Approximating logarithm using harmonic mean

Here is a function to approximate log10(x+1) for (x+1) < ~1.2:

a = 1.097
b = 0.085
c = 2.31

ans = 1 / (a - b*x + c/x)

It should look like that:

It works by adjusting harmonic mean to match log10, but the problem is in values of a, b, c.

The question is how to get just right a, b and c and how to make better approximation.

I made this code that can give a pretty good approximation for a, b, c, but my code wasn't able to make it any better.

import numpy as np

a = 1
b = 0.01
c = 2

def mlg(t):
    x = t
    if t == 0:
        x = 0.00000001
    x2 = x*x
    o = a - (b * x) + (c / x)
    return 1/o

def mlg0(t):
    x = t
    if t == 0:
        x = 0.00000001
    x2 = x*x
    o = a - (b * x) + (c / x)
    return o

for i in range(9000):
    n1 = np.random.uniform(0,1.19,1000)
    for i in range(1000):
        n = n1[i]
        o = np.log10(n+1)
        u = mlg(n) - o
        e = u ** 2

        de_da = 0 - 2 * (u) / (mlg0(n) ** 2)
        de_db = de_da * n 
        de_dc = de_da / n

        a -= de_da * 0.00001
        b -= de_db * 0.00001
        c -= de_dc * 0.00001

print(a,b,c)

How could the code be changed to generate better values?

I've used a method alike back propagation in NN, but it could not give me values any better.

Here is how the error is calculated:

Here are two approaches.

Method 1: series expansion in x (better for negative and positive x)

Method 2: fit the curve that passes through 3 points (here, x=0, ½ and 1)

Method 1.

If you expand them by Taylor series as powers of x then

Equating coefficients of x, x^2 and x^3 gives

In code:

import math
import numpy as np
import matplotlib.pyplot as plt

c = math.log( 10 )
a = c / 2
b = c / 12

print( "a, b, c = ", a, b, c )


x = np.linspace( -0.2, 0.2, 50 )
y = np.log10( 1 + x )
fit = x / ( a * x - b * x ** 2 + c )

plt.plot( x, y  , 'b-', label="Original" )
plt.plot( x, fit, 'ro', label="Fit"      )
plt.legend()
plt.show()

Output:

a, b, c =  1.151292546497023 0.19188209108283716 2.302585092994046

Method 2.

Fit to three points. Here we require

If we require this to fit at x=0, ½ and 1 we get (including L’Hopital’s rule for the limit at x=0)

This time I have used your interval x in [0,1] to plot the fit

import math
import numpy as np
import matplotlib.pyplot as plt

c = math.log( 10 )
a = c * ( 2/math.log(1.5) - 1/math.log(2) - 3 )
b = 2 * c * ( 1/math.log(1.5) - 1/math.log(2) - 1 )

print( "a, b, c = ", a, b, c )


x = np.linspace( 0.0, 1.0, 50 )
y = np.log10( 1 + x )
fit = x / ( a * x - b * x ** 2 + c )

plt.plot( x, y  , 'b-', label="Original" )
plt.plot( x, fit, 'ro', label="Fit"      )
plt.legend()
plt.show()

Output:

a, b, c =  1.1280638006656465 0.1087207987723298 2.302585092994046