Status Update – Summer 2016

by RoRi

June 30, 2016

Hi everyone. I thought I’d post a status update on the blog. Those of you who have been in touch know that my plan was to finish Apostol, Volume I on June 23. Since that has come and gone, and I’m still not finished, it looks like I’m falling behind schedule. There are still about 100 problems to do in Chapter 14, and I would like to clean up problems that I left incomplete in Chapters 8 and 13. My plan also included starting in on solving all of the problems in Hoffman & Kunze, which is getting delayed also.

Sadly, I don’t think I have time this summer to put into solving problems. I’ll be a first year grad student in the fall and have prelim exams that I would like to pass in August. Tentatively, my new plan is to finish up Apostol, and clean up the incomplete problems by the end of August, and then start in on Hoffman & Kunze (Linear Algebra) on September 1. I’ll probably also reduce the rate of problems from 150 or so per month to 60 per month. I think that’s a pace I can maintain (hopefully).

Anyway, sorry for not responding to comments and for falling behind. I suspect that this will be the new normal though, since I foresee being quite busy for the next two years. Feel free to keep leaving comments, and I’ll try to respond. Of course, if you see questions/comments from other users that you can answer, then feel free to chime in.

Sorry for being so slow! I’ll at least try to keep making forward progress.

Find a vector valued function satisfying given properties

by RoRi

June 1, 2016

If $A$ and $B$ are fixed vectors and $F$ is a vector valued function with

$F''(t) = tA + B,$

find $F(t)$ if $F(0) = D$ and $F'(0) = C$ .

We have

$\begin{align*} && F''(t) &= tA + B \\ \implies && \int F''(t) \,dt &= \int (tA+B) \, dt \\ \implies && F'(t) &= \frac{1}{2}t^2 A + tB + C \\ \implies && F(t) &= \frac{1}{6} t^3 A + \frac{1}{2}t^2 B + tC + D. \end{align*}$

Prove the zero derivative theorem for vector valued functions

by RoRi

June 1, 2016

Prove the zero derivative theorem for vector valued functions. In other words, show that if $F'(t) = 0$ for each $t$ in an open interval $I$ , then there exists a vector $C$ such that $F(t) = C$ for all $t \in I$ .

Proof. If $F'(t) = 0$ then we have

$\begin{align*} && (f_1'(t), \ldots, f_n'(t)) &= (0, \ldots, 0) \\ \implies && f_i'(t) &= 0 & \text{for all } i. \end{align*}$

By the zero derivative for real valued functions we then have $f_i (t) = c_i$ for some constant $c_i$ for each $i = 1, \ldots, n$ . Therefore we have

$(f_1(t), \ldots, f_n (t)) = (c_1, \ldots, c_n) \quad \implies \quad F(t) = C. \qquad \blacksquare$

Prove a condition for a vector valued function to be differentiable

by RoRi

June 1, 2016

If $F$ is a vector valued function, prove that $F$ is differentiable on the open interval $I$ if and only if for all $t \in I$ we have

$F'(t) = \lim_{h \to 0} \frac{1}{h} (F(t+h) - F(t)).$

Proof. First, assume $F$ is differentiable on $I$ . Then, by the definition of differentiability for vector valued functions we have

$F'(t) &= (f_1'(t), \ldots, f_n'(t))$

and each of the $f_i'(t)$ exists. Therefore by the usual definition of the derivative we have

$\begin{align*} F'(t) &= \left( \lim_{h \to 0} \frac{f_1 (t+h) - f_1(t)}{h}, \ldots, \lim_{h \to 0} \frac{f_n (t+h) - f_n (t)}{h} \right) \\[9pt] &= \lim_{h \to 0} \frac{1}{h} \left( (f_1 (t+h), \ldots, f_n (t+h)) - (f_1 (t), \ldots, f_n (t)) \right) \\[9pt] &= \lim_{h \to 0} \frac{1}{h} (F(t+h) - F(t)). \end{align*}$

Conversely, assume that

$F'(t) = \lim_{h \to 0} \frac{1}{h} (F(t+h) - F(t)).$

Then, the limit

$\lim_{h \to 0} \frac{1}{h} (f_i (t+h) - f_i(t))$

exists for each $i = 1, \ldots, n$ . Thus, $f_i'(t)$ exists for each $i$ . Therefore, $F'(t)$ exists $. \qquad \blacksquare$

Prove a property of limits of vector valued functions

by RoRi

June 1, 2016

Prove that

$\lim_{t \to p} F(t) = A \iff \lim_{t \to p} \lVert F(t) - A \rVert = 0.$

Proof. Let

$F(t) = (f_1(t), \ldots, f_n(t)), \qquad \text{and} \qquad A = (a_1, \ldots, a_n).$

Then,

$\begin{align*} \lim_{t \to p} F(t) = A && \iff && \lim_{t \to p} (f_1(t), \ldots, f_n (t)) &= (a_1, \ldots, a_n) \\ && \iff && \left( \lim_{t \to p} f_1(t), \ldots, \lim_{t \to p} f_n (t) \right) &= (a_1, \ldots, a_n). \end{align*}$

This holds if and only if

$\lim_{t \to p} f_i (t) = a_i \qquad \text{for } i = 1, \ldots, n.$

From the usual limit this is true if and only if

$\lim_{t \to p} | f_i (t) - a_i | = 0 \qquad \text{for } i = 1, \ldots, n.$

This is if and only if

$\begin{align*} &&\left( \lim_{t \to p} |f_1 (t) - a|, \ldots, \lim_{t \to p} |f_n (t) - a| \right) &= (0, \ldots, 0) \\ \iff && \lim_{t \to p} |F(t) - A| = 0. \qquad \blacksquare \end{align*}$

Compute G′ if G = F F′ x F′′

by RoRi

June 1, 2016

Let $G$ be the scalar triple product $F \cdot F' \times F''$ . Prove that $G' = F \cdot F' \times F'''$ .

Proof. We compute,

$\begin{align*} G = F \cdot F' \times F'' && \implies && G' &= (F \cdot F' \times F'')' \\ && \implies && G' &= F' \cdot (F' \times F'') + F \cdot (F' \times F'')' \\ && \implies && G' &= 0 + F \cdot (F'' \times F'' + F' \times F''') \\ && \implies && G' &= F \cdot F' \times F'''. \qquad \blacksquare \end{align*}$

Compute G′ if G = F x F′

by RoRi

June 1, 2016

Let $G = F \times F'$ . Compute $G'$ in terms of $F$ and $F'$ .

Using the product rule for the cross product we compute,

$\begin{align*} G = F \times F' && \implies && G' &= (F \times F')' \\ && \implies && G' &= F' \times F' + F \times F'' \\ && \implies && G' &= F \times F'' \end{align*}$

Since $F' \times F' = 0$ since a vector crossed with itself is 0.

Prove that F′′(t) has the same direction as a given F(t)

by RoRi

June 1, 2016

Let $A, B$ be nonzero vectors and

$F(t) = e^{2t} A + e^{-2t} B.$

Prove that $F''(t)$ has the same direction as $F(t)$ .

Proof. To show that $F(t)$ and $F''(t)$ have the same direction we must show that there exists a constant $c$ such that $F(t) = cF''(t)$ . So, we compute,

$\begin{align*} F(t) = e^{2t}A + e^{-2t}B && \implies && F'(t) &= 2e^{2t}A - 2 e^{-2t} B \\ && \implies && F''(t) &= 4 e^{2t} A + 4e^{-2t} B \\ && \implies && F(t) &= \frac{1}{4} F''(t). \qquad \blacksquare \end{align*}$

Prove that F”(t) is orthogonal to F'(t) under given conditions

by RoRi

June 1, 2016

Let $B$ be a nonzero vector and $F$ a vector valued function with $F(t) \cdot B = t$ for all $t$ , and such that the angle between $F'(t)$ and $B$ is constant. Prove that $F''(t)$ and $F'(t)$ are orthogonal.

Proof. Since $F(t) \cdot B = t$ we have $F'(t) \cdot B = c$ for some constant $c$ . Since the angle between $F'(t)$ and $B$ is constant we have

$\frac{F'(t) \cdot B}{\lVert F'(t) \rVert \lVert B \rVert} = d$

for some constant $d$ . Therefore, $\lVert F'(t) \rVert$ is constant. Hence, $\lVert F'(t) \rVert^2$ is constant, say $a$ . So, we have

$\begin{align*} && \lVert F'(t) \rVert^2 &= a \\ \implies && F'(t) \cdot F'(t) &= a \\ \implies && ( F'(t) \cdot F'(t)) &= 0 \\ \implies && 2F'(t) \cdot F''(t) &= 0 \\ \implies && F'(t) \cdot F''(t) &= 0. \qquad \blacksquare \end{align*}$

Compute the dot product of a vector and a vector valued integral

by RoRi

June 1, 2016

Let $A = 2 \mathbf{i} - 4 \mathbf{j} + \mathbf{k}$ and let

$B = \int_0^1 (te^{2t} \mathbf{i} + t \cosh (2t) \mathbf{j} + 2te^{2t} \mathbf{k}) \, dt.$

Compute the dot product $A \cdot B$ .

First, we evaluate the vector valued integral,

$\begin{align*} B &= \int_0^1 (te^{2t} \mathbf{i} + t \cosh (2t) \mathbf{j} + 2te^{2t} \mathbf{k}) \, dt \\[9pt] &= \left( \frac{1}{4} + \frac{1}{4}e^2, \frac{1}{4} + \frac{1}{2} \sinh 2 - \frac{1}{4} \cosh 2, \frac{1}{2} - \frac{3}{2} e^{-2} \right). \end{align*}$

Therefore, the dot product is

$\begin{align*} A \cdot B &= \frac{1}{2} + \frac{1}{2} e^2 - 1 - 2 \sinh 2 + \cosh 2 + \frac{1}{2} - \frac{3}{2} e^{-2} \\[9pt] &= \frac{1}{2}e^2 - e^2 + e^{-2} + \frac{1}{2}e^2 + \frac{1}{2} e^{-2} - \frac{3}{2} e^{-2} \\[9pt] &= 0. \end{align*}$