Report7 Lagrange 보간법, Newton 보간법, Cubic spline 2011, Feb 05 <br /> #include <stdio.h><br /> #include<br /> <math.h> <p>#define PI 3.14159265</p> <p>float xarr[6]={0.15,2.3,3.15,4.85,6.25,7.95};<br /> float yarr[6]={4.79867,4.49013,4.2243,3.47313,2.66674,1.51909};</p> <p>float want_x[14]={0.5,1,1.5,2,2.5,3,3.5,4,4.5,5,5.5,6,6.5,7};<br /> float want_y[14]={0,};</p> <p>float funtion(float x);</p> <p>int main(void)<br /> {<br /> for(int x=0;x<14;x++)<br /> {<br /> want_y[x]=0;<br /> for(int i=0;i<5;i++)<br /> {<br /> float temp = 1;<br /> for(int j=0;j<5;j++)<br /> {<br /> if(j!=i)<br /> {<br /> temp*=(want_x[x]-xarr[j])/(xarr[i]-xarr[j]);<br /> printf("*");<br /> }<br /> }<br /> want_y[x]+=temp*yarr[i];<br /> printf("%d^^%f//%f\n",i,want_y[x],temp);<br /> }<br /> printf("\n");<br /> }<br /> for(int i=0;i<14;i++)<br /> {<br /> printf("x = %f y = %f f(x) = %f f(x)-y = %f \n",want_x[i],want_y[i],funtion(want_x[i]),funtion(want_x[i])-want_y[i]);<br /> }</p> <p> return 0;<br /> }</p> <p>float funtion(float x)<br /> {</p> <p> return 4.8*cos(PI/20*x);<br /> }</p> </math.h></stdio.h> </div> <br /> #include <stdio.h><br /> #include<br /> <math.h> <p>#define PI 3.14159265</p> <p>float xarr[6]={0.15,2.3,3.15,4.85,6.25,7.95};<br /> float yarr[6]={4.79867,4.49013,4.2243,3.47313,2.66674,1.51909};</p> <p>float want_x[14]={0.5,1,1.5,2,2.5,3,3.5,4,4.5,5,5.5,6,6.5,7};<br /> float want_y[14]={0,};</p> <p>float d[6][6]={0,};</p> <p>float funtion(float x);</p> <p>int main(void)<br /> {</p> <p> for (int i=0; i<6; i++)<br /> {<br /> d[0][i]=yarr[i];<br /> }<br /> for (int k=1; k<=6-1; k++)<br /> for (int i=0; i<=6-1-k; i++)<br /> d[k][i]=(d[k-1][i+1]-d[k-1][i])/(xarr[k+i]-xarr[i]);</p> <p> for (int i=1; i<6; i++)<br /> {<br /> printf("%f ",d[i][0]);<br /> }</p> <p> for(int x=0;x<14;x++)<br /> {<br /> want_y[x]=0;<br /> for(int i=0;i<6;i++)<br /> {<br /> float temp=d[i][0];<br /> printf("%f\n",temp);<br /> for(int j=0;j<6;j++)<br /> {</p> <p> if(i==j)<br /> break;<br /> else<br /> {<br /> temp*=(want_x[x]-want_x[j]);<br /> }<br /> }<br /> want_y[x]+=temp;<br /> printf("%d^^%f//%f\n",i,want_y[x],temp);<br /> }<br /> printf("\n");<br /> }</p> <p> for(int i=0;i<14;i++)<br /> {<br /> printf("x = %f y = %f f(x) = %f f(x)-y = %f \n",want_x[i],want_y[i],funtion(want_x[i]),funtion(want_x[i])-want_y[i]);<br /> }</p> <p> return 0;<br /> }</p> <p>float funtion(float x)<br /> {</p> <p> return 4.8*cos(PI/20*x);<br /> }</p> </math.h></stdio.h> <br /> #include <stdio.h><br /> #include<br /> <math.h> <p>#define PI 3.14159265</p> <p>float xarr[6]={0.15,2.3,3.15,4.85,6.25,7.95};<br /> float yarr[6]={4.79867,4.49013,4.2243,3.47313,2.66674,1.51909};</p> <p>float want_x[14]={0.5,1,1.5,2,2.5,3,3.5,4,4.5,5,5.5,6,6.5,7};<br /> float want_y[14]={0,};</p> <p>float d[6][6]={0,};</p> <p>float funtion(float x);<br /> double CSpline(int n, double data);</p> <p>int main(void)<br /> {<br /> for (int i = 0; i < 14; i++)<br /> {<br /> printf("x = %f y = %f f(x) = %f f(x)-y = %f \n",want_x[i],CSpline(6,want_x[i]),funtion(want_x[i]),funtion(want_x[i])-CSpline(6,want_x[i]));<br /> }</p> <p> return 0;<br /> }</p> <p>double CSpline(int n, double data)<br /> {<br /> int k;<br /> double e[100],f[100],g[100],r[100],w[100];<br /> double A[100], B[100], C[100], D[100];<br /> double intPol;</p> <p> //form the tridiagonal system<br /> e[0]=0.;<br /> e[1]=0.;<br /> f[0]=0.;<br /> g[0]=0.;<br /> r[0]=0.;<br /> g[n-2]=0.;<br /> for (k = 1; k <= n-2; k++)<br /> {<br /> e[k+1] = xarr[k+1] - xarr[k];<br /> f[k] = 2.*(xarr[k+1] - xarr[k-1]);<br /> g[k] = xarr[k+1] - xarr[k];<br /> r[k] = 6.*((yarr[k+1] - yarr[k]) / (xarr[k+1] - xarr[k])<br /> + (yarr[k-1] - yarr[k]) / (xarr[k] - xarr[k-1]));<br /> }</p> <p> w[0] = 0.;<br /> w[n-1] = 0.;<br /> for (k = 2; k <= n-2; k++)<br /> {<br /> e[k] = e[k] / f[k-1];<br /> f[k] = f[k] - e[k]*g[k-1];<br /> } </p> <p> for (k = 2; k <= n-2; k++)<br /> {<br /> r[k] = r[k] - e[k]*r[k-1];<br /> } </p> <p> w[n-2] = r[n-2] / f[n-2];<br /> for (k = n-3; k >= 1; k--)<br /> {<br /> w[k] = (r[k] - g[k]*w[k+1]) / f[k];<br /> }</p> <p> for (k = 1; k <= n-1; k++)<br /> {<br /> A[k] = w[k-1] / (6.*(xarr[k] - xarr[k-1]));<br /> B[k] = w[k] / (6.*(xarr[k] - xarr[k-1]));<br /> C[k] = yarr[k-1] / (xarr[k] - xarr[k-1])<br /> - w[k-1] / 6.*(xarr[k] - xarr[k-1]);<br /> D[k] = yarr[k] / (xarr[k] - xarr[k-1])<br /> - w[k] / 6.*(xarr[k] - xarr[k-1]);<br /> } </p> <p> //보¬¢¬간Æ¡Ì값ƨ£ 계Æe산íe<br /> for (k = 1; k <= n-1; k++)<br /> {<br /> if(data >= xarr[k-1] && data <= xarr[k])<br /> {<br /> intPol = A[k] * pow(xarr[k] - data, 3)<br /> + B[k] * pow(data - xarr[k-1], 3)<br /> + C[k] * (xarr[k] - data)<br /> + D[k] * (data - xarr[k-1]);<br /> }<br /> }<br /> return(intPol);<br /> }</p> <p>float funtion(float x)<br /> {</p> <p> return 4.8*cos(PI/20*x);<br /> }</p> </math.h></stdio.h> Please enable JavaScript to view the comments powered by Disqus.
