复变函数问题为什么对于复数z(t)=x(t)+iy(t)在复平面C上光滑要求对任意的t(t在它自身的定义域内）x(t)的

复变函数问题
为什么对于复数z(t)=x(t)+iy(t)在复平面C上光滑要求对任意的t(t在它自身的定义域内）x(t)的导数平方与y(t)的平方的和不等于零? 餐餐吃鱼 1年前他留下的回答 已收到2个回答

luxfeliya 网友

该名网友总共回答了18个问题，此问答他的回答如下：采纳率：94.4%

就是说y(t),x(t)的导数不同时为0拉.你的那个z(t)=x(t)+iy(t)就可以看成是R到C的函数.它的smoothness 就相当于smoothness between manifold.
以下来自wiki,smooth function
第一个是smooth function的定义
第二个是smoothness between manifolds 的定义.
Consider an open set on the real line and a function f defined on that set with real values.Let k be a non-negative integer.The function f is said to be of class Ck if the derivatives f',f'',...,f(k) exist and are continuous (the continuity is automatic for all the derivatives except for f(k)).The function f is said to be of class C∞,or smooth,if it has derivatives of all orders.f is said to be of class Cω,or analytic,if f is smooth and if it equals its Taylor series expansion around any point in its domain.
Smooth functions between manifolds
Smooth maps between smooth manifolds may be defined by means of charts,since the idea of smoothness of function is independent of the particular chart used.If F is a map from an m-manifold M to an n-manifold N,then F is smooth if,for every scriptstyle p in M,there is a chart scriptstyle (U,varphi) in M containing p and a chart scriptstyle (V,psi) in N containing F(p) with scriptstyle F(U) subset V,such that scriptstylepsicirc F circ varphi^{-1} is a smooth from scriptstylevarphi(U) to scriptstylepsi(V) as a function from scriptstylemathbb{R}^m to scriptstylemathbb{R}^n.
Such a map has a first derivative defined on tangent vectors; it gives a fibre-wise linear mapping on the level of tangent bundles.

1年前他留下的回答

3

愤怒555 网友

该名网友总共回答了8个问题，此问答他的回答如下：

这个其实应该算是微分几何的问题。本来光滑的意思就是指导数连续。那么z(t)=x(t)+iy(t)光滑就可以当作是参数曲线来处理，基本理念是一个质点在曲线上运动，那么他要走不光滑的路线肯定要在不光滑的点上停下来。即在不光滑的点上速度[x(t)的导数平方与y(t)的平方的和]开根不为0。其实这是最低要求。例如x(t)=t^3,y(t)=t^2。那么在(0,0)点是不光滑的。...

1年前他留下的回答

2

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