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linear_approximation [2014/02/13 16:13]
127.0.0.1 external edit
linear_approximation [2014/03/21 11:11]
127.0.0.1 external edit
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 ===== Linear approximation ===== ===== Linear approximation =====
 ==== Set ==== ==== Set ====
-| @#88DDEE: $X,Y$ ... Banach spaces with topology | +| @#55CCEE: context ​    | @#55CCEE: $X,Y$ ... Banach spaces with topology | 
-| @#88DDEE: $\mathcal O$ ... open in $X$ | +| @#55CCEE: context ​    | @#55CCEE: $\mathcal O$ ... open in $X$ | 
-| @#88DDEE: $x\in\mathcal O$ | +| @#55CCEE: context ​    | @#55CCEE: $x\in\mathcal O$ | 
-| @#88DDEE: $f:\mathcal O\to Y$ |+| @#55CCEE: context ​    | @#55CCEE: $f:\mathcal O\to Y$ |
  
-| @#FFBB00: $J_x^f$ |+| @#FFBB00: definiendum ​| @#FFBB00: $J_x^f$ |
  
-| @#55EE55: $J_x^f$ ... bounded linear operator from $X$ to $Y$ | +| @#55EE55: postulate ​  | @#55EE55: $J_x^f$ ... bounded linear operator from $X$ to $Y$ | 
-| @#55EE55: $\mathrm{lim}_{h\to 0}\ \Vert f(x+h)-f(x)-J_x^f(h)\Vert / \Vert h\Vert\ =\ 0$ |+| @#55EE55: postulate ​  | @#55EE55: $\mathrm{lim}_{h\to 0}\ \Vert f(x+h)-f(x)-J_x^f(h)\Vert / \Vert h\Vert\ =\ 0$ |
  
 ==== Discussion ==== ==== Discussion ====
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   * The field $\mathbb C$ is $\mathbb R^2$ as a vector space. So to compute the linear approximation of a function $u(x,y)+i v(x,y)$, where $x,y$ and the functions $u,v$ are real, we must only identify it with the vector field $\langle u(x,​y),​v(x,​y)\rangle$.   * The field $\mathbb C$ is $\mathbb R^2$ as a vector space. So to compute the linear approximation of a function $u(x,y)+i v(x,y)$, where $x,y$ and the functions $u,v$ are real, we must only identify it with the vector field $\langle u(x,​y),​v(x,​y)\rangle$.
  
-Holomorphic functions: The complex numbers are an algebraic field and are hence equipped with the structure of complex multiplication map $\mathbb C\times \mathbb C\to\mathbb C$. If $j=j_1+i j_2$ and $d=d_1+i d_2$, then the multiplication is given by  +Much more on this in the article [[holomorphic function]].
- +
-$j\cdot d = (j_1\cdot d_1-j_2\cdot d_2)+i (j_2\cdot d_1+j_1\cdot d_2) \overset{\sim}{=} \langle j_1\cdot d_1+(-j_2)\cdot d_2,​j_2\cdot d_1+j_1\cdot d_2\rangle$ +
- +
-Let's compare ​this with the $2\times 2$ Jacobi matrix above, to see if the linear approximation at a point $z_0=x_0+i y_0$ of a complex function along $d=\langle d^1,​d^2\rangle$ can be formulated as a multiplication within $\mathbb C$, i.e. by multiplying a $d$-independent function $f'$ with the number $d^1+i d^2$. If it exists, it's given by +
- +
-$f'​(z_0) = \lim_{z \to z_0} {f(z) - f(z_0) \over z - z_0 }$ +
- +
-We call these functions //​holomorphic//:​ +
- +
-| @#EEEE55: $f:\mathcal O\to \mathbb C$ ... holomorphic on $\mathcal{O}\ \equiv\ \forall(z_0\in\mathcal O).\ \exists (f':​\mathcal O\to \mathbb C).\ J_{z_0}^f(d)=f'​(z_0)\cdot(d^1+i d^2) $ | +
-| @#EEEE55: $f:\mathcal O\to \mathbb C$ ... holomorphic on $\mathcal{O}\ \equiv\ \forall(z_0\in\mathcal O).\ \lim_{z \to z_0} {f(z) - f(z_0) \over z - z_0 }\in\mathbb C $ | +
- +
-Comparing ​the first components tells us that  +
- +
-$j_1=\frac{\partial u}{\partial x}$ +
- +
-$j_2=-\frac{\partial u}{\partial y}$  +
- +
-Comparing the second components then gives a strong necessary condition on $f$, which can also shown to be sufficient +
- +
-==Cauchy–Riemann equations== +
- +
-^ $\frac{ \partial u }{ \partial x } = \frac{ \partial v }{ \partial y }$ ^ +
-^ $\frac{ \partial u }{ \partial y } = -\frac{ \partial v }{ \partial x }$ ^ +
- +
-Hence +
- +
-^ $f$ ... holomorphic $\implies f'(x+i y)=\frac{\partial u}{\partial x}-i \frac{\partial u}{\partial y}$ ^ +
- +
-So for a holomorphic function ​$f$, the total change (a complex value) is determined by the real part (or alternatively complex part) of $f$ alone. +
- +
-=== Theorems === +
-The //​Looman–Menchoff theorem// gives a converse to the statement above: If $f$ is continuous, $u$ and $v$ have first partial derivatives (but not necessarily continuous),​ and they satisfy the Cauchy–Riemann equations, then $f$ is holomorphic. +
- +
-If a function is holomorphic,​ it's continuous. (That'​s not in general true for the derivative of functions $\mathbb R^n\to \mathbb R^n$ if $n>1$, e.g. consider the function $f(x,​y):​=xy/​(x^2+y^2)$ on $x,y\neq 0$ and else $f(0,​0):​=0$.) +
- +
-A holomorphic function, interpreted as a map from the complex plane to the complex plane, is (at each point with  at a point where $f'​(z)\ne 0$) conformal, i.e.preserves the angles of crossing curves. A holomorphism isomorphism is conformal. +
- +
-If $z\in\mathbb C$ and $b\in \mathbb R$, then $f(z):=z^b$ is holomorphic with $f'​(z)=b\ z^{b-1}$, except from maybe at $z=0$. By linearity, this extends to polynomials and infinite sums $\sum_k a_k\ z^{b_k}$. +
- +
-^ $ \mathrm{det}(J_z^f)=|f'​(z)|^2 $ ^+
  
 === Reference === === Reference ===
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