Julia語言 利用Lagrange方法求下列 滿足f(x)=ln (x) 的數據 計算P(1.5) 之值=??

Julia語言 利用Lagrange方法求下列
滿足f(x)=ln (x) 的數據
(1)
x = ([1.0,2.0,3.0,4.0])
f (x)= ([0.0,0.693,1.099,1.386])

(2)
x=[1.0 , 1.2 , 1.4 , 2.0]
f(x)=[0.0 , 0.182 , 0.336 , 0.693]

(3)
x=[1.0 , 1.2 , 1.7 , 2.0  , 2.2 ,  2.7  , 3.0 , 3.2  , 3.7  , 4.0]
f(x)=[0.0 , 0.182 , 0.531 , 0.693  , 0.788  ,  0.993 , 1.099 , 1.163 , 1.308 , 1.386]


計算P(1.5) 之值=??

程式
using Printf

#This function returns another function, which is the Lagrange Interpolant of the values xvals and yvals.
function LagrangeInterpolantGenerator(xvals,yvals)
    function LagrangeInterpolant(x)
        numvalstoevaluate = length(x)
        numvalstoevaluate == 1 ? output = 0 : output = zeros(numvalstoevaluate)
        for k = 1:numvalstoevaluate
            N = length(xvals)
            LagrangePolynomials = ones(N)
            for i in 1:N 
                for j in [1:i-1;i+1:N]     #Surprisingly, this works even in the i=1 and i=N cases.
                    LagrangePolynomials[i] = LagrangePolynomials[i].*(x[k]-xvals[j])./(xvals[i]-xvals[j])
                end
            end
            numvalstoevaluate == 1 ? output = sum(LagrangePolynomials.*yvals) : output[k] = sum(LagrangePolynomials.*yvals)
        end
        return output
    end
    return LagrangeInterpolant
end

#Examples
x= [ [1.0 ,2.0 , 3.0 , 4.0] ,
     [1.0 , 1.2 , 1.4 , 2.0] ,
     [1.0 , 1.2 , 1.7 , 2.0  , 2.2 ,  2.7  , 3.0 , 3.2  , 3.7  , 4.0]]
y= [[0.0,0.693,1.099,1.386],
    [0.0 , 0.182 , 0.336 , 0.693],
    [0.0 , 0.182 , 0.531 , 0.693  , 0.788  ,  0.993 , 1.099 , 1.163 , 1.308 , 1.386]]

for i = 1:3

    interpolantfunc = LagrangeInterpolantGenerator(x[i],y[i] )
    xa=1.5
    a=interpolantfunc(xa)   #returns  0.34650
    s = @sprintf("Lagrange插值法 P(%0.2f) = %0.5f " , float(xa) , float(a) )
    println(s)

end


輸出畫面

Lagrange插值法 P(1.50) = 0.39287 
Lagrange插值法 P(1.50) = 0.40447 
Lagrange插值法 P(1.50) = 0.40614