How to find variables in Julia using Linear Regression for Interpolation method?

Question

There is an unknown function , and there are unknown coefficients k, l. Task is to estimate k, l using linear regression, through the data table.

-2.0  1.719334581463762
-1.0  1.900158577875515
0.0   2.1
1.0   2.3208589279588603
2.0   2.5649457921363568

Till now mathematically I did like, taking logarithm on both sides

Then using the data table, 5 equations will be formed

Now apply the linear regressor, to this logarithm-transformed data, to estimate the coefficients k and l.

I have built a linear regresor,

using DataFrames, GLM

function LinearRegression(X)
    x = X[:,1]
    y = X[:,2]
    data = DataFrame(y = y, x = x)
    reg = lm(@formula(y ~ x), data)
    return coef(reg)[2], coef(reg)[1]
end

Any solution to how to find l and k values using this technique?

Answer 1

You're almost there, but I think you have a misconception mathematically in your code. You are right that taking the log of f(x) makes this essentially a linear fit (of form y = mx + b) but you haven't told the code that, i.e. your LinearRegression function should read:

function LinearRegression(X)
    x = X[:,1]
    y = X[:,2]
    data = DataFrame(y = log.(y), x = x)
    reg = lm(@formula(y ~ x), data)
    return coef(reg)[2], coef(reg)[1]
end

Note that I have written y = log.(y) to match the formula as otherwise you are fitting a line to exponential data. We don't take the log of x because it has negative values. Your function will then return the correct coefficients l and log(k) (so if you want just k itself you need to take the exponential) -- see this plot as proof that it fits the data perfectly!

Answer 2

You need to convert the intercept with exp and the slope keeps as it is.

using Statistics #mean

#Data
X = [-2.0  1.719334581463762
     -1.0  1.900158577875515
     0.0   2.1
     1.0   2.3208589279588603
     2.0   2.5649457921363568]

x = X[:,1]
y = X[:,2]
yl = log.(y)

#Get a and b for: log.(y) = a + b*x
b = x \ yl
a = mean(yl) - b * mean(x)

l = b
#0.10000000000000005
k = exp(a)
#2.1

k*exp.(l.*x)
#5-element Vector{Float64}:
# 1.719334581463762
# 1.900158577875515
# 2.1
# 2.3208589279588603
# 2.5649457921363568