# Tag Archives: M131A

teaching

## Basic Logic

I have written up a short essay that gives a brief and informal overview of the logical syntax used in mathematics. It covers the basic logical connectives and quantifiers, and gives a few examples. In particular, the final example (negation of the ε-δ definition of continuity) should be helpful in solving problem 2 in this week’s homework for Math 32AH.

Here is a link to the essay: Basic Logic. As usual, let me know in the comments below if anything in the essay is incorrect or unclear.

teaching

## Math 131A Review Questions

I have compiled a few final review questions for real analysis this quarter. They are available at

Solutions are now posted as well. As usual, let me know in the comments below if you notice any errors or if anything is unclear.

teaching

## A smooth but not analytic function

In the current assignment for real analysis, we consider the following function
$f(x) = \begin{cases} e^{-1/x^2} & x \neq 0\\ 0 & x = 0. \end{cases}$
We are asked to show that $$f^{(n)}(0) = 0$$ for all $$n$$, and that the Taylor series for $$f$$ at $$0$$ converges to $$f(x)$$ only for $$x \neq 0$$.

To compute the derivatives $$f^{(n)}(0)$$, it is easiest to use induction on $$n$$. For $$n = 1$$,
$f'(0) = \lim_{x \to 0} \frac{f(x) – f(0)}{x – 0} = \lim_{x \to 0} \frac 1 x e^{-1 / x^2}.$
To compute the limit, we first use the change of variables $$y = 1 / x$$ so that as $$y \to \pm \infty$$ we have $$x \to 0$$. Then
$\lim_{x \to 0} \frac 1 x e^{-1 / x^2} = \lim_{y \to \pm \infty} \frac{y}{e^{y^2}}.$
Using L’Hopital’s rule, it is easy to verify that the latter limit is zero. In fact, a similar argument shows that for any $$k$$
$\lim_{x \to 0} \frac{1}{x^k} e^{-1 / x^2} = 0.$
We will need this fact later on. At any rate, we’ve shown that $$f'(0) = 0$$.

For the inductive step, assume that $$f^{(n)}(0) = 0$$. Then
$f^{(n + 1)}(0) = \lim_{x \to 0} \frac{f^{(n)}(x) – f^{(n)}(0)}{x – 0} = \lim_{x \to 0} \frac{1}{x} f^{(n)}(x).$
After computing the first few derivatives, $$f'(x),\ f”(x), \ldots$$ for $$x \neq 0$$, you should be convinced that $$f'(x)$$ is of the form
$f^{(n)}(x) = \left (\frac{a_k}{x^k} + \frac{a_{k-1}}{x^{k-1}} + \cdots + a_0 \right ) e^{-1 / x^2}.$
Using the same trick as before, we can compute
$\lim_{x \to 0} \frac{a_i}{x^{i+1}} e^{-1 / x^2} = 0,$
hence
$f^{(n + 1)}(0) = \lim_{x \to 0} \frac{1}{x} f^{(n)}(x) = 0.$

Since all of the derivatives of $$f$$ are $$0$$ at $$x = 0$$, the Taylor series there is
$\sum_{k = 0}^\infty f^{(k)}(0) x^k = 0.$
However, $$f(x) \neq 0$$ for $$x \neq 0$$, so the Taylor series only agrees with $$f$$ at $$x = 0$$.

teaching

## Triangle Inequality and the Like

I didn’t finish a careful proof of the triangle inequality in class, so I am presenting a more polished argument here. I abandoned the text’s argument in favor of what I hope is a more intuitive (albeit longer) proof.

I’ve also had a number of requests for hints on a couple homework problems, specifically numbers 3.7 and 3.8. So here are my comments on those problems:

Problem 3.7 This problem asks us to prove that $$|b| < a$$ if and only if $$- a < b < a$$, and some variations on this result. Most often, it is easiest to prove "if and only if" statements in two parts: first prove that if $$|b| < a$$ then $$-a < b < a$$; then prove that if $$-a < b < a$$ then $$|b| < a$$. Notice that the statement $$a < b$$ is equivalent to $$a \leq b$$ and $$a \neq b$$. Therefore, you can use all of the order axioms for the strict inequality ($$<$$) except that O2 doesn't apply. You can prove the first implication by breaking the problem up into the cases where $$b$$ is positive and negative. Also, you can assume that $$a > 0$$, for otherwise the assumption $$|b| < a$$ is impossible. If $$b \geq 0$$ then $$|b| = b$$ so that $$|b| < a \Rightarrow b < a$$. On the other hand, we get $$-a < 0 < b$$. Combining this with the previous inequality gives $$-a < b < a$$, as desired. The case where $$b$$ is negative is similar. For the opposite implication (assuming $$-a < b < a$$), it is again helpful to break the proof up into cases depending on the sign of $$b$$. For example, if $$b \geq 0$$, then $$|b| = b < a$$, which does it. If $$b \leq 0$$, you'll have to appeal to the assumed inequality $$-a < b$$. Part (b) asks you to prove that $$|a - b| < c$$ if and only if $$b - c < a < b + c$$, and (c) asks you to prove the same thing replacing each $$<$$ with $$\leq$$. You can prove (b) by appealing to (a): just use $$a - b$$ for $$b$$ and $$c$$ for $$a$$ in part (a) and see what you get. For part (c), you can just argue that the same inequality as (a) holds with $$\leq$$, so the proof of (c) is essentially the same as the proof of (b). Don't worry about going through all the details again -- just convince me that you understand what is going on in you write-ups. Problem 3.8 In this problem, you are asked to prove that for $$a, b \in \mathbf{R}$$ if $$a \leq b_1$$ for every $$b_1 > b$$, then $$a \leq b$$.

I think the easiest way to approach this problem is to do a proof by contradiction. The idea is that in order to prove that a statement is true, you prove derive a contradiction from its negation. The negation of the statement we are trying to prove is, “for all $$b_1 > b$$, $$a \leq b_1$$ and $$a > b$$.”

If you think about what this statement is saying, it should be clear that it cannot possibly be true. For example, if $$a > b$$, then we can choose $$b_1 = (a + b) / 2$$ so that $$b < b_1 < a$$. But this choice of $$b_1$$ contradicts the assumption that for all $$b_1 > b$$ $$a \leq b_1$$. Since the negation of the statement we are trying to prove leads to a contradiction, the statement must be true.

teaching

## Binomial Theorem

I just finished writing up a proof of the Binomial Theorem that I wasn’t able to complete carefully in class. The write-up is available here. Let me know in the comments if anything us unclear or incorrect.