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Math Help - limits, continuity and Cauchy-Riemann equations

  1. #1
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    Question limits, continuity and Cauchy-Riemann equations

    1. a) If f is a real-valued function defined throughout R, except at the origin (0, 0), define what is meant by the statement: f (x, y)→ L as (x, y)→ (0, 0).



    2. a) Define what it means to describe a function f of two real variables as differentiable at (a, b)? Define (as limits) the partial derivatives df/dx and df/dy at (a, b) and prove that if f is differentiable at (a, b) then both these partial derivatives exist

    b) If g(x, y) = xy prove that g is not differentiable at (0, b) for any non-value of b

    1. a) The Cauchy-Riemann equation is the name given to the following pair of equations,
    ∂u/∂x=∂v/∂y and ∂u/∂y= -∂v/∂x which connects the partial derivatives of two functions u(x, y) and v(x, y)

    i) if u(x, y) =e^x cosy and v(x, y) =e^x siny, how do I prove that these functions satisfy the Cauchy-Riemann equations
    ii) if u(x, y) = In(x +y) and v(x, y) = tanֿ(y/x), how do I prove that theses functions satisfy the Cauchy-Riemann equations
    iii) if u and v are any functions that satisfy the Cauchy-Riemann equations,
    how do I prove that ∂u/∂x + ∂u/∂y=0

    b) If f is a real valued function of two variables, the set of points (x, y) for which f(x, y)=c, for some value of the constant c, is called a level curve( or contour line) of the function. How do I illustrate the level curves for the following functions:
    i) f(x, y) = x +y
    ii) g(x, y) = xy

    How would I calculate the gradient vectors of these functions and confirm in each case that the direction of this vector at any point is normal to the level curve passing through it
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  2. #2
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    Quote Originally Posted by freya81
    1. a) If f is a real-valued function defined throughout R, except at the origin (0, 0), define what is meant by the statement: f (x, y)→ L as (x, y)→ (0, 0).
    This is just book-work, you should be able to just look it
    up in any text book or your notes.

    \ f (\bold{x})\ \rightarrow\ L\ as\  \bold{x}\  \rightarrow\ 0\

    means that for all \varepsilon\ >\ 0 there exists \delta\ >\ 0 such that

    \|\bold{x}\|<\delta\ \Rightarrow\ \|f(\bold{x})-L\|\ <\ \varepsilon

    where \|\cdot\| in its first occurrence above is the 2-vector
    norm,and in its second occurrence is the absolute value, that is they
    denote the appropriate norms for the domain and range of f
    respectivly.

    RonL
    Last edited by CaptainBlack; November 22nd 2005 at 02:34 AM.
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  3. #3
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    1. a) For every E>0 there exists a D>0 such that if |(x,y)|<D then |f(x,y)-L|<E; usually, E is called "epsilon" and written as a lowercase epsilon, and D is called "delta" and written as a lowercase delta; also, in general D depends on E.

    2. a) f is differentiable at (a,b) iff there exists a linear mapping L such that lim((f(a+h,b+k)-f(a,b)-L(h,k))/|(h,k)|,(h,k)->(0,0))=0. ∂f/∂x(a,b)=lim((f(a+h,b)-f(a,b))/h,h->0), and ∂f/∂y(a,b)=lim((f(a,b+k)-f(a,b))/k,k->0).
    In fact, L=grad f; that is, L(h,k)=(∂f/∂x,∂f/∂y).(h,k)=h*∂f/∂x+k*∂f/∂y.
    Try breaking up the sum in the first expression, so that you get something that looks like the expressions for the partial derivatives.
    b) Did you copy this one correctly? If I recall, xy is differentiable (in fact, analytic) for all x and y.

    1. a)
    i-ii)Find the partial derivatives (considered as functions of real variables) and verify that they are the same.
    iii) Take the u-partial derivative of one equation and the v-partial of the other, and use the fact that for a twice-differentiable function, the "mixed partials" are equal.
    b) Set the functions equal to constants: In one case, you get circles, and in the other, you get hyperbolas.
    See above to calculate the gradient; to see that it is normal to the level curves, use "implicit differentiation" to get, say, dy/dx for the level curves, considered locally as graphs of functions of x. Then the tangent vector is (1,dy/dx). Now take the inner ("dot") product of the tangent and normal and you should get 0; note that for the level curve given by xy=0, you won't be able to do this at the origin because the curve is not differentiable there.
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