逆関数の補間について
こんにちは。前に質問させてもらい、計算については
理解できました。しかし、以下の文章のある部分の意味がわかりません。まず、文章は、
[Inverse Interpolation]
A process called inverse interpolation is often used to approximate an inverse function. Suppose that values {Yi}=f({Xi}) have been computed at X0,X1,...,Xn.
Using table
Y ; Y0 Y1 Y2 ......Yn
X ; X0 X1 X2 ......Xn
we form the interpolation polynomial
p(y)=Σ(i=1→n)CiΠ(j=0→i-1){Y－Yj}
The orijinal relationship, y=f(x), has an inverse, under certain conditions. This inverse is being approximated by x=p(y). Procedures Coef and Eval can be used to carry out the inverse interpolation by reversing the arguments x and y in the calling sequence for Coef.
Inverse interpolation can be used to find where a given functuin f has a root or zero. This means inverting the equation f(x)=0. We propose to do this by creating a table of values (f(Xi),Xi) and interpolating with a polynomial,p. Thus, p(Yi)=Xi. The points Xi should be chosen near the unknown root,r. The approximate root is then given by r ～p(0). For a concrete case, let the table of known values be
Y;-0.5789200,-0.3626370,-0.1849160,-0.0340642,0.0969858
X; 1.0 , 2.0 , 3.0 , 4.0 , 5.0
The nodes in this problem are the points in the row of the table headed y, and the function values being interpolated are in the x row. The resulting polynomial is
p(Y)=0.25Y^4＋1.2Y^3＋3.69Y^2＋7.39Y＋4.247470086
and p(0)=4.247470086. Only the last coefficient is shown with all the digits carried in the calculation, for it is the only one needed for the problem at hand.
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＜補足＞CoefとEvalについて
「 procedure; Coef(n,{Xi},{Yi},{Ai})
real array; {Xi}0:n, {Yi}0:n, {Ai}0:n
integer; i,j,n
for i=0 to n do
{Ai}←{Yi}
end for
for j=1 to n do
for i=n to j step -1 do
Ai←({Ai}－{Ai-1})/({Xi}－{Xi-j})
end for
end for
end procedure Coef 」
「 real function; Eval(n,{Xi},{Yi},{Ai})
real array; {Xi}0:n, {Ai}0:n
integer; i,n
real;t,temp
temp←An
for i=n-1 to 0 step -1 do
temp←(temp)(t-{Xi})＋{Ai}
end for
Eval←temp
end function Eval」
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です。この文章の
「The orijinal relationship, y=f(x), has an inverse, under certain conditions. This inverse is being approximated by x=p(y). Procedures Coef and Eval can be used to carry out the inverse interpolation by reversing the arguments x and y in the calling sequence for Coef.
Inverse interpolation can be used to find where a given functuin f has a root or zero. This means inverting the equation f(x)=0. We propose to do this by creating a table of values (f(Xi),Xi) and interpolating with a polynomial,p. Thus, p(Yi)=Xi. The points Xi should be chosen near the unknown root,r. The approximate root is then given by r ～p(0).」
という部分が理解できません。わかる方アドバイスお願いします（泣）
お礼
入れ替えを忘れていました。ありがとうございました。
補足
Determine the equation for g(x) where g is the graph of f(x) = | x | translated up 4, to the right 2, and compressed about the y-axis答えがg(x)=3|x-2|+4なんですが、自分でやるとg(x)=0.5|x+2|+4になってしまいます。