BilateralHypergeometricPFQ[{a1,…,ap},{b1,…,bq},z]
是双边超几何函数
.
BilateralHypergeometricPFQ
BilateralHypergeometricPFQ[{a1,…,ap},{b1,…,bq},z]
是双边超几何函数
.
更多信息
- 双边超几何级数的项与广义超几何级数具有类似的定义,但是它对所有整数求和,因此形成了一个双向无穷级数.
- 数学函数,适用于符号和数值操作.
具有级数展开式
,其中
是 Pochhammer 符号.- 如果
且
,双边超几何级数
是收敛的. - 当
时,双边超几何函数
使用 Borel 正则化计算. - 参数
不能为正整数,
不能为负整数. - BilateralHypergeometricPFQ 可以计算为任意数值精度.
- 对于某些特殊参数,BilateralHypergeometricPFQ 自动计算出精确值.
- BilateralHypergeometricPFQ 自动在列表上进行线程操作.
范例
打开所有单元 关闭所有单元基本范例 (3)
BilateralHypergeometricPFQ[{1 / 2, 3 / 4}, {1 / 4, 1 / 3}, 5.4]BilateralHypergeometricPFQ[{1 / 2, 3 / 4}, {1 / 4, -1 / 3}, 5.4]ReImPlot[BilateralHypergeometricPFQ[{1 / 2, 3 / 4}, {1 / 4, 1 / 3}, x], {x, 0, 0.9}]Series[ BilateralHypergeometricPFQ[{a1, a2, a3}, {b1, b2}, z], {z, 0, 2}]范围 (18)
数值运算 (4)
N[BilateralHypergeometricPFQ[{1 / 2, 3 / 4}, {1 / 4, 1 / 3}, 5], 30]BilateralHypergeometricPFQ[{1 / 2, 3 / 4}, {1 / 4, 1 / 3}, 5`20]BilateralHypergeometricPFQ[{1 / 2I, 3 / 4I}, {1 / 4I, 1 / 3I}, 5`20I]高精度高效地运算 BilateralHypergeometricPFQ:
BilateralHypergeometricPFQ[{1 / 2I, 3 / 4I}, {1 / 4I, 1 / 3I}, 5`100]//TimingBilateralHypergeometricPFQ[{1 / 2I, 3 / 4I}, {1 / 4I, 1 / 3I}, 5`2000];//TimingBilateralHypergeometricPFQ 在其第三个参数上逐元素线性作用于列表:
BilateralHypergeometricPFQ[{1 / 2, 3 / 4}, {1 / 4, 1 / 3}, {0.1, 0.3, 0.5}]具体值 (3)
对于某些参数,BilateralHypergeometricPFQ 自动运算为较简单的函数:
BilateralHypergeometricPFQ[{1 / 2}, {1}, z]BilateralHypergeometricPFQ[{1 / 2, 3 / 2}, {1, 5 / 2}, z]BilateralHypergeometricPFQ 在
处:
BilateralHypergeometricPFQ[{1 / 2, 1 / 3}, {1 / 4, 1 / 5, 1 / 6}, 1.]对于
的情况,BilateralHypergeometricPFQ 在
处:
BilateralHypergeometricPFQ[{a1, a2}, {b1, b2}, 0]积分 (2)
对 BilateralHypergeometricPFQ 积分:
Integrate[BilateralHypergeometricPFQ[{1 / 2, 3 / 4}, {1 / 4, 1 / 3}, z], z]BilateralHypergeometricPFQ 的定积分:
Integrate[BilateralHypergeometricPFQ[{0.5, 0.75}, {0.25, 0.3}, z], {z, 1 / 3, 1 / 2}]求导 (1)
特定 BilateralHypergeometricPFQ 的一阶导数:
D[BilateralHypergeometricPFQ[{1 / 2, 3 / 4}, {1 / 4, 1 / 3}, z], z]BilateralHypergeometricPFQ 的
阶导数:
D[BilateralHypergeometricPFQ[{1 / 2, 3 / 4}, {1 / 4, 1 / 3}, z], {z, n}]级数展开 (3)
计算 BilateralHypergeometricPFQ 在原点处的级数展开式:
Series[BilateralHypergeometricPFQ[{1 / 2, 3 / 4}, {1 / 4, 1 / 3}, z], {z, 0, 1}]计算 BilateralHypergeometricPFQ 在 Infinity 处的级数展开式:
Series[BilateralHypergeometricPFQ[{1 / 2}, {1 / 3}, z], {z, ∞, 1}]计算 BilateralHypergeometricPFQ 在通点处的级数展开:
Series[BilateralHypergeometricPFQ[{1 / 2, 3 / 4}, {1 / 4, 1 / 3}, z], {z, z0, 1}]可视化 (2)
ReImPlot[BilateralHypergeometricPFQ[{1 / 2, 3 / 4}, {1 / 4, 1 / 3}, z], {z, -1, 0.9}]ComplexContourPlot[Re[BilateralHypergeometricPFQ[{1 / 2, 3 / 4}, {1 / 4, 1 / 3}, z]], {z, -1 / 2 - 1 / 2I, 1 / 2 + 1 / 2I}]ComplexContourPlot[Im[BilateralHypergeometricPFQ[{1 / 2, 3 / 4}, {1 / 4, 1 / 3}, z]], {z, -1 / 2 - 1 / 2I, 1 / 2 + 1 / 2I}]函数性质 (3)
BilateralHypergeometricPFQ[{1 / 2, 3 / 2}, {9 / 4, 13 / 4}, 1.]BilateralHypergeometricPFQ[{a1, a2, a3}, {b1, b2}, z] == BilateralHypergeometricPFQ[{a3, a2, a1}, {b2, b1}, z]TraditionalForm[BilateralHypergeometricPFQ[{a1, a2, a3}, {b1, b2}, z]]应用 (1)
通过 BilateralHypergeometricPFQ 计算双向无穷和:
Sum[(Pochhammer[a, k]/Pochhammer[b, k])z^k, {k, -∞, ∞}]Sum[(Pochhammer[1 / 2, n]/Pochhammer[1, n])z^n, {n, -∞, ∞}]//AbsoluteTimingSum[(Pochhammer[1 / 3, n]/Pochhammer[1, n])z^n, {n, -∞, ∞}]//AbsoluteTimingSum[(Pochhammer[a1, n]Pochhammer[a2, n]/Pochhammer[b1, n]Pochhammer[b2, n])z^n, {n, -∞, ∞}]//AbsoluteTiming属性和关系 (2)
BilateralHypergeometricPFQ 可以写成两个 HypergeometricPFQ 之和:
BilateralHypergeometricPFQ[{1 / 2, 3 / 4}, {1 / 4, 1 / 3}, 3 / 10`20]HypergeometricPFQ[{1 / 2, 3 / 4, 1}, {1 / 4, 1 / 3}, 3 / 10`10] + HypergeometricPFQ[{1, 3 / 4, 2 / 3}, {1 / 2, 1 / 4}, 10 / 3] - 1BilateralHypergeometricPFQ 可以简化为初等函数:
BilateralHypergeometricPFQ[{1 / 2}, {1}, z]可能存在的问题 (1)
当
时,BilateralHypergeometricPFQ 使用 Borel 正则化,这可能很耗时:
BilateralHypergeometricPFQ[{1 / 2, 3 / 4}, {1 / 4, 1 / 3, 1 / 5}, 0.3]//AbsoluteTimingBilateralHypergeometricPFQ[{1 / 2, 3 / 4}, {1 / 4, 1 / 3}, 0.3]//AbsoluteTiming巧妙范例 (1)
BilateralHypergeometricPFQ 可以自动简化为更简单的特殊函数:
BilateralHypergeometricPFQlist = Inactivate[...];Grid[...]//TraditionalForm
文本
Wolfram Research (2024),BilateralHypergeometricPFQ,Wolfram 语言函数,https://reference.wolfram.com/language/ref/BilateralHypergeometricPFQ.html.
CMS
Wolfram 语言. 2024. "BilateralHypergeometricPFQ." Wolfram 语言与系统参考资料中心. Wolfram Research. https://reference.wolfram.com/language/ref/BilateralHypergeometricPFQ.html.
APA
Wolfram 语言. (2024). BilateralHypergeometricPFQ. Wolfram 语言与系统参考资料中心. 追溯自 https://reference.wolfram.com/language/ref/BilateralHypergeometricPFQ.html 年
BibTeX
@misc{reference.wolfram_2026_bilateralhypergeometricpfq, author="Wolfram Research", title="{BilateralHypergeometricPFQ}", year="2024", howpublished="\url{https://reference.wolfram.com/language/ref/BilateralHypergeometricPFQ.html}", note=[Accessed: 10-August-2026]}
BibLaTeX
@online{reference.wolfram_2026_bilateralhypergeometricpfq, organization={Wolfram Research}, title={BilateralHypergeometricPFQ}, year={2024}, url={https://reference.wolfram.com/language/ref/BilateralHypergeometricPFQ.html}, note=[Accessed: 10-August-2026]}