Mathematical Food for Thought


			Mathematical Food for Thought

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Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.

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To Converge Or Not To Converge. Topic: Real Analysis/S&S.	November 27th, 2006
Problem: (Stanford Math 51H, Cauchy Root Test) Suppose there exists a $\lambda \in (0,1)$ and an integer $N \ge 1$ such that $\|a_n\|^{\frac{1}{n}} \le \lambda$ for all $n \ge N$ . Prove that $\displaystyle \sum_{n=1}^{\infty} a_n$ converges. Solution: Well let’s just discard $\displaystyle \sum_{n=1}^{N-1} a_n$ because it is finite and obviously converges. It remains to show that $\displaystyle \sum_{n=N}^{\infty} a_n$ converges. But then we have $\|a_n\|^{\frac{1}{n}} \le \lambda \Rightarrow \|a_n\| \le \lambda^n$ . So $\displaystyle \sum_{n=N}^{\infty} a_n \le \sum_{n=N}^{\infty} \|a_n\| \le \sum_{n=N}^{\infty} \lambda^n$ . The last summation, however, is a geometric series with common ratio $0 < \lambda < 1$ , so it converges. Hence our sum does as well. QED. ——————– Comment: The root test is, in effect, a comparison to a geometric series. The hypothesis is that we can bound by a geometric series for all large , implying convergence. ——————– Practice Problem: (Stanford Math 51H) Discuss the convergence of $\displaystyle \sum_{n=1}^{\infty} \frac{\sin{\frac{1}{n}}}{n}$ . Posted in Real Analysis, Sequences & Series \|\| 3 Comments »
Limitten. Topic: Real Analysis.	November 23rd, 2006
Definition: A sequence $\{a_n\}$ has limit , where is a given real number, if for each $\epsilon > 0$ there is an integer $N \ge 1$ such that $\|a_n-L\| < \epsilon$ for all integers $n \ge N$ . ——————– Problem: (Stanford Math 51H) Prove that a sequence $\{a_n\}$ cannot have more than one limit. Solution: This is logically true, as usual, but a rigorous argument is much more fun. Suppose $\lim a_n = L_1$ and $\lim a_n = L_2$ . Letting $\epsilon = \frac{\|L_1-L_2\|}{2}$ , we know there exists such that $\|a_n-L_1\| < \epsilon$ for $n \ge N_1$ $\|a_n-L_2\| < \epsilon$ for $n \ge N_2$ . So the above two inequalities are true for $n \ge \max(N_1, N_2)$ . Adding the two inequalities together, we get $\|a_n-L_1\|+\|a_n-L_2\| < 2\epsilon = \|L_1-L_2\|$ . However, by the triangle inequality we know $\|a\|+\|b\| \ge \|a-b\|$ (by setting is the usual one), so $\|a_n-L_1\|+\|a_n-L_2\| \ge \|L_2-L_1\| = \|L_1-L_2\|$ . Then , a contradiction. Hence $\{a_n\}$ can have at most one limit. QED. ——————– Comment: The real definition of the limit is almost always complete overlooked in a regular calculus course (i.e. Calc AB and BC). But it’s pretty much the foundation of all of calculus so it is really quite nice to know and work with. ——————– Practice Problem: (Stanford Math 51H, Sandwich Theorem) If $\{a_n\}, \{b_n\}$ are given convergent sequences with $\lim a_n = \lim b_n$ , and if $\{c_n\}$ is any sequence such that $a_n \le c_n \le b_n \forall n \ge 1$ , prove that $\{c_n\}$ is convergent and $\lim c_n = \lim a_n = \lim b_n$ . Posted in Real Analysis \|\| No Comments »
Root Beer. Topic: Real Analysis.	November 22nd, 2006
Problem: (Stanford Math 51H) Prove that every positive real number has a positive square root (That is, for any , prove that there is a real number $\alpha$ such that $\alpha^2 = b$ ). [Usual properties of the integers are assumed.] Solution: Consider the set $S = \{x \in \mathbb{R}: x > 0, x^2 < b\}$ . We can show that is non-empty by selecting an integer large enough such that $\frac{1}{n^2} < b$ . Since is a real and the integers are unbounded, there exists a positive integer such that , thus $x^2 < k^2 \Rightarrow x < k$ so is bounded from above. Now there must exist $\alpha$ such that $\sup S = \alpha$ . We claim that $\alpha^2 = b$ . Suppose the contrary; then either $\alpha^2 < b$ or $\alpha^2 > b$ . CASE 1: If $\alpha^2 < b$ , then consider $\left(\alpha+\frac{1}{n}\right)^2 = \alpha^2+\frac{2\alpha}{n}+\frac{1}{n^2}$ . Choose large enough such that $\alpha^2+\frac{2\alpha}{n}+\frac{1}{n^2} < b$ which is definitely true for any $n > \frac{2\alpha+1}{b-\alpha^2}$ . But then $\alpha+\frac{1}{n} \in S$ and $\alpha+\frac{1}{n} > \alpha = \sup S$ , contradicting the fact that $\alpha$ is the supremum. CASE 2: If $\alpha^2 > b$ , then consider $\left(\alpha-\frac{1}{n}\right)^2 = \alpha^2-\frac{2\alpha}{n}+\frac{1}{n^2}$ . Again, choose large enough such that $\alpha^2-\frac{2\alpha}{n}+\frac{1}{n^2} > b$ which we know is true for $n > \frac{2\alpha}{\alpha^2-b}$ (furthermore, we impose the restriction $\alpha > \frac{1}{n}$ so our resulting real is positive). Then $x \le \alpha-\frac{1}{n}$ for all $x \in S$ and $\alpha-\frac{1}{n} < \alpha = \sup S$ , contradicting the fact that $\alpha$ is the supremum. Hence we know that $\alpha^2 = b$ , or that the square root of any positive real number exists and is a positive real number. QED. ——————– Comment: This is from the real analysis portion of a freshman honors calculus course, i.e. a rigorous treatment of the real numbers which is the basis for calculus like limits and stuff. Really understanding calculus involves really understanding how the real number system works. ——————– Practice Problem: (Stanford Math 51H) If $a, b \in \mathbb{R}$ with , prove: (a) There is a rational $r \in (a,b)$ . (b) There is an irrational $c \in (a,b)$ . (c) contains infinitely many rationals and infinitely many irrationals. Posted in Real Analysis \|\| 4 Comments »

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