範例1-2 通過下面3點的Lagrange內插法求P(1.5)之值?
x f(x)
==================
1.0 0.0
2.0 0.693
3.0 1.099
程式
#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
interpolantfunc = LagrangeInterpolantGenerator([1.0,2.0,3.0],[0.0,0.693,1.099])
xa=1.5
a=interpolantfunc(xa) #returns 0.38237
s = @sprintf("Lagrange插值法 P(%0.2f) = %0.5f " , float(xa) , float(a) )
println(s)
輸出畫面
Lagrange插值法 P(1.50) = 0.38237
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