000900N[f[n - 1], 50]99.999999999999999999999999999999999999999999942747

So

n = 15092688622113788323693563264538101449859497

A similar process can find the solution to

H_n = 1000

is

n = 110611511026604935641074705584421138393028001852577373936470952377218354575172401275457597579044729873152469512963401398362087144972181770571895264066114088968182356842977823764462179821981744448731785408629116321919957856034605877855212667092287520105386027668843119590555646814038787297694678647529533718769401069269427475868793531944696435696745559289326610132208504257721469829210704462876574915362273129090049477919400226313586033

For this calculation you'll need to increase the precision from 50 digits to something like 500 digits, something more than 435 becausen is a 435-digit number.

In case you're wondering whether my function for computing harmonic numbers is accurate enough, it's actually overkill, with error O(1/120n⁴).

The post Solving H_n = 100 first appeared on John D. Cook.