Taylor Series Expansions of Popular Functions
1. \(f(x) = (1 - x)\)
The valid range for this function is \(|x|
< 1\).
The Taylor Series expansion of \(f(x)\) around \(x
= 0\) is:
\[ f(x) = \sum_{n=0}^{\infty}
\frac{{f^{(n)}(0)}}{{n!}} x^n \]
where \(f^{(n)}(0)\) represents the
\(n\)-th derivative of \(f(x)\) evaluated at \(x = 0\).
For \(f(x) = (1 - x)\), the
derivatives are:
\[
\begin{align*}
f(x) &= (1 - x) \\
f'(x) &= -1 \\
f''(x) &= 0 \\
f'''(x) &= 0 \\
&\vdots \\
f^{(n)}(x) &= 0 \text{ for } n \geq 2
\end{align*}
\]
Thus, the Taylor Series expansion simplifies to:
\[ f(x) = 1 - x \]
2. \(f(x) = e^x\)
The valid range for this function is \(-\infty < x < \infty\).
The Taylor Series expansion of \(f(x)\) around \(x
= 0\) is:
\[ f(x) = \sum_{n=0}^{\infty}
\frac{{f^{(n)}(0)}}{{n!}} x^n \]
where \(f^{(n)}(0)\) represents the
\(n\)-th derivative of \(f(x)\) evaluated at \(x = 0\).
For \(f(x) = e^x\), the derivatives
are:
\[
\begin{align*}
f(x) &= e^x \\
f'(x) &= e^x \\
f''(x) &= e^x \\
f'''(x) &= e^x \\
&\vdots \\
f^{(n)}(x) &= e^x \text{ for all } n
\end{align*}
\]
Thus, the Taylor Series expansion is the Maclaurin series of \(e^x\), which is:
\[ f(x) = \sum_{n=0}^{\infty}
\frac{{x^n}}{{n!}} \]
3. \(f(x) = \ln(1 + x)\)
The valid range for this function is \(|x|
< 1\).
The Taylor Series expansion of \(f(x)\) around \(x
= 0\) is:
\[ f(x) = \sum_{n=0}^{\infty}
\frac{{f^{(n)}(0)}}{{n!}} x^n \]
where \(f^{(n)}(0)\) represents the
\(n\)-th derivative of \(f(x)\) evaluated at \(x = 0\).
For \(f(x) = \ln(1 + x)\), the
derivatives are:
\[
\begin{align*}
f(x) &= \ln(1 + x) \\
f'(x) &= \frac{{1}}{{1 + x}} \\
f''(x) &= -\frac{{1}}{{(1 + x)^2}} \\
f'''(x) &= \frac{{2}}{{(1 + x)^3}} \\
&\vdots \\
f^{(n)}(x) &= (-1)^{n-1} \frac{{(n - 1)!}}{{(1 + x)^n}} \text{ for }
n \geq 1
\end{align*}
\]
Thus, the Taylor Series expansion is:
\[ f(x) = \sum_{n=0}^{\infty} (-1)^{n-1}
\frac{{x^n}}{{n}} \]
4. \(f(x) = x^{1/2}\)
The valid range for this function is \(x
\geq 0\).
The Taylor Series expansion of \(f(x)\) around \(x
= a\) is:
\[ f(x) = \sum_{n=0}^{\infty}
\frac{{f^{(n)}(a)}}{{n!}} (x - a)^n \]
where \(f^{(n)}(a)\) represents the
\(n\)-th derivative of \(f(x)\) evaluated at \(x = a\).
For \(f(x) = x^{1/2}\), the
derivatives are:
\[
\begin{align*}
f(x) &= x^{1/2} \\
f'(x) &= \frac{{1}}{{2\sqrt{x}}} \\
f''(x) &= -\frac{{1}}{{4x^{3/2}}} \\
f'''(x) &= \frac{{3}}{{8x^{5/2}}} \\
&\vdots \\
f^{(n)}(x) &= \frac{{(-1)^{n-1}(2n - 3)!!}}{{2^n x^{(2n-1)/2}}}
\text{ for } n \geq 1
\end{align*}
\]
Thus, the Taylor Series expansion is:
\[ f(x) = \sum_{n=0}^{\infty}
\frac{{(-1)^{n-1}(2n - 3)!!}}{{2^n}} (x - a)^{n/2} \]
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