Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision Next revision | Previous revision | ||
|
linear_approximation [2014/02/13 16:13] 127.0.0.1 external edit |
linear_approximation [2014/04/11 16:42] nikolaj |
||
|---|---|---|---|
| Line 1: | Line 1: | ||
| ===== 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: definiendum | @#FFBB00: $J_x^f$ | | |
| - | | @#FFBB00: $J_x^f$ | | + | | @#55EE55: postulate | @#55EE55: $J_x^f$ ... bounded linear operator from $X$ to $Y$ | |
| - | + | | @#55EE55: postulate | @#55EE55: $\mathrm{lim}_{h\to 0}\ \Vert f(x+h)-f(x)-J_x^f(h)\Vert / \Vert h\Vert\ =\ 0$ | | |
| - | | @#55EE55: $J_x^f$ ... bounded linear operator from $X$ to $Y$ | | + | |
| - | | @#55EE55: $\mathrm{lim}_{h\to 0}\ \Vert f(x+h)-f(x)-J_x^f(h)\Vert / \Vert h\Vert\ =\ 0$ | | + | |
| ==== Discussion ==== | ==== Discussion ==== | ||
| Line 45: | Line 43: | ||
| * 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 === | ||
