Mathematical Appendix
- 4 minutes read - 823 wordsThe mathematical appendix contains some remainders of useful mathematical results of general interest, remainders of frequently used calculus rules, and frequently used symbols of the material.
Used limit results
- Let \(f\), \(g\), \(h\) be functions.
- Let \(c\), \(L\) be real numbers, and \(\alpha>0\).
Continuous mapping theorem
- If \(f\) is continuous at \(x_{0}\),
- then \({\displaystyle \lim _{x\to x_{0}}{f(x)}=f'(x)}\)
Squeeze theorem
- If \({\displaystyle \lim _{x\to c}f(x)=\lim _{x\to c}h(x)=L}\) and \({\displaystyle f(x)\leq g(x)\leq h(x)}\),
- then \({\displaystyle \lim _{x\to c}g(x)=L}\).
L’Hôpital’s rule
- If \({\displaystyle \lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0{\text{ or }}\pm \infty ,}\) and \({\displaystyle g'(x)\neq 0}\), and \({\displaystyle \lim _{x\to c}{\frac {f'(x)}{g'(x)}}}\) exists,
- then \({\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}}\).
Limits of powers
\begin{align*} & {\displaystyle \lim _{x\to x_{0}} c = c} \\ & {\displaystyle \lim _{x\to \infty} x^{\alpha} = \infty} \\ & {\displaystyle \lim _{x\to \infty} x^{-\alpha} = \lim _{x\to \infty} \frac{1}{x^{\alpha}} = 0} \\ & {\displaystyle \lim _{x\to c} x^{\alpha} = c^{\alpha}} \\ & {\displaystyle \lim _{x\to c} x^{-\alpha} = c^{-\alpha}} \end{align*}
Limits of exponentials
\begin{align*} & {\displaystyle \lim _{x\to \infty} \alpha^{x} = \left\{ \begin{matrix} &\infty & \alpha > 1 \\ &1 & \alpha = 1 \\ &0 & \alpha < 1 \end{matrix} \right.} \\ & {\displaystyle \lim _{x\to \infty} \alpha^{-x} = \lim _{x\to \infty} \frac{1}{\alpha^{x}} = \left\{ \begin{matrix} &0 & \alpha > 1 \\ &1 & \alpha = 1 \\ &\infty & \alpha < 1 \end{matrix} \right.} \end{align*}
Used derivative results
- Let \(f\), \(g\) be functions.
- Let \(c\) be any real number, and \(\alpha>0\).
Definition of the derivative
- \({\displaystyle \lim _{h\to 0}{f(x+h)-f(x) \over h}=f'(x)}\)
Inverse function theorem
- If \(f\) is a continuously differentiable function with \(f'(x_{0})\) at the point \(x_{0}\), then
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\(f\) is invertible in a neighborhood of \(x_{0}\),
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the inverse is continuously differentiable, and
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the derivative of the inverse function at \(y_{0}=f(x_{0})\) is the reciprocal of the derivative of \(f\) at \(x_{0}\), i.e.
\begin{align*} \left(f^{-1}\right)'(y_{0}) = \frac{1}{f'(x_{0})} = \frac{1}{f'\left(f^{-1}(y_{0})\right)}. \end{align*}
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Implicit function theorem
- Let \(f\) be a continuously differentiable function of two variables and a \((\hat x_{1}, \hat x_{2})\) be a point so that \(f(\hat x_{1}, \hat x_{2}) = \bar f\). If \(f_{x_{2}}(\hat x_{1}, \hat x_{2}) \neq 0\), then
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there is a neighborhood of \((\hat x_{1}, \hat x_{2})\) and an implicit function \(g\) such that
\begin{align*} f(x_{1}, g(x_{1})) = c \end{align*}
for all \(x_{1}\) in the neighborhood, and
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the derivative of the implicit function is given by
\begin{align*} g'(x_{1}) = \frac{\mathrm{d} x_{2}}{\mathrm{d} x_{1}} = - \frac{f_{x_{1}}(x_{1}, g(x_{1}))}{f_{x_{2}}(x_{1}, g(x_{1}))} \end{align*}
for all \(x_{1}\) in the neighborhood.
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Basic differentiation rules
\begin{align*} &{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(f+g)={\frac {\mathrm {d} f}{\mathrm {d} x}}+{\frac {\mathrm {d} g}{\mathrm {d} x}}} \\ &{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(c\cdot f)=c\cdot {\frac {\mathrm {d} f}{\mathrm {d} x}}}\\ & {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}(f\cdot g)=f\cdot {\frac {\mathrm {d} g}{\mathrm {d} x}}+g\cdot {\frac {\mathrm {d} f}{\mathrm {d} x}}}\\ & {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\left({\frac {f}{g}}\right)={\dfrac {-f\cdot {\dfrac {\mathrm {d} g}{dx}}+g\cdot {\dfrac {\mathrm {d} f}{\mathrm {d} x}}}{g^{2}}}} \\ & {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}[f(g(x))]={\frac {\mathrm {d} f}{\mathrm {d} g}}\cdot {\frac {\mathrm {d} g}{\mathrm {d} x}}=f'(g(x))\cdot g'(x)} \\ & {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\left({\frac {1}{f}}\right)=-{\frac {f'}{f^{2}}}} \end{align*}
Differentiation of powers
\begin{align*} & {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}( c)=0} \\ & {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}{\sqrt {x}}={\frac {1}{2{\sqrt {x}}}}} \\ & {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}x=1} \\ & {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}{\frac {1}{x}}=-{\frac {1}{x^{2}}}} \end{align*}
And more generally
\begin{align*} & {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}x^{c}=c x^{c-1}} \end{align*}
Differentiation of exponentials and logarithms
\begin{align*} & {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\mathrm{e}^{x}=\mathrm{e}^{x}} \\ & {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}a^{x}=a^{x}\ln(a)} \\ & {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\ln(x)={\frac {1}{x}}} \end{align*}
Elasticity
- For a differentiable function \(f\neq 0\), the elasticity is defined by \[ e^{f}(x) = f_{x}(x)\frac{x}{f(x)} = \frac{\mathrm{d}\, f(x)}{\mathrm{d}\, x}\frac{x}{f(x)}. \]
- For appropriate functions \(y = f(x)\), by the inverse function theorem, we also have \[ e^{f}(x) = \frac{1}{f^{-1}_{y}(y)}\frac{x}{f(x)} = \frac{1}{\frac{\mathrm{d}\, x}{\mathrm{d}\, y}} \frac{x}{f(x)}. \]
Table of integrals
- Let \(f\), \(g\) be differentiable functions and \(f'\), \(g'\) be their derivatives.
- Let \(c\) be any real number, and \(\alpha>0\).
Definite and indefinite integration
- Indefinite integral (is a function)
\[ f(x) = \int_{0}^{x} f'(z) \mathrm{d}\, z, \]
- also usually denoted as \[ \int f'(x) \mathrm{d}\, z + K, \] where \(K\) is a constant.
- Definite integral (is a number) \[\int_{x_{0}}^{x_{1}} f'(x) \mathrm{d}\, x = f(x_{1}) - f(x_{0}).\]
Basic integration rules
\begin{align*} & \int [f(x) + g(x)] \mathrm{d}\, x = \int f(x) \mathrm{d}\, x + \int g(x) \mathrm{d}\, x \\ & \int c f(x) \mathrm{d}\, x = c \int f(x) \mathrm{d}\, x \\ & \int f(x) g'(x) \mathrm{d}\, x = f(x) g(x) - \int f'(x) g(x) \mathrm{d}\, x \end{align*}
Integration of powers
\begin{align*} & \int c\ \mathrm{d}\, x = c x + K \\ & \int x^{c} \mathrm{d}\, x = \frac{x^{c + 1}}{c + 1} + K \\ \end{align*}
Integration of exponentials
\begin{align*} & \int e^{x} \mathrm{d}\, x = e^{x} + K \\ \end{align*}