zOla. nb Clear["Global*" f[n] 1=P1ot[f[n0=3],f[n0=5],f[n0=7]},【x,-1,1}, PlotRange→{0,4 Dashing[o. 00])) PlotLabe1→f[n] PlotLegends→ ced[LineLegend[[style[n=3",Italic, 10] style["n=7",Italic, 10]],LegendMarkerSize+[30,10)],[0.8,0.7]]; fn1:=1 n 丌丌(n2x2+1) g2=P1ot[f[n0=5],f[n0=15],f[n0=45]},{x,-1,1}, PlotRange→{0,5}, Plotstyle+[Red],[Green, Dashed], [Blue, Dashing[o0]]1, PlotLabel→f[n] Placed[LineLegend[[style["n = 5", Italic, 10], style["n =15",Italic, 101 style["n=45", Italic, 101), LegendMarkerSize+[30,10]],[0.8,0.7]]i Sin[xn] f[n_] 93=Plot[{f[n0=5],f[n0=10],f[n0=20]},{x,-1,1},P1 stRange→{0,7 Plotstyle+[[Red], [Green, Dashed],[Blue, Dashing[o 0]11 Placed [LineLegend[[style["n =5", Italic, 10], Style["n=10", Italic, 10] style["n 20", Italic, 10]], LegendMarkerSize-[30,101],[o8,0.7]i 十 选取其中之一加以证明 丌x-+E e-0r(x2+2 r(r2+e) ■ Heaviside阶跃函数Clear["Global`*"] f[n_] := n π -n2 x2 ; g1 = Plot[{f[n0 = 3], f[n0 = 5], f[n0 = 7]}, {x, -1, 1}, PlotRange  {0, 4}, PlotStyle  {{Red}, {Green, Dashed}, {Blue, Dashing[0.00]}}, PlotLabel  f[n], PlotLegends  Placed[LineLegend[{Style["n = 3", Italic, 10], Style["n = 5", Italic, 10], Style["n = 7", Italic, 10]}, LegendMarkerSize  {30, 10}], {0.8, 0.7}]]; f[n_] := 1 π n π (n2 x2 + 1) ; g2 = Plot[{f[n0 = 5], f[n0 = 15], f[n0 = 45]}, {x, -1, 1}, PlotRange  {0, 5}, PlotStyle  {{Red}, {Green, Dashed}, {Blue, Dashing[0.0]}}, PlotLabel  f[n], PlotLegends  Placed[LineLegend[{Style["n = 5", Italic, 10], Style["n = 15", Italic, 10], Style["n = 45", Italic, 10]}, LegendMarkerSize  {30, 10}], {0.8, 0.7}]]; f[n_] := Sin[x n]2 π x2 n ; g3 = Plot[{f[n0 = 5], f[n0 = 10], f[n0 = 20]}, {x, -1, 1}, PlotRange  {0, 7}, PlotStyle  {{Red}, {Green, Dashed}, {Blue, Dashing[0.0]}}, PlotLabel  f[n], PlotLegends  Placed[LineLegend[{Style["n = 5", Italic, 10], Style["n = 10", Italic, 10], Style["n = 20", Italic, 10]}, LegendMarkerSize  {30, 10}], {0.8, 0.7}]]; Grid[ {{g1, g2, g3}}] n = 3 n = 5 n = 7 -1.0 -0.5 0.0 0.5 1.0 1 2 3 4 n -n2 x2 π n = 5 n = 15 n = 45 -1.0 -0.5 0.0 0.5 1.0 1 2 3 4 5 n π2 n2 x2 + 1 n = 5 n = 10 n = 20 -1.0 -0.5 0.0 0.5 1.0 1 2 3 4 5 6 7 sin2 (n x) π n x2 选取其中之一加以证明 。 lim ε0 ε πx2 + ε2 = lim ε0 ε lim ε0 π x2 + ε2 = 0 x ≠ 0 lim ε0 ε πε2 = ∞ x = 0 ⟹ lim ε0-∞ +∞ ε πx2 + ε2 x = lim ε0 1 π tg-1 x ε -∞ +∞ = 1 ◼ Heaviside 阶跃函数 4 z07a.nb