等距點插值公式

1、2012-2013 (1)專業(yè)課程實踐論文等距節(jié)點插值公式柳希元，0818180127, R數(shù)學08-1班、算法理論將Newton差商插值多項式中各階差商用相應差分替代，就可得到各種形 式的等距節(jié)點插值公式。如果節(jié)點込F +馳=0丄 ,.要計算_附近點X的函數(shù)- 值,可令護眼專腌腮L于是Nn(x + =禹 + tA/a + 耳®+- + - " + - A"/fl為Newton前插公式。其中也=&mfk - AraffcAfjr = Ar+1 一 fk如果要求表示函數(shù)在工屛付近的值此時應用Newton插值公式，插值點應按'-| ；：? i 1 &q

2、uot;4 L的次序排列，有叫(刃=x91) + /xB,x_vx_2(z - XJ(z-Xn_t) + f 北和場i-W衍仗一?！?(工 寓1)做變換応口徨/感*廣農(nóng)住灼;:鏈帶入公式得” vt(t - 1) ,t(t 一 1) ft - n +1)凡仗+tft) = 4 +岡人+七+計曠啟為Newton后插公式。其中嚴人=% -嚴人負二 /fc+i- fk二、算法框圖三、算法程序class Interpolationpublic :Interpolation(int num, double x1, double x2, double func);double ComputeForwardV

3、alue( double x);/ compute forward interpolation valueln terpolati on();private :void GetForwardTable(); / get the forward differential tableprivate :int m_num; / the number of interpolation pointsdouble m_x1, m_x2;/ the first point m_x1 and last point m_x2double m_step; / the in terpolati on stepdou

4、ble * m_func; / the function value of interpolation pointsdouble * m_ftable; / the forward differential table;#in clude <iostream>#i nclude <limits>using namespacestd;#define NUM 11/上l?面?輸o?入？需要°&多0少|(zhì)個?樣本a?#define MIN 0/上|?面?輸o?入？區(qū)？間？的1最 a?小?值| 1#define MAX 10/上|?面?輸o?入？區(qū)？間？的1最a

5、大？值| 1int main()/下？面?輸o?入?y的1值| 1double funcNUM=0,1,4,9,16,25,36,49,64,81,100；/上I?面？輸o?入?y的值| 1double x1=MIN, x2=MAX, x;int num=NUM;char flag= 'Y'In terpolati on test (n um, x1, x2, fun c);while (flag= Y )cout<< "I nput x:"cin»x;if (!cin)/ checking failure stateci n.clea

6、r(); / clear failure tagcin .ig nore(n umeric_limits<int >:max(),'n' ); / clear in put buffercontinue ;if (x<x1 | x>x2)cout<<"-I nvalid in put: "<<x<<"-" <<e ndl;<<e ndl;cout<<"Only the number between " vvx1<v&q

7、uot; and " <<x2<<" is valid."elsecout<< "Forward in terpolati on value:"<<test.ComputeForwardValue(x+0.001)<<e ndl;cout<<endl<< "Do you want to process? please input(Y/N):"<<endl;cin> >flag;return 0;In terpolati

8、o n:l nterpolatio n(int num, double x1, double x2, double func)m_num = num;m_x1 = x1;m_x2 = x2;m_step = (m_x2-m_x1)/( num-1); m_func = new double m_num; m_ftable = new double m_num;for ( int i=0; i<m_num; +i) m_fu nci = fun ci; m_ftablei = fun ci;GetForwardTable();In terpolatio n:l nterpolatio n(

9、)delete m_func;delete m_ftable;void Interpolation:GetForwardTable()/ get the forward differe ntial tableint i, j;for (i=1; i<m_num; +i)for (j=m_num-1; j>=i; -j) m_ftablej = m_ftablej-m_ftablej-1;double x) double In terpolati on:ComputeForwardValue( / compute forward in terpolati on valuedouble

10、 * coef; /coefficient talbedouble result, t;int i;coef = new double m_num;/compute the coefficie nt tablet = (x-m_x1)/m_step;for (i=1, coef0=1; i<m_num; +i)coefi = coefi-1*(t-i+1)/i;for (i=0, result=0; i<m_num; +i) result += m_ftablei*coefi;delete coef;return result;四、算法實現(xiàn)例i當心時求.在范圍m內(nèi)任意他刑的值解：(

11、1) 在程序中輸入11個樣本：0,1,4,9,16,25,36,49,64,81,100且輸入MIN : 0, MAX : 10,樣本容量：11。(2) 在運行程序中輸入范圍內(nèi)的任意.的值，如：1.2, 1.4等。(3) 程序提示答案和提示是否繼續(xù)，是則輸入大寫丫，否則輸入大寫NfE3 C A'ind q-a system32cmd,exeInput x： 1_ 2Forward interpolation value = 1-4424Do you uant to process? please input<¥/N>:Y1nput x： 1.4Forward in

12、terpolation ualue: 1.9628Do you want to process? please injiut (¥/N): NPress any key to cont inue , 例2.當給出15個亠:1.38, 1.48, 1.58, 1.69, 1.81,1.94, 2.10, 2.28, 2.50, 2.76,3.06, 3.41, 3.83, 4.33, 4.93,且給出范圍： m 二二二求出任意廠對的值。解：(1) 給出15個題目上顯示的值：1.38, 1.48, 1.58, 1.69, 1.81,1.94, 2.10, 2.28, 2.50, 2.76

13、,3.06, 3.41, 3.83, 4.33, 4.93,且輸入 MIN : 4000, MAX : 11000,樣本容量：15。(2) 在運行程序中輸入范圍內(nèi)的任意的值，如：1001，5200等。(3) 程序提示答案和提示是否繼續(xù)，是則輸入大寫 丫，否則輸入大寫Nral C ：W i ndowssystem3 2cmd .exeInput x: 4001 Forward interpolation value =3?4?Do sroti viant to process? please input <V/N>:*Input x： 5200Forward inteppolat ion value: 1.61375Douant