Mathematical Appendix

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The 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
- \(f\) is invertible in a neighborhood of \(x_{0}\),
- the inverse is continuously differentiable, and
- 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*}

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
- 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
- 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.

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*}