數(shù)學(xué)試驗(yàn)報(bào)告 數(shù)據(jù)擬合

實(shí)驗(yàn)報(bào)告實(shí)驗(yàn)10數(shù)據(jù)擬合與曲線擬合一．實(shí)驗(yàn)解讀本次試驗(yàn)分別通過(guò)線性擬合和非線性擬合的方法來(lái)模擬兩個(gè)變量的取值。對(duì)練習(xí)1用線性模擬的方法來(lái)解決，觀察散點(diǎn)圖與模擬函數(shù)圖形的接近程度。對(duì)練習(xí)2用非線性模擬的方法，其中可以分別用指數(shù)函數(shù)、對(duì)數(shù)函數(shù)、雙曲函數(shù)、三角函數(shù)、分式有理多項(xiàng)式函數(shù)和分段函數(shù)等初等函數(shù)擬合，觀察散點(diǎn)圖與模擬函數(shù)圖形的接近程度，選擇最適合的模擬函數(shù)。二．實(shí)驗(yàn)計(jì)劃（一）線性擬合練習(xí)1程序1biao={{xl,yl},{x2,y2},…,{xn,yn}}f[x_]:=a*x+bft1=Fit[lianxi1biao,{1,x},x]gp=Plot[ft1,{x,x1,xn},PlotStyle->{RGBColor[1,0,0]}]fp=ListPlot[lianxi1biao,PlotStyle->{PointSize[a],RGBColor[0,0,1]}]Show[fp,gp]a=;b=;f[x_]=a*x+b;dareta=Sum[（biao[[i,2]]-f[biao[[i,1]]]）A2,{i,1,n}]練習(xí)1為研究某一化學(xué)反應(yīng)過(guò)程中溫度x對(duì)合格率的影響，測(cè)得數(shù)據(jù)如下x100110120130140150160170180190y45515461667074788589試求擬合曲線程序如下lianxi1biao={{100,45},{110,51},{120,54},{130,61},{140,66},{150,70},{160,74},{170,78},{180,85},{190,89}}ft1=Fit[lianxi1biao,{1,x},x]gp=Plot[ft1,{x,100,190},PlotStyle->{RGBColor[1,0,0]}]fp=ListPlot[lianxi1biao,PlotStyle->{PointSize[0.05],RGBColor[0,0,1]}]Show[fp,gp]實(shí)驗(yàn)思路：1、先對(duì)練習(xí)1的十組數(shù)據(jù)線性擬合，并從形與量看擬合效果；2、對(duì)練習(xí)1的十組數(shù)據(jù)中的九組數(shù)據(jù)線性擬合，并從形與量看擬合效果；3、對(duì)練習(xí)1的十組數(shù)據(jù)中的八組數(shù)據(jù)線性擬合，并從形與量看擬合效果；4、對(duì)練習(xí)1的十組數(shù)據(jù)中的六組數(shù)據(jù)線性擬合，并從形與量看擬合效果；5、對(duì)練習(xí)1的十組數(shù)據(jù)增加五組數(shù)據(jù){200,94},{210,99},{220,104},{230,109},{240,114}線性擬合，并從形與量看擬合效果。（二）非線性擬合練習(xí)2在某一化學(xué)反應(yīng)中，由實(shí)驗(yàn)得到生物濃度與時(shí)間的關(guān)系如下，求濃度與時(shí)間的關(guān)系曲線t（分）12345678y46.48.08.49.289.59.79.86t（分）910111213141516y1.010.210.3210.4210.510.5510.5810.6作散點(diǎn)圖，程序如下lianxi2biao={{1,4},{2,6.4},{3,8.0},{4,8.4},{5,9.28},{6,9.5},{7,9.7},{8,9.86},{9,10.0},{10,10.2},{11,10.32},{12,10.42},{13,10.5},{14,10.55},{15,10.58},{16,10.6}}gp=ListPlot[lianxi2biao,PlotStyle->{RGBColor[0,1,0],PointSize[0.04]}]Show[gp]a用指數(shù)函數(shù)和曲線擬合，程序如下先用指數(shù)函數(shù)擬合fx[x_]:=1/xfy[y_]:=Log[y]lianxi2biao={{1,4},{2,6.4},{3,8.0},{4,8.4},{5,9.28},{6,9.5},{7,9.7},{8,9.86},{9,10.0},{10,10.2},{11,10.32},{12,10.42},{13,10.5},{14,10.55},{15,10.58},{16,10.6}};nb=Table[{fx[lianxi2biao[[i,1]]],fy[lianxi2biao[[i,2]]]},{i,1,15}];ft3=Fit[nb,{1,x},x]f4=Exp[2.46274-1.155/x]t1=Plot[f4,{x,1,18},PlotStyle->{RGBColor[1,0,0]}]t2=ListPlot[lianxi2biao,PlotStyle->{RGBColor[0,1,0],PointSize[0.05]}]Show[%,%%]再用曲線擬合，并比較兩者的效果dareta=Sum[(lianxi2biao[[i,2]]-f[lianxilbiao[[i,l]]])人2,{i,l,16}]g[y_]:=1/yb=Table[{lianxi2biao[[i,l]],g[lianxi2biao[[i,2]]]},{i,l,l6}]ft5=Fit[b,{l,l/x},x]f5=l/ft5t3=Plot[f5,{x,l,l6},PlotStyle->{RGBColor[0,0,l]}]Show[tl,t2,t3]b用多項(xiàng)式函數(shù)擬合，程序如下lianxi2biao={{l,4},{2,6.4},{3,8.0},{4,8.4},{5,9.28},{6,9.5},{7,9.7},{8,9.86},{9,l0.0},{l0,l0.2},{ll,l0.32},{l2,l0.42},{l3,l0.5},{l4,l0.55},{l5,l0.58},{l6,l0.6}}gp=ListPlot[lianxi2biao,PlotStyle->{RGBColor[0,l,0],PointSize[0.04]}]ft2=Fit[lianxi2biao,Table[xAi,{i,0,3}],x]fp=Plot[ft2,{x,0,l6},PlotStyle->{RGBColor[l,0,0]}]Show[gp,fp]n=l6;f[x_]=Fit[lianxi2biao,Table[xAi,{i,0,3}],x];dareta=Sum[(lianxi2biao[[i,2]]-f[lianxi2biao[[i,l]]])A2,{i,l,l6}]實(shí)驗(yàn)思路：1、改變多項(xiàng)式最高次數(shù)分別為4,5,8。觀察圖形與殘差值。2、對(duì)十六組數(shù)據(jù)中的l0組數(shù)進(jìn)行擬合。觀察圖形與殘差值。c用三角函數(shù)擬合，程序如下lianxi2biao={{1,4},{2,6.4},{3,8.0},{4,8.4},{5,9.28},{6,9.5},{7,9.7},{8,9.86},{9,10.0},{10,10.2},{11,10.32},{12,10.42},{13,10.5},{14,10.55},{15,10.58},{16,10.6}}x[x_]:=Sin[x]nb=Table[{fx[lianxi2biao[[i,1]]],lianxi2biao[[i,2]]},{i,1,16}]ft6=Fit[nb,{1,Sin[x]},x]h1=Plot[ft6,{x,1,16},PlotStyle->{RGBColor[1,0,1]}]h2=ListPlot[lianxi2biao,PlotStyle->{RGBColor[0,1,1],PointSize[0.05]}]Show[h1,h2]d用分段函數(shù)擬合，程序如下lianxi2biao={{1,4},{2,6.4},{3,8.0},{4,8.4},{5,9.28},{6,9.5},{7,9.7},{8,9.86},{9,10.0},{10,10.2},{11,10.32},{12,10.42},{13,10.5},{14,10.55},{15,10.58},{16,10.6}}f[x_]:=xA1.5+l;xv=10f[x_]:=10+0.2*x;x>10h3=ListPlot[lianxi2biao,PlotStyle->{RGBColor[0,1,1],PointSize[0.05]}]h4=Plot[f[x],{x,1,16},PlotStyle->{RGBColor[1,0,0]}]Show[h3,h4]三實(shí)驗(yàn)過(guò)程與結(jié)果練習(xí)1輸入以下程序lianxi1biao={{100,45},{110,51},{120,54},{130,61},{140,66},{150,70},{160,74},{170,78},{180,85},{190,89}}ft1=Fit[lianxi1biao,{1,x},x]gp=Plot[ft1,{x,100,190},PlotStyle->{RGBColor[1,0,0]}]fp=ListPlot[lianxi1biao,PlotStyle->{PointSize[0.05],RGBColor[0,0,1]}]Show[fp,gp]f[x_]=Fit[lianxi1biao,{1,x},x]dareta=Sum[(lianxilbiao[[i,2]]-f[lianxi1biao[[i,1]]])A2,{i,1,10}]運(yùn)行結(jié)果為7.22424運(yùn)行結(jié)果為{100,45},{110,51},{120,54},{130,61},{140,66},{150,70},{160,74},{170,78},{180,85},{190,89}}-2.73939+0.48303x90807060120140160十組數(shù)據(jù)中的九組數(shù)據(jù)線性擬合。運(yùn)行結(jié)果為-2.77778+0.483333x7.8125十組數(shù)據(jù)中的組八數(shù)據(jù)線性擬合。運(yùn)行結(jié)果為-1.75+0.475x十組數(shù)據(jù)中的六組數(shù)據(jù)線性擬合。運(yùn)行結(jié)果為-5.38095+0.505714x3.27619十組數(shù)據(jù)中增加五組數(shù)據(jù){200,94},{210,99},{220,104},{230,109},{240,114}線性擬合運(yùn)行結(jié)果為-3.40238+0.487857x7.60476實(shí)驗(yàn)觀察：線性擬合函數(shù)隨著數(shù)據(jù)的變化而變化，當(dāng)數(shù)據(jù)的組數(shù)越多時(shí)，殘差值越小，線性擬合函數(shù)與實(shí)際數(shù)據(jù)的接近度越大。練習(xí)2-a輸入以下程序lianxi2biao={{1,4},{2,6.4},{3,8.0},{4,8.4},{5,9.28},{6,9.5},{7,9.7},{8,9.86},{9,10.0},{10,10.2},{11,10.32},{12,10.42},{13,10.5},{14,10.55},{15,10.58},{16,10.6}}gp=ListPlot[lianxi2biao,PlotStyle->{RGBColor[0,1,0],PointSize[0.04]}]Show[gp]運(yùn)行結(jié)果為2.46109652.557.5109652.557.51012.5再輸入程序fx[x_]:=1/xfy[y_]:=Log[y]lianxi2biao={{1,4},{2,6.4},{3,8.0},{4,8.4},{5,9.28},{6,9.5},{7,9.7},{8,9.86},{9,10.0},{10,10.2},{11,10.32},{12,10.42},{13,10.5},{14,10.55},{15,10.58},{16,10.6}};nb=Table[{fx[lianxi2biao[[i,1]]],fy[lianxi2biao[[i,2]]]},{i,1,15}];ft3=Fit[nb,{1,x},x]f4=Exp[2.46274-1.155/x]t1=Plot[f4,{x,1,18},PlotStyle->{RGBColor[1,0,0]}]t2=ListPlot[lianxi2biao,PlotStyle->{RGBColor[0,1,0],PointSize[0.05]}]Show[%,%%]運(yùn)行結(jié)果為再輸入程序g[y_]:=1/yb=Table[{lianxi2biao[[i,1]],g[lianxi2biao[[i,2]]]},{i,1,16}]ft5=Fit[b,{1,1/x},x]f5=1/ft5t3=Plot[f5,{x,1,16},PlotStyle->{RGBColor[0,0,1]}]Show[t1,t2,t3]實(shí)驗(yàn)觀察：指數(shù)函數(shù)與該數(shù)據(jù)的吻合度較高。運(yùn)行結(jié)果為練習(xí)2-b輸入以下程序lianxi2biao={{1,4},{2,6.4},{3,8.0},{4,8.4},{5,9.28},{6,9.5},{7,9.7},{8,9.86},{9,10.0},{10,10.2},{11,10.32},{12,10.42},{13,10.5},{14,10.55},{15,10.58},{16,10.6}}gp=ListPlot[lianxi2biao,PlotStyle->{RGBColor[0,1,0],PointSize[0.04]}]ft2=Fit[lianxi2biao,Table[xAi,{i,0,3}],x]fp=Plot[ft2,{x,0,16},PlotStyle->{RGBColor[1,0,0]}]Show[gp,fp]f[x_]=Fit[lianxi2biao,Table[xAi,{i,0,3}],x];dareta=Sum[(lianxi2biao[[i,2]]-f[lianxi2biao[[i,1]]])A2,{i,1,16}]運(yùn)行結(jié)果為1.22919改變多項(xiàng)式最高次項(xiàng)為4運(yùn)行結(jié)果如下改變多項(xiàng)式最高次項(xiàng)為5運(yùn)行結(jié)果如下改變多項(xiàng)式最高次項(xiàng)為8改變多項(xiàng)式最高次項(xiàng)為8運(yùn)行結(jié)果如下468468對(duì)十六組數(shù)據(jù)中的10組數(shù)進(jìn)行擬合運(yùn)行結(jié)果如下實(shí)驗(yàn)觀察：在不斷增大最高次項(xiàng)次數(shù)的過(guò)程中，非線性擬合函與實(shí)際數(shù)據(jù)的吻合度越高。練習(xí)2-c輸入程序如下lianxi2biao={{1,4},{2,6.4},{3,8.0},{4,8.4},{5,9.28},{6,9.5},{7,9.7},{8,9.86},{9,10.0},{10,10.2},{11,10.32},{12,10.42},{13,10.5},{14,10.55},{15,10.58},{16,10.6}}fx[x_]:=Sin[x]nb=Table[{fx[lianxi2biao[[i,1]]],lianxi2biao[[i,2]]},{i,1,16}]ft6=Fit[nb,{1,Sin[x]},x]h1=Plot[ft6,{x,1,16},Pl