TruncateSum[sexpr,n]
对 sexpr 中的 Sum 进行截断,最多包含 n 项.
TruncateSum[sexpr,{m,n,…}]
对 sexpr 中的多个 Sum 进行截断,使用迭代指定 {m,n,…}.
TruncateSum
TruncateSum[sexpr,n]
对 sexpr 中的 Sum 进行截断,最多包含 n 项.
TruncateSum[sexpr,{m,n,…}]
对 sexpr 中的多个 Sum 进行截断,使用迭代指定 {m,n,…}.
更多信息和选项
- TruncateSum 通常用于将涉及无限和的符号解截断为有限和,从而可以轻松地对此类近似进行数值计算.
- 和表达式 sexpr 可由未计算的和与 Inactive 和进行任意组合而形成.
- TruncateSum 将根据以下条件对大型正的或负的求和极限进行截断:
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, if 



, if 


- 可给出以下选项:
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ActivateResult True 是否 Activate 结果 WorkingPrecision Automatic 内部计算使用的精度
范例
打开所有单元 关闭所有单元基本范例 (2)
TruncateSum[Sum[f[i], {i, 1, ∞}], 12]expr = Inactive[Sum][1 / x^i, {i, 1, ∞}]TruncateSum[expr, 5]TruncateSum[Inactive[Sum][x^(-i), {i, 1, Infinity}], 5, ActivateResult -> False]Activate[%]范围 (13)
基本用法 (3)
expr = Sum[Sin[x^i], {i, 1, Infinity}]TruncateSum[expr, 10]expr = Inactive[Sum][Exp[-Sin[x]], {x, 1, Infinity}]TruncateSum[expr, m]对于未计算的和,TruncateSum 直接计算截断的和:
expr = Sum[Cos[i], {i, 3, Infinity}]TruncateSum[expr, 7]有限和 (3)
TruncateSum[Inactive[Sum][Tanh[x]^i, {i, 1, 1000}], 5]expr = Sum[Sin[x^i], {i, 1, n}]TruncateSum[expr, 10]expr = Inactive[Sum][(-x)^n, {n, -100, 100}];TruncateSum[expr, 10, ActivateResult -> False]无限和 (4)
expr = Sum[1 / n, {n, 1, Infinity}]TruncateSum[expr, 14]TruncateSum[Inactive[Sum][Sin[x^i], {i, -Infinity, Infinity}], 10, ActivateResult -> False]Activate[%]TruncateSum[Inactive[Sum][Cos[n + x], {n, -Infinity, 0}], 5]pol = x^2 Inactive[Sum][Sin[i], {i, 1, Infinity}] + x Inactive[Sum][Cos[j], {j, 1, Infinity}] + Inactive[Sum][E^(-k), {k, 1, Infinity}];N[TruncateSum[pol, 4]]多项和 (2)
expr = Inactive[Sum][Sin[i*j], {i, 1, Infinity}, {j, 1, Infinity}];TruncateSum[expr, 5]TruncateSum[expr, {3, 4}]TruncateSum[expr, {4, All}]TruncateSum[expr, {All, 4}]dsum = Underoverscript[∑, i = 1, ∞]Underoverscript[∑, j = 1, 10](i^2 + j^3);TruncateSum[dsum, 5]TruncateSum[dsum, {3, 4}]处于未激活状态的和 (1)
对于未激活的和,TruncateSum 尝试计算截断和:
expr = Inactive[Sum][i^2, {i, 1, Infinity}]TruncateSum[expr, n]用 ActivateResultFalse 避免激活处于未激活状态的和:
TruncateSum[expr, n, ActivateResult -> False]Activate[%]选项 (2)
WorkingPrecision (1)
应用 (10)
微分方程 (4)
weqn = D[u[x, t], {t, 2}] == D[u[x, t], {x, 2}];
ic = {u[x, 0] == x ^ 2(Pi - x), Derivative[0, 1][u][x, 0] == 0};
bc = {u[0, t] == 0, u[Pi, t] == 0};sol = DSolveValue[{weqn, ic, bc}, u[x, t], {x, t}]从 Inactive 和中提取前三项:
asol = TruncateSum[sol, 3]Plot3D[asol, {x, 0, Pi}, {t, 0, 4Pi}, ColorFunction -> Hue]weqn = D[u[x, y, t], {t, 2}] == Laplacian[u[x, y, t], {x, y}];ic = {u[x, y, 0] == (1 / 10)(x - x ^ 2)(2y - y ^ 2), Derivative[0, 0, 1][u][x, y, 0] == 0};bc = {u[x, 0, t] == 0,
u[0, y, t] == 0, u[1, y, t] == 0, u[x, 2, t] == 0};(sol = FullSimplify[DSolveValue[{weqn, ic, bc}, u[x, y, t], {x, y, t}], K[1]∈Integers && K[3]∈Integers && K[1] ≥ 1 && K[3] ≥ 1])//TraditionalForm从 Inactive 和中提取前几项:
h[x_, y_, t_] = TruncateSum[sol, 5]Animate[Plot3D[h[x, y, t], {x, 0, 1}, {y, 0, 2}, Ticks -> False, PlotRange -> {-1 / 36, 1 / 36}, MeshStyle -> Red, PerformanceGoal -> "Quality"], {t, 0, 8}, SaveDefinitions -> True, DefaultDuration -> 12, AnimationRepetitions -> 1]heqn = D[u[x, t], t] == D[u[x, t], {x, 2}];
ic = u[x, 0] == x(3 - x) ^ 2;
bc = {u[0, t] == 0, u[3, t] == 0};sol = DSolveValue[{heqn, ic, bc }, u[x, t], {x, t}]截断 Inactive 和:
asol = TruncateSum[sol, 3]Plot3D[asol//Evaluate, {x, 0, 3}, {t, 0, 3 / 2}, Exclusions -> None, PlotRange -> All]eqn = I D[ψ[x, t], t] == -2 D[ψ[x, t], {x, 2}];
f[x_] := -350 + 155 x - 22 x^2 + x^3
sol = DSolveValue[{eqn, ψ[5, t] == 0, ψ[10, t] == 0, ψ[x, 2] == f[x]}, ψ, {x, t}]u[k_Integer] = TruncateSum[sol, k]Table[eqn /. ψ -> u[k], {k, 4}]//SimplifyTable[{u[k][5, t] == 0, u[k][10, t] == 0}, {k, 4}]只有 u∞ 的情况下满足初始条件,但在 t==2 时快速收敛:
Grid@Partition[Table[Plot[{u[k][x, 2], f[x]}, {x, 5, 10}, ImageSize -> 200], {k, 4}], 2]差分方程 (1)
渐近 (3)
pse = Asymptotic[Sin[x], {x, 0, Infinity}]TruncateSum[pse, 7]Asymptotic[E ^ x, {x, 1, Infinity}]TruncateSum[%, 5]截断 Hypergeometric1F1 函数的级数:
h1f1[a_, b_, z_] = Inactive[Sum][(Pochhammer[a, i]z^i/Pochhammer[b, i]i!), {i, 0, Infinity}]f[a_, b_, z_, n_] = TruncateSum[h1f1[a, b, z], n + 1]与内置的 Hypergeometric1F1 函数相比较:
Table[(Hypergeometric1F1[a, b, z] - f[a, b, z, n]) /. {a -> 1.1, b -> 1.2, z -> 2.3}, {n, 5, 20, 5}]Plot[{Hypergeometric1F1[-1.1, 2, z], f[-1.1, 2, z, 5], f[-1.1, 2, z, 10], f[-1.1, 2, z, 15]}, {z, 0, 17}, PlotLegends -> {"Exact", "n=5", "n=10", "n=15"}]逆拉普拉斯变换 (2)
ilt = InverseLaplaceTransform[1 / (s Sinh[3 s] Sinh[5 s]), s, t]Plot[TruncateSum[ilt, 45], {t, 0, 60}]ilt = InverseLaplaceTransform[((Exp[s] - 1) Sin[π / 5]) / (s (Exp[2 s] - 2 Exp[s] Cos[π / 5] + 1)), s, t]Plot[TruncateSum[ilt, 6], {t, 0, 15}]属性和关系 (4)
TruncateSum[expr,n] 将表达式中的每个和截断为最多包含 n 项:
TruncateSum[Underoverscript[∑, i = 1, ∞]f[i] + Underoverscript[∑, j = 1, ∞]g[j], 4]TruncateSum 截断双无限和的两个极限:
TruncateSum[Underoverscript[∑, i = -∞, ∞]f[i], 4]TruncateSum 只将和的下限从 –∞ 截断为有限值:
TruncateSum[Underoverscript[∑, i = -∞, 0]f[i], 4]TruncateSum 激活表达式中所有处于未激活状态的和:
expr = Inactive[Sum][f[i], {i, 1, 100}]TruncateSum[expr, 5]ActivateResult False 可用于避免激活:
TruncateSum[expr, 5, ActivateResult -> False]相关指南
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文本
Wolfram Research (2024),TruncateSum,Wolfram 语言函数,https://reference.wolfram.com/language/ref/TruncateSum.html.
CMS
Wolfram 语言. 2024. "TruncateSum." Wolfram 语言与系统参考资料中心. Wolfram Research. https://reference.wolfram.com/language/ref/TruncateSum.html.
APA
Wolfram 语言. (2024). TruncateSum. Wolfram 语言与系统参考资料中心. 追溯自 https://reference.wolfram.com/language/ref/TruncateSum.html 年
BibTeX
@misc{reference.wolfram_2026_truncatesum, author="Wolfram Research", title="{TruncateSum}", year="2024", howpublished="\url{https://reference.wolfram.com/language/ref/TruncateSum.html}", note=[Accessed: 02-November-2026]}
BibLaTeX
@online{reference.wolfram_2026_truncatesum, organization={Wolfram Research}, title={TruncateSum}, year={2024}, url={https://reference.wolfram.com/language/ref/TruncateSum.html}, note=[Accessed: 02-November-2026]}