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Find the Taylor polynomial of (1 + x)α

Show that

    \[ T_n ((1+x)^{\alpha}) = \sum_{k=0}^n \binom{\alpha}{k} x^k, \]


    \[ \binom{\alpha}{k} = \frac{\alpha (\alpha - 1) \cdots (\alpha - k +1)}{k!}. \]

Since we do not have the binomial theorem for real powers (we have proved a formula for (a+b)^n for integers n, but in this case the power we have is a real number \alpha), we use induction to determine the kth derivative of the function f(x) = (1+x)^{\alpha}. First, we compute a few derivatives of f(x),

    \begin{align*}  f'(x) &= \alpha (1+x)^{\alpha - 1} \\  f''(x) &= \alpha(\alpha - 1)(1+x)^{\alpha - 2} \\  f'''(x) &= \alpha (\alpha - 1)(\alpha - 2) (1+x)^{\alpha - 3}.  \end{align*}

So, we conjecture

    \[ f^{(n)} (x) = n! \cdot \binom{\alpha}{n} (1+x)^{\alpha - n}.\]

We have shown this is true for n =1, and if we suppose it is true for a positive integer k then we have

    \begin{align*}   f^{(k+1)}(x) = (f^{(k)}(x))' &= \left( k! \binom{\alpha}{k} (1+x)^{\alpha -k} \right)' \\[9pt]  &= k! \binom{\alpha}{k} (\alpha - k) (1+x)^{\alpha - (k+1)} \\[9pt]  &= (k+1)! \binom{\alpha}{k+1} (1+x)^{\alpha - (k+1)}. \end{align*}

Therefore, the formula holds for all positive integers. Now we can compute the Taylor polynomial directly,

    \begin{align*}  T_n (1+x)^{\alpha} &= \sum_{k=0}^n \frac{f^{(k)}(0)}{k!} x^k \\[9pt]  &= \sum_{k=0}^n \binom{\alpha}{k} x^k. \end{align*}

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