Finding the Maclaurin Expansion for \(f(x) =(1+x) ^{m}\)

Find the Maclaurin series for \(f(x) =(1+x) ^{m},\) where \(m\) is any real number.

Solution We begin by finding the derivatives of \(f\) at \(0\): \[ \begin{array}{@{\hspace{-8pc}}ll@{\qquad}rll} f(x)&=&(1+x)^{m} && f(0)&=&1 \\ f^\prime (x)&=&m(1+x)^{m-1} && f^{\prime} (0)&=&m \\ f^{\prime \prime} (x)&=&m(m-1)(1+x)^{m-2} && f^{\prime \prime} (0)&=&m(m-1) \\ \vdots & & & & & \vdots & & & &\\ f^{(n)}(x)&=&m(m-1)(m-2)\cdots (m-n+1)(1+x)^{m-n} && f^{(n)}(0) &=& m(m-1)(m-2)\,\cdots \,(m-n+1) \end{array} \]

The Maclaurin expansion for \(f\) is \[ \begin{eqnarray*} (1+x)^{m} &=&1+m\,x+\frac{m(m-1)}{2!}x^{2}+\cdots \\[4pt] && +\, \frac{m(m-1)(m-2)\,\cdots \,(m-n+1)}{n!}x^{n}+\cdots \\[4pt] &=&{m\choose 0}+{m\choose 1}x+{m\choose 2}x^{2}+\cdots +{m\choose n}x^{n}+\cdots =\sum_{k\,=\,0}^{\infty }{m\choose k}\,x^{k} \end{eqnarray*} \]

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where \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ {m\choose 0} =1 \quad\hbox{and}\quad {m\choose k} =\frac{m (m-1) (m-2) \,\cdots\, (m-k+1) }{k!} }} \]