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12

Suppose $f: \mathbf{C} \rightarrow \mathbf{C}$ is analytic near $a \in \mathbf{C}$. Geometrically, this can be interpreted at saying that the effect of $f$ on a small vector $h$ emanating from $a$ is just multiplication by the complex number $f'(a)$: $$ f(a+h) \approx f(a) +f'(a)h,$$ where the error tends to zero sublinearly as $|h| \rightarrow 0$. That is, ...


5

Hint: What happens on the curves $$(x(t), y(t)) = (1/t \pm e^{-t}, 1/t), \quad t > 0?$$ Here is a little animation of the surface in question. It was surprisingly challenging to obtain a smooth plot in the neighborhood of $(0,0)$, but I managed to find a nice parametrization. The red and green curves are the ones I described, but it is quite clear ...


5

You've had some complex analysis, so you know what a harmonic function is. Take the gradient of any harmonic function. They also have harmonic functions in three dimensions, same example. You said not to do that. Life is tough. Two dimensional, we can take harmonic function $x^2-y^2,$ which is the real part of $(x+yi)^2,$ to get vector field $$ (2x, -2y). ...


4

First, your desired statement is true, provided that the following conditions are fulfilled $f \geq 0$, $f$ is measurable (w.r.t. the product $\sigma$-algebra), $X,Y$ are $\sigma$-finite measure spaces. The exact statemement (sometimes called Tonelli's theorem, or the Fubini-Tonelli-theorem) is that if $$ \int \int f(x,y) \, dx\, dy < \infty, $$ ...


4

There are two parts to the (weak) Whitney embedding theorem: 1) Any abstract manifold can be embedded in $\mathbb{R}^N$ for some $N$. 2) Any $k$-dimensional submanifold of $\mathbb{R}^N$ can in fact be embedded in $\mathbb{R}^{2k+1}$. They prove part (2) of this. For a proof of (1), there's a nice exposition in Lee, Smooth Manifolds. Here's the idea for ...


4

For all $(x,y)\neq (0,0)$ such that $x\neq y$ one has $f(x,y)=\dfrac{2x^3}{x-y}-x^2-xy-y^2$, so if the limit exists, due to $\lim \limits_{(x,y)\to(0,0)}\left(x^2-xy-y^2\right)$ existing, so does $\lim \limits_{(x,y)\to (0,0)}\left(\dfrac{2x^3}{x-y}\right)$, but this last limit can easily be seen to be $k$-dependent if $y=x-kx^3$. So consider the paths ...


3

This is basically a question of how one converts back and forth between vector calculus and differential forms notation. I suggest using an intermediary notation--the notation of clifford algebra and geometric calculus--which has the simplicity of vector calculus notation and the power of differential forms operations. Let's do it starting with ...


3

You are correct: $\sqrt[3]{x}$ is an odd function, so $\int_{-y}^y\sqrt[3]{x}\,dx$ must be zero. The reason Maple is giving a nonsensical answer is because, viewed as a complex function, $x^{1/3}$ is multivalued, and they used a branch of $x^{1/3}$ which is positive real when $x$ is, and complex when $x$ is negative. It turns out this is the natural choice ...


3

if you wanted to go the long way..by setting $F = -y/x$ then you would have $$ y'' = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}y' $$ or $$ y'' = \frac{y}{x^2} +\left(-\frac{1}{x}\right)\left(-\frac{y}{x}\right) = \frac{2y}{x^2} $$ now the problem is that you have the original result wrong it should not have a minus sign


3

Using geometric calculus--the calculus of clifford algebra--we can write any vector field $F$ in terms of its value on a boundary curve $\partial M$ and its divergence and curl within a region $M$. $$iF(p) = \oint_{\partial M} G(p-p') \, d\ell' \, F(p') + \int_M G(p-p') \, dA' \, \nabla F|_{p'}$$ where $G(p) = p/2\pi p^2$ is the 2d Green's function for ...


3

The difference seems to be between $f'(x) = 0$ at a point $x \in Domain(f)$ vs. $f'(x) = 0$ for all $x \in Domain(f)$ or at least on an open interval in the domain. In the first case, it doesn't mean $f$ is constant. E.g., $f(x) = x^2$ at $x = 0$. However in the second case it does.


3

The equation $$F(cx - az, cy-bz) = 0$$ defines implicitly a function $(x,y)\mapsto z(x,y)$ under reasonable hypothesis (see http://en.wikipedia.org/wiki/Implicit_function_theorem). Applying the chain rule to $$\pmatrix{x\cr y}\longmapsto \pmatrix{cx-az(x,y)\cr cy-bz(x,y)}\longmapsto F(cx - az(x,y), cy-bz(x,y))$$ and using that $F(cx - az(x,y), cy-bz(x,y))=0$ ...


2

Consider $f(x,y)$ as a sort of contour map, with the height $z$ in 3 dimensions equal to $f(x,y)$. The function defines a continuous surface in that space. Now stare at that function from far off in the "South-East" direction, looking North-West. Then if $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}$, the projection of that surface that ...


2

HINT: The set of points, where $f(x,y)=1$ is dense in $[0,1]\times[0,1]$, but not (relatively) dense in $[0,1]$'s on the axes.


2

Here's an example of a function with two maxima and no saddle points. I saw it on a mathematics professor's home page once—I think, someone who was even a prominent user here—and I simply cannot remember who they were. I hope someone will let me know in the comments so I can credit them. Take a function with two maxima and a saddle point between them, say ...


2

That «the curl and divergence form a basis» does not really mean anything. The curl and the divergence are operators acting on vector fields, and they do not form a basis in any sense. The contemplation of any counterexample to your claim should provide ample food for thought...


2

You should know that $\nabla f(1,1,1)$ is a normal vector to the surface at said point. Hence, the normal line is: $$L: \quad {\bf X}(t) = (1,1,1) + t \nabla f (1,1,1), \quad t \in \Bbb R.$$ To show the claim, you must find $t_0 \in \Bbb R$ such that ${\bf X}(t_0) = (3,1,0)$. Ok?


2

We split up $\partial D$ into $C_1, C_2, C_3$, where: \begin{align*} C_1:\qquad (f, g) &= (2(1 - t), 4(1 - t)) &\text{where } t \in [0, 1] \\ C_2:\qquad (f, g) &= (2t, 0) &\text{where } t \in [0, 1] \\ C_3:\qquad (f, g) &= (2, 4t) &\text{where } t \in [0, 1] \\ \end{align*} Note that: \begin{align*} \int_{C_1} f \, dx + g \, dy &= ...


2

Write a change of variables, $u = x + y$, $v = x - y$. Then the Jacobian $J = \partial(x,y)/\partial(u,v) = 1/2$ and hence $$\iint_D f(x+y) \ dx \ dy = \iint_D f(u) \ J \ du \ dv = \frac{1}{2} \int_{-1}^1\int_{-1}^1 f(u) \ du \ dv$$ Can you take it from here?


2

As pointed out in the comments, there is indeed a nice geometric interpretation in Tristan Needham's "Visual Complex Analysis". See, especially, the material on the "Polya Vector field". In case you cannot access that book, I've thrown in some point form notes below. Let $f = u + iv$ be a complex-valued function, defined on some open domain $D \subset ...


2

If functions $u$ and $v$ satisfy the Cauchy-Riemann equations then theirs gradients are orthogonal and equal in length: $\nabla u \perp \nabla v$, $|\nabla u|=|\nabla v|$. And this means that level curves of $u$ and $v$ are orthogonal where the gradients do not vanish.


2

$$J(x) = x^TAx = \sum_{i,j} A_{ij}x_ix_j \implies \dfrac{dJ(x)}{dx_k} = \sum_{i,j} A_{ij} \dfrac{d(x_ix_j)}{dx_k}$$ We have $$\dfrac{d(x_ix_j)}{dx_k} = \delta_{ik}x_j + \delta_{jk}x_i$$ Hence, \begin{align} \dfrac{dJ(x)}{dx_k} & = \sum_{i,j} A_{ij} \left(\delta_{ik}x_j + \delta_{jk}x_i\right) = \sum_{j}A_{kj}x_j + \sum_{i}A_{ik}x_i = \left((Ax)_k + ...


1

You are given equations in Cartesian coordinates. At first you need to seek intersection in the same system for handling convenience of given equations. Later on, go in for cylindrical coordinates. But do not a priori choose to adopt cylindrical coordinates. To handle the total situation, choose a proper parameter arising out of given Cartesian equations ...


1

I would say $\int_{\frac{-3\pi}{4}}^{\frac{\pi}{4}} \int_{0}^{2} \int_{0}^{r(\cos\theta-\sin\theta)}r\, dz\,dr\,d\theta=\int_{\frac{-3\pi}{4}}^{\frac{\pi}{4}} \int_{0}^{2} r^2(\cos\theta-\sin\theta)\, dr\,d\theta=\frac{8}{3}\int_{\frac{-3\pi}{4}}^{\frac{\pi}{4}} (\cos\theta-\sin\theta)\,d\theta=$ ...


1

$V = \displaystyle \int_{0}^{\pi/6} \int_{0}^1 \int_{0}^{\pi} xy\sin (yz) dxdydz$


1

This is really a modification of the proof of inverse function theorem, using contraction mapping theorem (fixed point theorem). Note that the matrix $Df$ is uniformly closed to the identity matrix. (I am just imitating the proof of IFT in Rudin's Principles of Mathematical Analysis). For each $y \in \mathbb R^2$, let $\phi :\mathbb R^2 \to \mathbb R^2$ be ...


1

The version of the Cauchy-Riemann equations that I prefer is (assuming we are using $x,y$ as coordinates for the real and imaginary part on the domain) $$ i\partial_xf = \partial_y f $$ This has a direct geometrical interpretation: the partial derivative with respect to the real part rotated by $\pi/2$ (which is what multiplying by $i$ does to a complex ...


1

I personally think of Cauchy Riemann equations in terms of differential geometry. A function $f: \mathbb C \to \mathbb C $ can be thought of as a function from $\mathbb{R} ^2 \to \mathbb{R}^2$, say $f= (u(x,y) , v(x,y))$. Now conformality, (for local diffeomorphisms) is ensured by the fact that First Fundamental form of any surface patch $\sigma$ of ...


1

As others have already stated, Maple is interpreting $\sqrt[3]{x}$ as a complex function. Here's how it's being calculated $$ \int_0^1\int_{-y}^y\sqrt[3]{x}\ dxdy= \int_0^1\int_{-y}^y x^{\frac{1}{3}}\ dxdy $$ $$ = \int_0^1\left[\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]_{-y}^y dy = \int_0^1\left[\frac{x^{\frac{4}{3}}}{\frac{4}{3}}\right]_{-y}^y dy $$ $$ ...


1

It would be better to say, that antiderivative is $(\sqrt[3]{x})^4$, but in the rest you are right.



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