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Status Update – Summer 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.

Prove a condition for a vector valued function to be differentiable

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