Fall 2014

1. Every real number is the value of \(f\) at some real number.

2. There are two distinct real numbers at which \(f\) takes the same value.

3. For every positive \(\varepsilon\) there is a positive \(\delta\) such that whenever two real numbers differ by at most \(\delta\), their images under \(f\) differ by at most \(\varepsilon\).

4. There is a positive \(\varepsilon\) such that for every positive \(\delta\) there are two real numbers differing by at most \(\delta\) whose images under \(f\) differ by more than \(\varepsilon\).

Base case: for \(n=1\), we have \(a_1 = 1 \leq a_2 = \frac{16} 3 \leq 10.\) Suppose this holds for \(k < n\), then \[ a_{n-1} \leq a_n = \frac{a_{n-1}}{3} + 5 \implies 3a_{n-1} \leq a_{n-1} + 15 \implies a_{n-1} \leq \frac{15}{2} \]

and thus we have \[ a_{n+1} = \frac{a_n}{3} + 5 = \frac{1}{3}(a_n + 15) \\ = \frac{1}{3}((\frac{a_{n-1}}{3} + 5) + 15) \\ = \frac{a_{n-1} + 60}{9} \\ \leq \frac{\frac{15}{2} + 60}{9} \\ = \frac{150}{18} \\ < \frac{180}{18} = 10, \]

and \(a_{n+1} \leq 10\). Moreover, note that the relation \(a_{n+1} = \frac{a_n}{3} + 5\) can be rewritten as \[ a_n = 3a_{n+1} - 15, \\ a_{n-1} = 3a_n - 15. \] Using the inductive hypothesis \(a_{n-1} \leq a_n\), we can thus write \[ 3a_n - 15 = a_{n-1} \leq a_n = 3a_{n+1} - 15, \] from which we get \(3a_{n} - 15 \leq 3a_{n+1} - 15\) and thus \(a_{n} \leq a_{n+1}\).

To compute \(\lim_{n\to\infty}a_n\), perhaps there are easier ways, but we can just use generating functions. Note that the limit exists by the Monotone Convergence Theorem. Let \(A(x) = \sum_{n=0}^\infty a_n x^n\) where \(a_0 = 0\). Then applying the magic sauce, we have \[\begin{align*} a_n = \frac{1}{3}a_{n-1} + 5 &\implies \sum_{n=1}^\infty a_nx^n = \frac{1}{3}\sum_{n=1}^\infty a_{n-1}x^n + 5\sum_{n=1}^\infty x^n \\ &\implies A(x) - a_0 = \frac 1 3 xA(x) + 5\left( \frac 1 {1-x} - 1\right) \\ &\implies A(x)\left(1 - \frac x 3\right) = 5\left( \frac x {1-x}\right) \\ &\implies A(x) = 15\left(\frac 1 {3-x} \right)\left(\frac x {1-x} \right) \\ &\implies A(x) = \frac{15x}{(3-x)(1-x)} \\ &\implies A(x) = \frac{-\frac{45}{2}}{3-x} + \frac{\frac{15}{2}}{1-x} \\ &\implies A(x) = \frac 3 2 \left(-5 \left( \frac{1}{1-\frac x 3} \right) + 5\left( \frac 1 {1-x}\right) \right) \\ &\implies A(x) = \frac{15}{2} \sum_{n=0}^\infty \left(1 - \left( \frac 1 3\right)^n\right)x^n \\ &\implies a_n = \frac {15} 2 \left(1 - \left( \frac 1 3\right)^n\right) \end{align*}\]

and so we find \[ \lim_{n\to\infty}a_n = \lim_{n\to\infty}\frac {15} 2 \left(1 - \left( \frac 1 3\right)^n\right) = \frac{15}{2}. \qed \] > Alternatively: \[ a_{n+1} = \frac 1 3 a_n + 5 \implies \lim_{n\to\infty} a_{n+1} = \lim_{n\to\infty} \frac 1 3 a_n + 5 \\ \implies L = \frac 1 3 L + 5 \implies \frac 2 3 L = 5 \implies L = \frac {15} 2 \]

Suppose \(\exists M_g \suchthat \forall x,~ g(x) < M\). Then let \(\varepsilon > 0\) be arbitrarily chosen; we want to show that there exists a \(\delta\) such that \(\abs{x} \leq \delta \implies \abs{f(x)g(x)} \leq \varepsilon\). Since \(\lim_{x\to 0} f(x) = 0\), choose a \(\delta_f\) such that \(\abs{x} \leq \delta_f \implies \abs{f(x)} \leq \frac{\varepsilon}{M_g}\). So letting \(\delta = \delta_f\), we have \[ \abs{x} \leq \delta \implies \abs{f(x)g(x)} = \abs{f(x)} \abs{g(x)} \leq {\frac{\varepsilon}{M_g}}\abs{g(x)} \leq \frac{\varepsilon}{M_g}M_g = \varepsilon. \qed \] Let \(f(x) = x\) and \(g(x) = \frac{1}{x}\). Note that \(g(x)\) is unbounded in any neighborhood of 0, and \(f(x)g(x) = 1 \not\to 0\).

Write \(w_1 = v_1+v_2\), \(w_2 = v_2-v_3\), \(w_3 = v_2+2v_3\). In the ordered basis \((v_1,v_2,v_3)\) these have coordinate matrix \[ A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 0 & -1 & 2 \end{pmatrix}. \] A short expansion gives \(\det A = 3 \neq 0\), so \(A\) is invertible. Thus \((w_1,w_2,w_3)\) is obtained from a basis by an invertible linear change of coordinates, hence is itself a basis of \(V\).

We first note that we can rewrite the equation of the region to obtain something more familiar: \(x^2 + y^2 = 2x \implies (x-1)^2 + y^2 = 1\), which is a translated circle. Integrating over this region will be easy compared to the line integral, so we apply Green’s theorem: \[ \int_C xe^x ~dx + ye^y +x^2 ~dy = \iint_D 2x ~dA. \]

We can parameterize this region as \[ D = \theset{x^2+y^2-2x = 0 \suchthat (x,y) \in \RR^2, y \geq 0} = \theset{(r(1+ \cos\theta), r\sin\theta) \suchthat \theta \in [0, \pi), r\in [0, 1]}. \]

Noting that \(dA = r~dr~d\theta\), we can then integrate \[ \iint_D 2x ~dA = \int_0^{\pi} \int_0^1 2(r(1 + \cos\theta)) r ~dr ~d\theta \\ = 2\int_0^{\pi} \int_0^1 r^2(1+\cos \theta) ~dr ~d\theta \\ = 2\int_0^{\pi} \frac 1 3 r^3(1+\cos \theta) \bigg\rvert_0^1 ~d\theta\\ = \frac 2 3 \int_0^{\pi} (1+\cos \theta) ~d\theta\\ = \frac 2 3 ( \theta + \sin \theta) \bigg\rvert_0^\pi \\ = \frac 2 3 [(\pi + 0) - (0 + 0)] = \frac 2 3 \pi. \qed \]

If \(A\) has two distinct eigenvalues, we will have \(A = PD\inverseof{P}\) where \(P\) is the matrix of eigenvectors and \(D\) has eigenvalues on the diagonal. We can compute the characteristic polynomial \[ p_\chi(x) = x^2 - (\Tr A)x + \det A = x^2 - 7x + 6 = (x-6)(x-1), \]

and so \(\spec(A) = \theset{6,1}\). Computing the kernel of \(A-\lambda I\) for each of these yields \[ \vector v_1 = \left[ \begin{array} { c } { 1 } \\ { - 2 } \end{array} \right], \vector v_2 = \left[ \begin{array} { c } { 2 } \\ { 1 } \end{array} \right], \]

And so we can write and check \[ P = \left[ \begin{array} { c c } { 1 } & { 2 } \\ { - 2 } & { 1 } \end{array} \right] \\ D = \left(\begin{array}{rr} 6 & 0 \\ 0 & 1 \end{array}\right) \\ \]

We can compute \(PP^T = \mathrm{diag}(5,5)\), so \(P\) can be made orthogonal by replacing \(P\) with \((1/\sqrt 5) P\). With this replacement, a quick computation shows that \(PDP^T = A\).

We will use the fact that \(A = PD\inverseof{P}\) where since \(A\) is symmetric and \(P\) is orthogonal. We can write \[ \inner{A\vector x}{\vector x} = \vector x^T A^T \vector x = \vector x^T (PDP^T)^T \vector x = (P^T \vector x)^T D (P^T\vector x) = \vector y^T D \vector y = \inner{\vector y}{ D\vector y}, \] where \(\vector x\in S^2 \implies P^T \vector x \definedas \vector y \in S^2\) since \(P^T\) is both orthogonal and full-rank (and thus a bijection \(S^2 \selfmap\)).

We can now expand \[ \inner{\vector y}{D \vector y} = \sum_{i=1}^2 y_i \lambda_i y_i = \sum_{i=1}^2 \lambda_i y_i^2 \]

We now note that we can take \(\vector y = \thevector{0, 1}\), in which case \(D\vector y = \thevector{0, \lambda_2} = \thevector{0, 1}\) and thus \(\inner{\vector y}{D \vector y} = 1\) is a candidate minimum.

We can write this as the constrained optimization problem \[ \text{Minimize } f(y_1, y_2) = 6y_1^2 + 1 y_2^2\\ \text{subject to } g(y_1, y_2) = y_1^2 + y_2^2 = 1 \]

where we note that this constraint is equivalent to the original \(\norm{\vector y} = \sqrt{y_1^2 + y_2^2} = 1\).

This can be approached with Lagrange multipliers, i.e. looking at where \(\nabla f = \lambda \nabla g\). This yields \[ \thevector{12y_1, 2y_2} = \lambda \thevector{2y_1, 2y_2} \implies \\ 6y_1 = \lambda y_1, ~ y_2 = \lambda y_2. \]

The second condition forces \(y_2 \in \theset{0,1}\), and solving for \(\lambda\) in this expression yields \(\lambda = 1\) and so the first condition forces \(y_1 = 0\). This leaves only one possibility, \(\vector y = \thevector{0, 1}\), which is indeed the candidate from above. Thus the minimum value is 1. \(\qed\)

We need to show that \(R\) is reflexive, transitive, and symmetric. 1. Reflexive: this would say that \(x\sim x \iff x^2-4x = x^2-4x\), which is true. 2. Transitive: suppose \(x\sim y\) and \(y\sim z\), we want to show \(x\sim z\). But we have \[ x^2 - 4x = y^2-4y ~\&~ y^2-4y = z^2-4z \implies x^2-4x = y^2-4y = z^2-4z \] 1. We want to show \(x\sim y \implies y \sim x\), which follows because \(x^2-4x = y^2-4y \iff y^2-4y = x^2-4x\). 2. The equivalence classes: \[\begin{align*} x^2-4x &= 0: &\theset{0, 4}\\ x^2-4x &= -3: &\theset{1,3} \\ x^2-4x &= -4: &\theset{2} \\ x^2-4x &= 5: &\theset{5} \end{align*}\]

A function \(f: \RR^2 \to \RR\) is totally differentiable at \(\vector x\) if there exists a linear map \(D: \RR^2 \to \RR\) such that \[f(\vector x + \vector h)- f(\vector x) = D(\vector x)\vector h + \varepsilon(\vector h) \text{ where } \frac{\varepsilon(\vector h)}{\norm{\vector h}} \to 0. \] Equivalently, it is when the following limit exists and satisfies \[ \lim_{\vector h \to \vector 0} \abs{\frac{f(\vector x + \vector h) - f(\vector x) - D(\vector x)\vector h}{\norm{\vector h}}} = 0, \] in which case we write \(f' = D\), or in this special case, \(f' = \nabla f\).

Substituting in $\vector x = \vector 0$, being differentiable at zero requires that $\exists D$ such that
$$
f(\vector h) - f(\vector 0) = D(\vector 0)\vector h + \varepsilon(\vector h), ~\frac{\varepsilon(\vector h)}{\norm{\vector h}} \to 0, 
\\ \text{ or } \\ 
\lim_{\vector h \to \vector 0} \abs{\frac{f(\vector h) - f(\vector 0) - D(\vector 0)\vector h}{\norm{\vector h}}} = 0.
$$
We'll make use the mean value theorem. The general statement is
$$
f' \text{ exists and continuous on } (a,b)
\implies \exists A \in [a,b] \suchthat \\
f(b) - f(a) = f'(c)(b-a)
$$
and here we'll use
$$
f_x \text{ exists and continuous on } (0, t) 
\implies \exists A \in [0, t] \suchthat \\ 
f(t, y) - f(0, y) = f_x(A, y)(t - 0) = tf_x(A, y) \\
\implies \color{purple} f(t, y) - f(0, y) = tf_x(A, y)
$$
where moreover 
$$
\lim_{t\to 0} f_x(A, y) = f_x(\lim_{t\to 0} A, y) = f(0, y)
$$ 
since $f_x$ was assumed to be continuous.

We'll also use the "best linear approximation" definition of differentiability. In general, it states
$$
f' \text{ exists on } (a,b) \\
\implies f(x) = f(a) + f'(a)(x-a) + o(x-a),~ \frac{o(x-a)}{x-a} \to 0 \\
\implies f(x) - f(a) = f'(a)(x-a) + o(x-a),~ \frac{o(x-a)}{x-a} \to 0 \\
\implies f(b) - f(a) = f'(a)(b-a) + o(b-a),~ \frac{o(b-a)}{b-a} \to 0
$$
and here we'll take $a=0, b=t$, so $x-a = x$ and $b-a = t$, and use the variant
$$
f_y \text{ exists on } (0, t) \\ 
\implies f(x, t) - f(x, 0) = f_y(x, 0)(t - 0) + o(t),~ \frac{o(t)}{t} \to 0 \\
\implies \color{green}f(x, t) - f(x, 0) = tf_y(x,0) + o(t), ~\frac{o(t)}{t} \to 0.
$$
We can then write
$$\begin{align*}
f(h_1, h_2) - f(0, 0)
&= {\color{green}f(h_1, h_2) - f(h_1, 0)} + {\color{purple}f(h_1, 0) - f(0, 0)} \\
&= {\color{green}h_2f_y(0, 0) + o(h_2)} + {\color{purple}h_1f_x(A, 0)}\\
&= (h_1 f_x(0, 0) - {\color{red}h_1 f_x(0, 0)}) + {\color{purple}h_1f_x(A, 0)} + {\color{green}h_2f_y(0, 0) + o(h_2)} \\
&= h_1 f_x(0, 0) + {\color{green}h_2 f_y(0, 0)} + h_1({\color{purple}f_x(A,0)} - {\color{red}f_x(0,0)}) + {\color{green}o(h_2)}
\end{align*}$$

Now, since $f_x$ is continuous, ${\color{purple}f_x(A,0)} - {\color{red}f_x(0,0)} \to 0$ and so $h_1(f_x(A,0) - f_x(0,0)) = o(h_1)$. We thus have
$$\begin{align*}
f(h_1, h_2) - f(0, 0)
&= h_1 f_x(0, 0) + h_2 f_y(0, 0) + o(h_1) + o(h_2) \\
&= h_1 f_x(0, 0) + h_2 f_y(0, 0) + o(\norm{\vector h}) \\ \\
&\implies 
f(h_1, h_2) - f(0, 0) - h_1 f_x(0, 0) - h_2 f_y(0, 0) = o(\norm{\vector h})
\end{align*}$$

and so if we define 
$$D(\vector h) = h_1 f_x(0, 0) + h_2 f_y(0, 0),
$$ 

we can take absolute values to find that 
$$
\abs{f(\vector h) - f(\vector 0) - D(\vector h)} = o(\norm{\vector h}) \\
\text{ and so } \\
\lim_{\vector h \to \vector 0} \abs{\frac{f(\vector h) - f(\vector 0) - D(\vector h)}{\norm{\vector h}}} = 0
$$

which forces $f$ to be totally differentiable at $\vector 0$ with derivative $D$.

Goal: Prove that if the partial derivatives \(f_x\) and \(f_y\) of \(f: \mathbb{R}^2 \to \mathbb{R}\) exist in an open neighborhood of \((0,0)\) and \(f_x\) is continuous at \((0,0)\), then \(f\) is (totally) differentiable at \((0,0)\) provided \(f_y(0,0)\) exists.

<1>1. Definition: \(f\) is differentiable at \(\mathbf{0} = (0,0)\) with total derivative \(L = (f_x(\mathbf{0}), f_y(\mathbf{0}))\) if and only if \[\lim_{\mathbf{h} \to \mathbf{0}} \frac{|f(h_1, h_2) - f(0,0) - h_1 f_x(0,0) - h_2 f_y(0,0)|}{\|\mathbf{h}\|} = 0.\]

<1>2. Decomposition of the difference: For small \(\mathbf{h} = (h_1, h_2)\), \[f(h_1, h_2) - f(0,0) = [f(h_1, h_2) - f(0, h_2)] + [f(0, h_2) - f(0,0)].\]

<1>3. Application of the single-variable Mean Value Theorem: There exists \(\theta \in (0, 1)\) such that \(f(h_1, h_2) - f(0, h_2) = h_1 f_x(\theta h_1, h_2)\).

<1>4. Linear approximation along the \(y\)-axis: \(f(0, h_2) - f(0,0) = h_2 f_y(0,0) + r(h_2)\) where \(\lim_{h_2 \to 0} \frac{r(h_2)}{h_2} = 0\).

<1>5. Bounding the remainder: \[|f(h_1, h_2) - f(0,0) - h_1 f_x(0,0) - h_2 f_y(0,0)| \le |h_1| |f_x(\theta h_1, h_2) - f_x(0,0)| + |r(h_2)|.\]

<1>6. The limit as \(\|\mathbf{h}\| \to 0\) vanishes: Proof: <2>1. Since \(|h_1| \le \|\mathbf{h}\|\), \(\frac{|h_1| |f_x(\theta h_1, h_2) - f_x(0,0)|}{\|\mathbf{h}\|} \le |f_x(\theta h_1, h_2) - f_x(0,0)|\). <2>2. As \(\mathbf{h} \to \mathbf{0}\), \((\theta h_1, h_2) \to (0,0)\). Since \(f_x\) is continuous at \((0,0)\), \(\lim_{\mathbf{h} \to \mathbf{0}} |f_x(\theta h_1, h_2) - f_x(0,0)| = 0\). <2>3. Since \(|h_2| \le \|\mathbf{h}\|\), \(\frac{|r(h_2)|}{\|\mathbf{h}\|} \le \frac{|r(h_2)|}{|h_2|} \to 0\) as \(h_2 \to 0\) (for \(h_2 \neq 0\); for \(h_2 = 0\), \(r(h_2) = 0\)). <2>4. Combining these bounds shows the quotient in <1>1 tends to \(0\).

<1>7. Conclusion: \(f\) is totally differentiable at \((0,0)\).