Clean up LaTeX in old calculus class notes

analysis--real-analysis-sequences-series.tex

@@ -9,13 +9,15 @@ \subsection{Definition of sequences and series}
 
 A sequence is often defined by giving the closed-form expression for the $\nth$ item, e.g. $a_n = \frac{1}{n}$. This refers to the sequence
 
-$$\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, ...$$
+\begin{align*}
+  \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, ...
+\end{align*}
 
 The corresponding series is
 
-$$
-\sumn \frac{1}{n} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ...
-$$
+\begin{align*}
+  \sumn \frac{1}{n} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ...
+\end{align*}
 
 A sequence may have a limit: $\lim_{n \rightarrow \infty} a_n$. If it does then the sequence \textbf{converges}. If not, it \textbf{diverges}.
 
@@ -25,45 +27,45 @@ \subsubsection{Typical questions about sequences (11.1)}
 \textbf{Determine whether the sequence converges or diverges. If it converges, find the limit.}
 
 - \textbf{Easier} (11.1.24)
-  $$
+\begin{align*}
   a_n = \frac{n^3}{n^3 + 1}
-  $$
-  It's asking whether the sequence converges, so we want to try to compute the limit of the sequence as $n \rightarrow \infty$. If the limit exists then it converges. Rearrange so that all $n$s are on the bottom of fractions:
-  $$
+\end{align*}
+It's asking whether the sequence converges, so we want to try to compute the limit of the sequence as $n \rightarrow \infty$. If the limit exists then it converges. Rearrange so that all $n$s are on the bottom of fractions:
+\begin{align*}
   a_n = \frac{1}{1 + \frac{1}{n^3}}
-  $$
-  So now
-  $$
+\end{align*}
+So now
+\begin{align*}
   \limn a_n = \frac{1}{1 + \limn\frac{1}{n^3}} = \frac{1}{1 + 0} = 1.
-  $$
+\end{align*}
 
 - \textbf{Easier} (11.1.26)
-  $$
+\begin{align*}
   a_n = \frac{n^3}{n + 1}
-  $$
-  With this one, we can't get the $n$s to appear only on the bottom of fractions:
-  $$
+\end{align*}
+With this one, we can't get the $n$s to appear only on the bottom of fractions:
+\begin{align*}
   a_n = \frac{n^2}{1 + \frac{1}{n}}
-  $$
-  So
-  $$
+\end{align*}
+So
+\begin{align*}
   \limn a_n = \frac{\limn n^2}{1 + \limn\frac{1}{n}} = \frac{\limn n^2}{1 + 0} = \limn n^2.
-  $$
-  But $n^2$ increases without bound as $\ninfty$. People write that as $\limn n^2 = \infty$. In any case, the limit does not exist and this sequence diverges.
+\end{align*}
+But $n^2$ increases without bound as $\ninfty$. People write that as $\limn n^2 = \infty$. In any case, the limit does not exist and this sequence diverges.
 
 - \textbf{Harder} (11.1.56)
-  $$
+\begin{align*}
   a_n = \frac{(-3)^n}{n!}
-  $$
-  The negative factor means that the terms of the sequence alternate between being negative and positive. Rewrite it as
-  $$
+\end{align*}
+The negative factor means that the terms of the sequence alternate between being negative and positive. Rewrite it as
+\begin{align*}
   a_n = (-1)^n\frac{3}{1}\frac{3}{2}\frac{3}{3}\frac{3}{4}...\frac{3}{n-1}\frac{3}{n}
-  $$
-  Now we use two theorems in the book. The first ("Theorem 6") says that if $\limn |a_n|$ is zero, then so is $\limn a_n$. So that means we can see what happens while ignoring the -1 term. The second theorem is the "Squeeze Theorem": if we can show that the quantity we're studying lies between 0 and something else whose limit is zero, then we've proved that the limit of our quantity is 0. To do that we notice that all the terms from $\frac{3}{4}$ to $\frac{3}{n-1}$ are less than 1. Therefore
-  $$
+\end{align*}
+Now we use two theorems in the book. The first ("Theorem 6") says that if $\limn |a_n|$ is zero, then so is $\limn a_n$. So that means we can see what happens while ignoring the -1 term. The second theorem is the "Squeeze Theorem": if we can show that the quantity we're studying lies between 0 and something else whose limit is zero, then we've proved that the limit of our quantity is 0. To do that we notice that all the terms from $\frac{3}{4}$ to $\frac{3}{n-1}$ are less than 1. Therefore
+\begin{align*}
   0 < |a_n| < \frac{9}{2}\frac{3}{n} = \frac{27}{2n}
-  $$
-  But $\limn \frac{27}{2n} = 0$ so this (together with the Theorem 6 argument about the $(-1)^n$) proves that the sequence converges and that its limiting value is 0.
+\end{align*}
+But $\limn \frac{27}{2n} = 0$ so this (together with the Theorem 6 argument about the $(-1)^n$) proves that the sequence converges and that its limiting value is 0.
 
 \subsubsection{Series}
 
@@ -80,9 +82,9 @@ \subsubsection{Series}
 \textbf{Geometric series}
 Several questions require recognizing a geometric series. A geometric series is a series where each term differs by a constant multiplier:
 
-$$
-s = a + ar^1 + ar^2 + ar^3 + ... = \sumn ar^{n-1}
-$$
+\begin{align*}
+  s = a + ar^1 + ar^2 + ar^3 + ... = \sumn ar^{n-1}
+\end{align*}
 
 Theorem: if $|r| < 1$ then the geometric series converges. Its value is $\frac{a}{1-r}$. For $|r| \geq 1$ it diverges.
 
@@ -91,21 +93,21 @@ \subsubsection{Typical questions about series}
 \textbf{Find the values of x for which the series converges. Find the sum of the series for those values of x.}
 
 - 11.2.59
-  $$
+\begin{align*}
   \sum_{n=0}^{\infty} \frac{(x-2)^n}{3^n}
-  $$
-  We have to recognize that this is a geometric series. Rewrite it as
-  $$
+\end{align*}
+We have to recognize that this is a geometric series. Rewrite it as
+\begin{align*}
   \sum_{n=0}^{\infty} \Big(\frac{x-2}{3}\Big)^n
-  $$
-  and it's clear that it is a GS with $a=1$ and $r=\frac{x-2}{3}$. Therefore it converges if
-  $$
+\end{align*}
+and it's clear that it is a GS with $a=1$ and $r=\frac{x-2}{3}$. Therefore it converges if
+\begin{align*}
   -1 < \frac{x-2}{3} < 1 \Leftrightarrow -1 < x < 5
-  $$
-  and when x is in that range, its sum is
-  $$
+\end{align*}
+and when x is in that range, its sum is
+\begin{align*}
   \frac{a}{1-r} = \frac{1}{1 - \frac{x-2}{3}} = \frac{3}{5-x}.
-  $$
+\end{align*}
 
 \subsubsection{Test for divergence of series}
 If the \textbf{sequence} does not converge to zero then the \textbf{series} (the sum of the infinite sequence) is divergent. That's fairly intuitive: if the sequence doesn't converge to zero then the series is going to be summing infinitely many non-zero terms and so will not converge.
@@ -120,24 +122,24 @@ \subsubsection{Integral test for convergence}
 \textbf{Integral test}: If $\int_1^\infty f(x) dx$ is convergent (has a finite value) then the series is convergent. And if the integral is divergent then the series is divergent.
 
 For example
-$$
-\int_1^\infty \frac{1}{x^2} dx = \frac{-1}{x} \Big|_1^\infty = 1.
-$$
+\begin{align*}
+  \int_1^\infty \frac{1}{x^2} dx = \frac{-1}{x} \Big|_1^\infty = 1.
+\end{align*}
 
 so that shows that $\sumn \frac{1}{n^2}$ converges (to something, we don't know what). On the other hand
 
-$$
-\int_1^\infty \frac{1}{x} dx = \ln x \Big|_1^\infty = \lim_{x \to \infty} \ln x - 0
-$$
+\begin{align*}
+  \int_1^\infty \frac{1}{x} dx = \ln x \Big|_1^\infty = \lim_{x \to \infty} \ln x - 0
+\end{align*}
 
 which is divergent (not a finite value). So that shows that $\sumn \frac{1}{n}$ does not converge.
 
 \textbf{Remainder theorem for integral test}
 Because of the details of the location of the rectangle lines in the approximation, the size of the remainder lies between these two integrals:
 
-$$
-\int_{x=n+1}^\infty f(x) dx \leq R_n \leq \int_{x=n}^\infty f(x) dx
-$$
+\begin{align*}
+  \int_{x=n+1}^\infty f(x) dx \leq R_n \leq \int_{x=n}^\infty f(x) dx
+\end{align*}
 
 
 \subsubsection{p-series}
@@ -157,9 +159,9 @@ \subsubsection{Comparison to known-convergent series}
 \subsubsection{Limit comparison test}
 This involves looking at the limit of the \textbf{ratio} of the terms from two different series. Suppose you have a series $\sumn b_n$ which you know either converges or diverges, and you have a series $\sumn a_n$ which you don't know the behavior of. If you can show that their ratio has a finite limit:
 
-$$
-\limn \frac{a_n}{b_n} = c > 0
-$$
+\begin{align*}
+  \limn \frac{a_n}{b_n} = c > 0
+\end{align*}
 then either \textbf{both converge} or \textbf{both diverge}. Seeing as you know how $b$ behaves, that tells you whether $a$ converges or not.
 
 
@@ -191,13 +193,13 @@ \subsection{Ratio and root tests}
 \subsubsection{Ratio test}
 Write down the ratio of successive terms in the series $\frac{a_{n+1}}{a_n}$, take the absolute value and take the limit of that. So $\limn \Big|\frac{a_{n+1}}{a_n}\Big|$. There are 3 possible outcomes:
 
-$$
-\begin{cases}
-0 < \text{limit} < 1&Convergent\\
-\text{limit} = 1&Inconclusive\\
-\text{limit} > 1 ~\text{or}~ \text{limit} = \infty&Divergent\\
-\end{cases}
-$$
+\begin{align*}
+  \begin{cases}
+    0 < \text{limit} < 1&Convergent\\
+    \text{limit} = 1&Inconclusive\\
+    \text{limit} > 1 ~\text{or}~ \text{limit} = \infty&Divergent\\
+  \end{cases}
+\end{align*}
 
 If the original series was alternating, then these conclusions of "convergent" and "divergent" apply to the alternating series too, as well as the series of absolute values.
 
@@ -208,13 +210,14 @@ \subsubsection{Root test}
 Evaluate the limit of the absolute value of the $\nth$ root of the $\nth$: $|\sqrt[n]{a_n}|$. The cases are the same as for the ratio test:
 
 
-$$
-\begin{cases}
-0 < \text{root} < 1&Convergent\\
-\text{root} = 1&Inconclusive\\
-\text{root} > 1 ~\text{or}~ \text{root} = \infty&Divergent\\
-\end{cases}
-$$
+\begin{align*}
+  \begin{cases}
+    0 < \text{root} < 1&Convergent\\
+    \text{root} = 1&Inconclusive\\
+    \text{root} > 1 ~\text{or}~ \text{root} = \infty&Divergent\\
+  \end{cases}
+\end{align*}
+
 
 
 \subsection{Strategy}
@@ -238,9 +241,9 @@ \subsubsection{Maclaurin series derivation}
 
 Suppose that any function of a real number $f(x)$ can be represented by a "power series" with certain coefficients $c_i$
 
-$$
-f(x) = c_0 + c_1x^1 + c_2x^2 + c_3x^3 + c_4x^4 + ...
-$$
+\begin{align*}
+  f(x) = c_0 + c_1x^1 + c_2x^2 + c_3x^3 + c_4x^4 + ...
+\end{align*}
 
 If that is so, there are two questions:
 
@@ -252,71 +255,71 @@ \subsubsection{Maclaurin series derivation}
 
 To answer (1), firstly, without differentiating at all, we see that $c_0 = f(0)$. Then differentiate once:
 
-$$
-f'(x) = c_1 + 2c_2x^1 + 3c_3x^2 + 4c_4x^3 + ...
-$$
+\begin{align*}
+  f'(x) = c_1 + 2c_2x^1 + 3c_3x^2 + 4c_4x^3 + ...
+\end{align*}
 
 and we see that evaluating the first derivative at $x=0$ gives the coefficient
 $c_1 = f'(0)$. Differentiating again yields $c_2$:
 
 \begin{align*}
-f''(x) &= (2\cdot 1)c_2 + (3\cdot 2)c_3x^1 + (4\cdot 3)c_4x^2 + ... \\
-\frac{f''(0)}{2} &= c_2
+  f''(x) &= (2\cdot 1)c_2 + (3\cdot 2)c_3x^1 + (4\cdot 3)c_4x^2 + ... \\
+  \frac{f''(0)}{2} &= c_2
 \end{align*}
 
 Continuing like that it becomes clear that each coefficient is equal to an $\nth$ derivative evaluated at zero, divided by a factorial term which results from the repeated differentiation:
 
 \begin{align*}
-f'''(x) &= (3\cdot2)c_3 + (4\cdot3\cdot2)c_4x^1 + ... \\
-\frac{f'''(0)}{3!} &= c_2
+  f'''(x) &= (3\cdot2)c_3 + (4\cdot3\cdot2)c_4x^1 + ... \\
+  \frac{f'''(0)}{3!} &= c_2
 \end{align*}
 
 and in general $c_n = \frac{f^{(n)}(0)}{n!}$.
 
 For example, take $f(x) = e^x$. The derivatives are all the same of course: $f^{(n)}(x) = e^x$. And $e^0 = 1$, so
 
-$$
-e^x = 1 + \frac{x}{1} + \frac{x^2}{2!} + \frac{x^3}{3!} + ... = \sum_{i=0}^\infty \frac{x^i}{i!}
-$$
+\begin{align*}
+  e^x = 1 + \frac{x}{1} + \frac{x^2}{2!} + \frac{x^3}{3!} + ... = \sum_{i=0}^\infty \frac{x^i}{i!}
+\end{align*}
 
 \section{Exercises}
 
 - \textbf{11.1: 3, 25, 37}
 - \textbf{11.2: (14, 29, 42)}
-    - One trick to find sums: use partial fractions and then write first few terms of series out and identify a telescoping sum (q. 14)
+- One trick to find sums: use partial fractions and then write first few terms of series out and identify a telescoping sum (q. 14)
 
 - \textbf{11.3: (7, 16, 25, 26) Determine whether convergent or divergent (Integral test)}
-    - Use integral test (if $\int_1^\infty f(x) dx$ is finite then converges, otherwise diverges)
-    - Although it seems less work to use limit comparison test for some
+- Use integral test (if $\int_1^\infty f(x) dx$ is finite then converges, otherwise diverges)
+- Although it seems less work to use limit comparison test for some
 
 - \textbf{11.4: (18, 25, 26) Determine whether convergent or divergent}
-    - Questions concern series with positive terms
-    - Use Comparison Test or Limit Comparison Test
-    - $\frac{1}{n}$ diverges. So if you can show that terms are larger than this, or that limit of ratio with this exists, then it must also be divergent.
-    - $\frac{1}{n^2}$ converges. If you can show that terms are smaller than this, or that limit of ratio with this exists, then it converges.
+- Questions concern series with positive terms
+- Use Comparison Test or Limit Comparison Test
+- $\frac{1}{n}$ diverges. So if you can show that terms are larger than this, or that limit of ratio with this exists, then it must also be divergent.
+- $\frac{1}{n^2}$ converges. If you can show that terms are smaller than this, or that limit of ratio with this exists, then it converges.
 
 - \textbf{11.5 (5, 8, 11) Determine whether convergent or divergent}
-    - Questions typically concern alternating series
-    - Alternating series test: show that successive values are decreasing \textbf{and} that limit of sequence is 0.
-    - If it's not obviously decreasing, convert $a_n$ to continuous function $f(x)$ and show that derivative is negative or becomes negative for large values of $x$.
+- Questions typically concern alternating series
+- Alternating series test: show that successive values are decreasing \textbf{and} that limit of sequence is 0.
+- If it's not obviously decreasing, convert $a_n$ to continuous function $f(x)$ and show that derivative is negative or becomes negative for large values of $x$.
 
 - \textbf{11.6 (5, 6, 29) Absolutely convergent | Conditionally convergent | Divergent}
-    - Questions typically concern alternating series
-    - Use Ratio Test on absolute values to check if convergent or divergent. If it is convergent on the Ratio Test then it is "absolutely convergent".
-    - If Ratio Test is inconclusive (limit $= 1$) then the Alternating Series Test might show that it is Conditionally Convergent.
+- Questions typically concern alternating series
+- Use Ratio Test on absolute values to check if convergent or divergent. If it is convergent on the Ratio Test then it is "absolutely convergent".
+- If Ratio Test is inconclusive (limit $= 1$) then the Alternating Series Test might show that it is Conditionally Convergent.
 
 - \textbf{11.8 (10, 12) Find radius / interval of convergence}
-    - Typically, use the ratio test (converges for values of $x$ that make $\limn |\frac{a_{n+1}}{a_n}|$ less than 1)
-    - If they ask for interval, then you need to test the endpoints for convergence separately.
+- Typically, use the ratio test (converges for values of $x$ that make $\limn |\frac{a_{n+1}}{a_n}|$ less than 1)
+- If they ask for interval, then you need to test the endpoints for convergence separately.
 
 - \textbf{11.9 (3, 6) Find a power series representation for a given function, and radius/interval of convergence.}
-    - Match up the function to the formula for the sum of a geometric series ($a/(1-r)$) and thus construct a geometric series that it's equal to.
-    - Use ratio test to determine interval of convergence
-    - A variant is: you have to differentiate the function before it looks like $a/(1-r)$. Then find the power series, and integrate the power series to get something equal to the original function.
+- Match up the function to the formula for the sum of a geometric series ($a/(1-r)$) and thus construct a geometric series that it's equal to.
+- Use ratio test to determine interval of convergence
+- A variant is: you have to differentiate the function before it looks like $a/(1-r)$. Then find the power series, and integrate the power series to get something equal to the original function.
 
 - \textbf{11.10 (8, 9, 17) Find a Taylor / Maclaurin series for a given function, and radius/interval of convergence}
-    - Memorize definition of general Taylor series (Maclaurin is $a=0$)
-    - Compute derivatives and evaluate them at zero
-    - Use those to write out first few terms of Taylor series
-    - If there's a pattern, represent the sum using sigma notation
-    - Use the ratio test to assess radius of convergence
+- Memorize definition of general Taylor series (Maclaurin is $a=0$)
+- Compute derivatives and evaluate them at zero
+- Use those to write out first few terms of Taylor series
+- If there's a pattern, represent the sum using sigma notation
+- Use the ratio test to assess radius of convergence