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当据a href="//www.parkandroid.com/wiki/binomial-theorem-n-choose-k/" class="wiki_link" title="二项式定理" target="_blank">二项式定理据/a>对于正整数指数据span class="katex"> N.据/span>可以概括为负整数指数。这产生了几个熟悉的据a href="//www.parkandroid.com/wiki/taylor-series/" class="wiki_link" title="Maclaurin系列" target="_blank">Maclaurin系列据/a>在微积分和其他数学领域的许多应用。据/p>
F.据/span>(据/span>X.据/span>)据/span>=据/span>(据/span>1据/span>+据/span>X.据/span>)据/span>-据/span>3.据/span>不是多项式。虽然积极的力量据span class="katex">
1据/span>+据/span>X.据/span>可以扩展到多项式中,例如多项式。据span class="katex">
(据/span>1据/span>+据/span>X.据/span>)据/span>3.据/span>=据/span>1据/span>+据/span>3.据/span>X.据/span>+据/span>3.据/span>X.据/span>2据/span>+据/span>X.据/span>3.据/span>那据span class="katex">
F.据/span>(据/span>X.据/span>)据/span>不能,因此不等于的单体术语的有限总和据span class="katex">
F.据/span>(据/span>X.据/span>)据/span>。但如果允许无限的总和,有一种方法可以恢复相同类型的扩展。据/p>
作为第一个近似,自从据span class="katex">
F.据/span>'据/span>(据/span>0.据/span>)据/span>=据/span>-据/span>3.据/span>由这件事据a href="//www.parkandroid.com/wiki/power-rule/" class="wiki_link" title="电力规则" target="_blank">电力规则据/a>,切线线据span class="katex">
X.据/span>=据/span>0.据/span>是据span class="katex">
y据/span>=据/span>1据/span>-据/span>3.据/span>X.据/span>。所以小小的据span class="katex">
X.据/span>那据span class="katex-display">
(据/span>1据/span>+据/span>X.据/span>)据/span>3.据/span>1据/span>≈据/span>1据/span>-据/span>3.据/span>X.据/span>。据/span>该近似已经非常有用,但是可以使用系列更仔细地仔细地近似函数。据/p>
拓展据span class="katex">
(据/span>1据/span>+据/span>X.据/span>)据/span>3.据/span>1据/span>作为Maclaurin系列。据/p>
Maclaurin系列据span class="katex">
F.据/span>(据/span>X.据/span>)据/span>,无论它收敛,都可以表示为据/p>
F.据/span>(据/span>X.据/span>)据/span>=据/span>F.据/span>(据/span>0.据/span>)据/span>+据/span>F.据/span>'据/span>(据/span>0.据/span>)据/span>X.据/span>+据/span>2据/span>!!据/span>F.据/span>'据/span>'据/span>(据/span>0.据/span>)据/span>X.据/span>2据/span>+据/span>⋯据/span>+据/span>K.据/span>!!据/span>F.据/span>(据/span>K.据/span>)据/span>(据/span>X.据/span>)据/span>X.据/span>K.据/span>+据/span>⋯据/span>。据/span> 让我们据span class="katex">
F.据/span>(据/span>X.据/span>)据/span>=据/span>(据/span>1据/span>+据/span>X.据/span>)据/span>3.据/span>1据/span>。应用这一点据a href="//www.parkandroid.com/wiki/power-rule/" class="wiki_link" title="电力规则" target="_blank">电力规则据/a>反复,我们有据span class="katex-display">
F.据/span>(据/span>X.据/span>)据/span>=据/span>(据/span>1据/span>+据/span>X.据/span>)据/span>-据/span>3.据/span>F.据/span>'据/span>(据/span>X.据/span>)据/span>=据/span>-据/span>3.据/span>(据/span>1据/span>+据/span>X.据/span>)据/span>-据/span>4.据/span>F.据/span>'据/span>'据/span>(据/span>X.据/span>)据/span>=据/span>(据/span>-据/span>3.据/span>⋅据/span>-据/span>4.据/span>)据/span>(据/span>1据/span>+据/span>X.据/span>)据/span>-据/span>5.据/span>F.据/span>(据/span>K.据/span>)据/span>(据/span>X.据/span>)据/span>=据/span>-据/span>3.据/span>⋅据/span>-据/span>4.据/span>⋯据/span>(据/span>-据/span>3.据/span>-据/span>K.据/span>+据/span>1据/span>)据/span>(据/span>1据/span>+据/span>X.据/span>)据/span>-据/span>3.据/span>-据/span>K.据/span>⟹据/span>F.据/span>(据/span>0.据/span>)据/span>=据/span>1据/span>⟹据/span>F.据/span>'据/span>(据/span>0.据/span>)据/span>=据/span>-据/span>3.据/span>⟹据/span>F.据/span>'据/span>'据/span>(据/span>0.据/span>)据/span>=据/span>-据/span>3.据/span>⋅据/span>-据/span>4.据/span>⋮据/span>⟹据/span>F.据/span>(据/span>K.据/span>)据/span>(据/span>0.据/span>)据/span>=据/span>-据/span>3.据/span>⋅据/span>-据/span>4.据/span>⋯据/span>(据/span>-据/span>3.据/span>-据/span>K.据/span>+据/span>1据/span>)据/span>。据/span> 所以Maclaurin系列变成了据/p>
F.据/span>(据/span>X.据/span>)据/span>=据/span>1据/span>-据/span>3.据/span>X.据/span>+据/span>2据/span>!!据/span>-据/span>3.据/span>⋅据/span>-据/span>4.据/span>X.据/span>2据/span>+据/span>⋯据/span>+据/span>K.据/span>!!据/span>-据/span>3.据/span>⋅据/span>-据/span>4.据/span>⋯据/span>(据/span>-据/span>3.据/span>-据/span>K.据/span>+据/span>1据/span>)据/span>X.据/span>K.据/span>+据/span>⋯据/span>。据/span> 这会融合据span class="katex">
|据/span>X.据/span>|据/span>据据/span>1据/span>由这件事据a href="//www.parkandroid.com/wiki/convergence-ratio-test/" class="wiki_link" title="比率测试" target="_blank">比率测试据/a>。据span class="katex">
□据/span>
以上示例立即推广所有负整数指数据span class="katex"> α.据/span>。让我们据span class="katex"> α.据/span>是一个实数和据span class="katex"> K.据/span>一个正整数。界定据/p>
(据/span>K.据/span>α.据/span>)据/span>=据/span>K.据/span>!!据/span>α.据/span>(据/span>α.据/span>-据/span>1据/span>)据/span>......据/span>(据/span>α.据/span>-据/span>K.据/span>+据/span>1据/span>)据/span>=据/span>K.据/span>!!据/span>(据/span>α.据/span>-据/span>K.据/span>)据/span>!!据/span>α.据/span>!!据/span>那据/span>
然后与示例中的分析相同据/p>
让我们据span class="katex"> N.据/span>是一个正整数。然后据span class="katex-display"> (据/span>1据/span>+据/span>X.据/span>)据/span>N.据/span>1据/span>=据/span>1据/span>-据/span>N.据/span>X.据/span>+据/span>2据/span>(据/span>-据/span>N.据/span>)据/span>(据/span>-据/span>N.据/span>-据/span>1据/span>)据/span>X.据/span>2据/span>+据/span>⋯据/span>=据/span>K.据/span>=据/span>0.据/span>σ.据/span>∞据/span>(据/span>K.据/span>N.据/span>+据/span>K.据/span>-据/span>1据/span>)据/span>(据/span>-据/span>1据/span>)据/span>K.据/span>X.据/span>K.据/span>对于据span class="katex"> |据/span>X.据/span>|据/span>据据/span>1据/span>。据p>
让我们据span class="katex"> N.据/span>是一个正整数。然后据span class="katex"> (据/span>N.据/span>-据/span>N.据/span>)据/span>=据/span>__________据/span>。据/span>
对于据span class="katex"> N.据/span>=据/span>2据/span>, 我们有据/p>
(据/span>K.据/span>-据/span>2据/span>)据/span>=据/span>K.据/span>!!据/span>(据/span>-据/span>2据/span>)据/span>(据/span>-据/span>3.据/span>)据/span>......据/span>(据/span>-据/span>K.据/span>-据/span>1据/span>)据/span>=据/span>(据/span>-据/span>1据/span>)据/span>K.据/span>(据/span>K.据/span>+据/span>1据/span>)据/span>。据/span>
所以据/p>
(据/span>1据/span>+据/span>X.据/span>)据/span>2据/span>1据/span>=据/span>K.据/span>=据/span>0.据/span>σ.据/span>∞据/span>(据/span>K.据/span>-据/span>2据/span>)据/span>X.据/span>K.据/span>=据/span>K.据/span>=据/span>0.据/span>σ.据/span>∞据/span>(据/span>K.据/span>+据/span>1据/span>)据/span>(据/span>-据/span>X.据/span>)据/span>K.据/span>。据/span>
堵塞据span class="katex"> X.据/span>=据/span>-据/span>2据/span>1据/span>例如,给出据/p>
4.据/span>=据/span>K.据/span>=据/span>0.据/span>σ.据/span>∞据/span>2据/span>K.据/span>K.据/span>+据/span>1据/span>。据/span>
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