Title: Part IA / Analysis I
Tags: Integration Limits_And_Convergence Differentiability Power_Series Series Continuity Limits Complex_Differentiation Absolute_Convergence Taylor's_Theorem

Q: Definition. Least upper bound axiom.
A: Every non-empty set bounded above has a supremum

Q: Definition. Supremum.
A: If \(S \subset \mathbb{R}, S \neq \emptyset\) we day that \(K = \text{sup }S\) if \(x \leq K, \forall x \in S\)  and given \(\varepsilon > 0, \exists x \in S, x > K - \varepsilon\)

Q: Lemma. The limit is unique. If \(a_n \to a\) and \(a_n \to b\), then \(a=b\)
A: Given \(\varepsilon > 0\), \(\exists n_1\) such that \(|a_n - a| < \varepsilon\) for all \(n \geq n_1\) and also some \(n_2\) such that \(|a_n - b| < \varepsilon\) for all \(n \geq n_2\). Then let \(N = \text{max}(n_1,n_2)\) so for all \(n \geq N\), \(|a-b| \leq |a_n-a| + |a_n - b| < 2\varepsilon\). So \(a=b\)

Q: Lemma. If \(a_n \to a\) as \(n \to \infty\) and \(n_1 < n_2 , \dots \), then \(a_{n_j} \to a\) as \(j \to \infty\)
A: Given \(\varepsilon > 0, \exists N\) such that \(|a_n - a| < \varepsilon\) for all \(n \geq N\). Since \(n_j \geq j\) by induction, \(|a_{n_j} - a| < \varepsilon\) for all \(j \geq N\)

Q: Lemma. If \((z_n)\) is a complex sequence, then \(z_n \to z\) if and only if \(\text{Re}(z_n) \to \text{Re}(z)\) and \(\text{Im}(z_n) \to \text{Im}(z)\)
A: Note that \(\forall w \in \mathbb{C}: \max(|\text{Re}(w)|, |\text{Im}(w)|) \leq |w| \leq |\text{Re}(w)| + |\text{Im}(w)|\)
If \(z_n \to z\), then \(\forall \varepsilon, \exists N\) such that \(n \geq N \implies |z_n - z| < \varepsilon \implies |\text{Re}(z_n) - \text{Re}(z_n)<\varepsilon| \text{ and } |\text{Im}(z_n) - \text{Im}(z)| < \varepsilon\).
If \(\text{Re}(z_n) \to \text{Re}(z)\) and \(\text{Im}(z_n) \to \text{Im}(z)\) then \[\forall \varepsilon, \exists N_1, N_2 \text{ s.t. } n \geq N_1 \implies |\text{Re}(z_n) - \text{Re}(z)| < \varepsilon \text{ and the same for the imaginary part.}\]
Pick the largest of \(N_1,N_2\) and then for n bigger than this \[|z_n - z| \leq |\text{Re}(z_n) - \text{Re}(z)| + |\text{Im}(z_n) - \text{Im}(z)| < 2\varepsilon\]

Q: Theorem. Bolzano-Weierstrass. If \(x_n\) is a sequence of real numbers and there exists some k such that \(|x_n|<k\) for all n, then we can find \(n_1 < n_2 < n_3 < \dots \) and \(x \in \mathbb{R}\) such that \(x_{n_j} \to x\) as \(j \to \infty\). Any bounded sequence has a convergent subsequence
A: Let \([a_1, b_1]\) be the range of the sequence, then let the midpoint be \(c_1\). Either there are infinitely many values in \(x_n\) in the first half or the second half. Now we repeat this process and half our interval size every time to get \([a_k, b_k]\) where \(a_k = a_{k-1}, b_k = \frac{b_{k-1} + a_{k-1}}{2}\) or \(a_k = \frac{a_{k-1} + b_{k-1}}{2}, b_k = b_{k-1}\) so \(b_k - a_k = \frac{1}{2}(b_{k-1} - a_{k-1})\). Now by construction \(a_k \leq a_{k+1} < b_{k+1} \leq b_k\), so \((a_k)\) is increasing and bounded above by \(b_1\) and \((b_k)\) is decreasing and bounded below by \(a_1\). So \(a_k \to a \in [a_1,b_1]\) and \(b_k \to b \in [a_1,b_1]\). Which means \((b-a) = \frac{1}{2}(b-a) \implies b = a\). Construct \(n_j\) as follows. Pick \(n_1 = 1\), then pick \(n_{j+1}\) to be the smallest integer greater than \(n_j\) such that \(x_{n_j} \in [a_j, b_j]\) which we can do as there are infinitely many terms in each interval. Now \(a_j \leq x_{n_j}\leq b_j \implies x_{n_j} \to a\).

Q: Definition. Cauchy Sequence. A sequence \(a_n\) is called a Cauchy sequence [...]
A: if given \(\varepsilon > 0\) there exists \(N > 0\) such that \(|a_n -a_m| < \varepsilon\) for all \(n,m \geq N\). Informally, the terms of the sequence grow ever closer together such that there are infinitely many consecutive terms within a small region.

Q: Lemma. If a sequence converges, it is a Cauchy sequence
A: If \(a_n \to a\), given \(\varepsilon > 0\) then \(\exists N\) such that \(\forall n \geq N, |a_n - a| < \varepsilon\). Then take \(m,n \geq N\), and we have \[|a_n - a_m| \leq |a_n - a| + |a_m - a| < 2 \varepsilon\]

Q: Theorem. Every Cauchy sequence convereges.
A: Take \(\varepsilon = 1\), so \(N = N(1)\) in the Cauchy property and for all \(m,n \geq N, |a_n - a_m| < 1\).
So by triangle inequality \(|a_m| \leq ||a_m - a_N| + |a_N| < 1 + |a_N|\). So the sequence after this point is bounded.
Then we pick the maxmimum of the remaining term to produce a bound k for all n.
By Bolzano-Weierstrass this has a convergent subsequence.
Given \(\varepsilon > 0\), there exists \(j_0\) such that \(|a_{n_j} - a| < \varepsilon\) for all \(j \geq j_0\). Also \(\exists N(\varepsilon)\) such that \(|a_m - a_n| < \varepsilon \) for all \(m,n \geq N(\varepsilon)\). Combining these, we take j such that \(n_j \geq \max(N(\varepsilon), n_{j_0})\). So for \(n \geq N(\varepsilon)\), using the triangle inequality \[|a_n - a| \leq |a_n - a_{n_j}| + |a_{n_j} - a| < 2\varepsilon\]

Q: Lemma. Suppose \(\exists N\) such that \(a_j = b_j\) for all \(j \geq N\). Then either \(\sum_{j=1}^{\infty}a_j\) and \(\sum_{n=1}^{\infty}b_j\) both converge, or they both diverge. In other words, the inital terms do not matter for considering convergence
A: For any \(n \geq N\), write \(s_n = \sum_{j=1}^na_j = \sum_{j=1}^{N-1}a_j + \sum_{j=N}^na_j, d_n = \sum_{j=1}^nb_j = \sum_{j=1}^{N-1}b_j + \sum_{j = N}^n b_j\). Taking the difference we get \[s_n - s_n = \sum_{j=1}^{N-1}a_j - \sum_{j=1}^{n-1}b_j\] which is finite. So \(s_n\) converges if and only if \(d_n\) also converges.

Q: Lemma. If \(\sum_{j=1}^{\infty} a_j\) converges, then \(\lim_{j \to \infty}a_j = 0\)
A: Given \(s_n = \sum_{j=1}^{\infty} a_j\), we have \(a_n = s_n - s_{n-1}.\) If \(s_n \to a\), then \(a_n \to 0\) since \(s_{n-1}\) also tends to a.

Q: Theorem. Comparison Test. Suppose \(0 \leq b_n \leq a_n\) for all n. If \(\sum_{j=1}^{\infty}a_j\) converges, then \(\sum_{j=1}^{\infty}b_j\) converges.
A: Let \(s_N\) be the Nth partial sum over the \(a_n\), and let \(d_N\) be the Nth partial sum over the \(b_n\).
Since \(b_n \leq a_n, d_N \leq s_N\). But \(s_N \to s\), so \(d_N \leq s_N \leq s\). So \(d_N\) is an increasing sequence that is bounded above by s, so it converges.

Q: Theorem. Cauchy's Root Test. Suppose we have a sequence of non-negative terms \(a_n\). Suppose that \(a_n^{1/n} \to a\) as \(n \to \infty\). Then if \(a < 1\), the series \(\sum a_n\) converges. If \(a > 1\), the series \(\sum a_n\) diverges.
A: If \(a < 1\), let us choose an r such that \(a < r < 1\). By the definition of the limit, \(\exists N\) such that \(\forall n \geq N, a_n^{1/n} < r\). This implies that \(a_n < r^n\). The geometric series \(\sum r^n\) converges. By comparison the series \(a_n\) converges.
If \(a > 1\), for all \(n \geq N\), \(a_n^{1/n} > 1\) which implies \(a_n > 1\), thus \(\sum a_n\) diverges, since \(a_n\) does not tend to 0.

Q: Theorem. D'Alembert's Ratio Test. Suppose \(a_n > 0\), and \(\frac{a_{n+1}}{a_n} \to l\). If \(l < 1\), then the series \(\sum a_n\) converges. If \(l > 1\), then the series \(\sum a_n\) diverges.
A: Suppse \(l < 1\), we can choose \(l < r < 1\), \(\exists N\) such that \(\forall n \geq N\), \(\frac{a_{n+1}}{a_n} < r\). Therefore \(a_n < r^{n-N}a_N\). Hence, \(a_n < kr^n\) where k is independent of n. Applying the comparison test, the series must converge.
If \(l > 1\), we can choose \(l > r > 1\). Then \(\exists N\) such that \(\forall n \geq N\), \(\frac{a_{n+1}}{a_n} > r\). As before, \(a_n > r^{n-N}a_N\). But the \(r^{n-N}\) diverges, so the original series diverges.

Q: Theorem. Cauchy's Condensation Test. Let \(a_n\) be a decreasing sequence of positive terms. Then \(\sum_{n=1}^{\infty}a_n\) converges if and only if \(\sum_{n=1}^{\infty} 2^n a_{2^n}\)  converges.
A: If \(a_n\) is decreasing then \(a_{2^k} \leq a_{2^{k-1}+1} \leq a_{2^{k-1}}; 1 \leq i \leq 2^{k-1}; k \geq 1\)
Assume \(\sum a_n\) converges to A. Then \[2^{n-1}a_{2^n} = a_{2^n} + \dots + a_{2^n} \leq a_{2^{n-1}+1} + a_{2^{n-1}+2} + \dots + a_{2^n} = \sum_{m = 2^{n-1}+1}^{2^n}a_m\]Thus, \(\sum_{n=1}^N 2^{n-1}a_{2^n} \leq \sum_{n=1}^N \sum_{m=2^{n-1}+1}^{2^n} a_m = \sum_{n=2}^{2^N}a_m.\)
Therefore \(\sum_{n=1}^N 2^na_{2^n} \leq 2\sum_{n=2}^{2^N}a_m \leq 2(A-a_1)\)
Thus \(\sum_{n=1}^N2^na_{2^n}\) converges, since it is increasing and bounded above. For the converse assume that \(\sum 2^n a_{2^n}\) converges to B.
Then \(\sum_{m=2^{n-1}}^{2^n} a_m = a_{2^{n-1}} + a_{2^{n-1}+1} + \dots + a_{2^n} \leq a_{2^{n-1}} + a_{2^{n-1}} + \dots + a_{2^{n-1}} = 2^{n-1}a_{2^{n-1}}\).
So we have \(\sum_{m = 2}^{2^N} a_m = \sum_{n=1}^N \sum_{2^{n-1}+1}^{2^n} a_m \leq \sum_{n=1}^N 2^{n-1}a_{2^{n-1}} \leq \frac{1}{2}B\).
So \(\sum_{m=1}^N\) is a bounded, increasing sequence and hence converges.

Q: Theorem. Alternating Series Test. If \(a_n\) decreases and tends to zero as \(n \to \infty\), then the alternating series \[\sum_{n=1}^{\infty} (-1)^{n+1} a_n\] converges.
A: Let us consider the partial sum \[s_n = a_1 - a_2 + a_3 - \dots +(-1)^{n+1}a_n\]In particular, \[s_{2n} = (a_1 - a_2) + (a_3 - a_4) + \dots + (a_{2n-1} - a_{2n})\]Since the sequence is decreasing, each parenthesised block is positive. Then \(s_{2n} \geq 2_{2n - 2}\). We can also write the partial sum as \[s_{2n} = a_1 - (a_2 - a_3) - (a_4 - a_5) - \dots -(a_{2n-2}- a_{2n-1}) - a_{2n}\]
Each parenthesised block is negative so \(s_{2n} \leq a_1\). So \(2_{2n}\) is increasing and bounded above, so it must converge. Now note that \[s_{2n+1} = s_{2n} + a_{2n+1} \to s_{2n}\]
since \(a_{2n+1} \to 0\). So \(s_{2n+1}\) also converges, in fact to the same limit. Hence \(s_n\) converges to this same limit.

Q: Theorem. If \(\sum_{n+1}^{\infty} a_n\) is absolutely convergent, then it is convergent
A: Let \(s_n = \sum_{k=1}^n a_k\) and \(\overline{s_n} = \sum_{k=1}^n |a_k|\)
Suppose \(n \leq m\), then \[|s_m - s_n| = \left | \sum_{k=n+1}^m a_k\right | \leq \sum_{k=n+1}^m |a_k| = \overline{s_m} - \overline{s_n}\]
\((\overline{s_n})\) is convergent by assumption, hence Cauchy, so \(s_n\) is also Cauchy and hence convergent.

Q: Theorem. If \(\sum_1^{\infty}a_n\) is absolutely convergent, then every rearrangement of this series converges to the same value.
A: Consider the real case. Let \(t_n = \sum_1^n a_n'\) be a rearrangement of \(s_n = \sum_1^n a_n\). Let \(s_n \to s\). Suppose that \(a_n \geq 0\).
Given any \(n \in \mathbb{N}\), we can find some \(q \in \mathbb{N}\) such that \(s_q\) contains every term of \(t_n\). Since the \(a_n \geq 0\), \[t_n \leq s_q \leq s\]As \(n \to \infty\), the \(t_n\) are increasing and bounded above, so it must tend to some \(t \leq s\).
This argument is symmetric so we can also get that \(s \leq t \implies s = t\).
Now if \(a_n \not \geq 0\).
Consider \(v_n = \frac{|a_n| + a_n}{2}\) and \(w_n = \frac{|a_n|-a_n}{2}\).
Since \(\sum |a_n|\) converges, both \(\sum v_n, \sum w_n\) converge.
Since all \(v_n, w_n \geq 0\), we can deduce that \(\sum v_n = \sum v_n'\) and \(\sum w_n = \sum w_n'\). The claim follows since \(a_n = v_n - w_n\).
If \(a_n \in mathbb{C}\), write \(a_n = x_n + iy_n\) and the same arguments apply as the series \(\sum x_n, \sum y_n\) are absolutely convergent.

Q: Definition. \(f : E \to \mathbb{C}\) is continous at a if for every sequence \(z_n \in E\) that converges to a, we have \([...]\)
A: f(z_n) \to f(a)

Q: Definition. \(f: E \to \mathbb{C}\) is continous at \(a\) if given \(\varepsilon > 0, \exists \delta > 0\) such that for every \(z \in E\), if [...]
A: \(|z-a| < \delta\), then \(|f(z) - f(z)| < \varepsilon\)

Q: Claim. The \(\varepsilon - \delta\) definition of continuity imples the sequence convergence definition.
A: We know that given \(\varepsilon > 0, \exists \delta > 0\) such that for all \(z \in E, |z - a| < \delta\) implies \(|f(z) - f(a)| < \varepsilon\). Let \(z_n \to a\), then by the definition of the limit of the sequence then there exists \(n_0\) such that for all \(n \geq n_0\) we have \(|z_n - a| < \delta\). But this implies that \(|f(z_n) - f(z)| < \varepsilon\), i.e. \(f(z_n) \to f(z)\).

Q: Claim. The convergence definition of continuity implies the \(\varepsilon - \delta\) definition.
A: We know that for every sequence \(z_n \in E\) that converges to a, \(f(z_n) \to f(a)\).
Suppose f is not continous as a, according to the \(\varepsilon - \delta\) definition.
Then there exists some \(\varepsilon\) such that for all \(\delta > 0\), there exists \(z \in E\) such that \(|z-a| < \delta\) but \(|f(z) - f(z)| \geq \varepsilon\).
So, let us contruct a sequence of \(\delta\) values to substitute into this definiton.
Let \(\delta = 1/n\). Then the \(z_n\) given by this \(\delta\) is such that \(|z_n - a| < 1/n\) and \(|f(z_n) - f(z)| \geq \varepsilon\).
Clearly, \(z_n \to a\), but \(f(z_n)\) does not tend to \(f(a)\) because the difference between the two is always greater than \(\varepsilon\).
This is a contradiction since we assumed that f is continous by the convergence defintion.

Q: Proposition. Let \(g,f : E \to \mathbb{C}\) be continous functions at a point \(a \in E\). Then all of the functions \[f(z) + g(z)\] \[f(z)g(z)\] \[\lambda f(z), \lambda \in \mathbb{R}\]are all continous. In addition, if \(f(z) \neq 0\) everywhere in E, then \(\frac{1}{f}\) is a continuous function at a.
A: Using the convergence definition of continuity, this is obvious using the fact that limits of sequences behave analogously

Q: Theorem. Let \(f : A \to \mathbb{C}\) and \(g : B \to \mathbb{C}\) where \(A,B \subseteq \mathbb{C}\) be two functions that can be composed, i.e. \(f(A) \subseteq B\). If \(f\) is continuous at \(a \in A\) and \(g\) is continuous at \(f(a) \in B\), then \(g \circ f: A \to \mathbb{C}\) is continous at a.
A: Take any sequence \(z_n \to a\). By assumtpion, \(f(z_n) \to f(z)\).
Now let us define a new sequence \(w_n = f(z_n)\). Then \(w_n \in B\) and \(w_n \to f(a)\).
Thus, \(g(f(z_n)) = g(w_n) \to g(f(a))\) by continuity, as required.

Q: Definition. Let \(E \subseteq \mathbb{C}, a \in \mathbb{C}\). \(a\) is a limit points of E if [...]
A: for any \(\delta > 0\), there exists \(z \in E\) such that \(0 < |z-a| < \delta\)

Q: Definition. Let \(f: E \subseteq \mathbb{C} \to \subseteq{C}\), and let \(a \in \mathbb{C}\) be a limit point of E. We say that \(f \to l\) as \(z \to a\), if [...]
A: given \(\varepsilon > 0\) there exists \(\delta > 0\) such that whenever \(0 < |z-a| < \delta\) and \(z \in E, |f(z) - l| < \varepsilon\). Equivalently, \(f(z_n) \to l\) for every sequence \(z_n \in E\), such that \(z_n \to a\) but \(z_n \neq a\).

Q: Theorem. Intermediate Value Theorem. Let \(f: [a,b] \to \mathbb{R}\) be a continuous function where \(f(a) \neq f(b)\). Then f takes all values in the interval \([f(a), f(b)]\).
A: WLOG assume \(f(a) < f(b)\). Take an \(\eta \) such that \(f(a) < \eta < f(b)\).
WTP that exists some \(c \in [a,b]\) with \(f(c) = \eta\)
Let \(S = \{x \in [a,b]:f(x) < \eta\}\)
\(a \in S\) so it is non-empty, it is bounded above by b so there is a supremum.
Let \(\sup S = c, c\leq b \).
By definition of supremum, given \(n\) there exists \(x_n \in S\) such that \(c - \frac{1}{n} < x_n \leq c\).
So the sequence \(x_n\) tends to c. We know that \(f(x_n) < \eta\) for all \(x_n\) by definition of the set S. By the continuity of f, \(f(x_n) \to f(c)\).
Thus \(f(c) \leq \eta\).
Now consider the fact that \(c \neq b\), if \(c = b\), then \(f(b) \leq \eta\) which is a contradiction since \(\eta < f(b)\).
For large n, we can ensure that \(c + \frac{1}{n} \in [a,b]\). So by continuity of the function \(f(c + \frac{1}{n}) \to f(c)\).
But \(f(c + \frac{1}{n}) \geq \eta\) because c is the supremum of s.
Thus \[f(c) \geq \eta \implies f(c) = \eta\]

Q: Theorem. Let \(f: [a,b] \to \mathbb{R}\) be continuous. Then the function is bounded, i.e. there exists \(k \in \mathbb{R}\) such that \(|f(x)| \leq k\) for every point \(x \in [a,b]\).
A: Suppose that such a function is not bounded.
So for any integer \(n \geq 1\), there exists \(x_n \in [a,b]\) such that \(|f(x_n)| > n\).
By Bolzano-Weierstrass, the sequence \(x_n\), which is bounded by \(a \leq x_n \leq b\), has a convergent subsequence \(x_{n_j} \to x\) such that \(x \in [a,b]\).
Then by continuity of f, \(f(x_{n_j}) \to f(x)\). But \(|f(x_{n_j})| > n_j \to \infty\).
This is a contradiction.

Q: Theorem. Suppose \(f: [a,b] \to \mathbb{R}\) is a continuous function. Then there exists \(x_1, x_2 \in [a,b]\) such that \[f(x_1) \leq f(x) \leq f(x_2)\] for all \(x \in [a,b]\). In other words, a continuous function on a closed bounded interval is bounded and attains its bounds.
A: Let \(A = \{f(x) : x \in [a,b]\}\) be the image of \([a,b]\) under f.
A must be bounded, it is also non-empty and hence has a supremum \(M = \sup A\).
Then by definition of supremum, given some integer \(n \geq 1\) there exists \(x_n \in [a,b]\) such that \(M - \frac{1}{n} < f(x_n) \leq M\).
By Bolzano-Weierstass theorem, there exists a convergent subsequence, \(x_{n_j} \to x \in [a,b]\).
Since \(f(x_{n_j}) \to M\), then by continuity, \(f(x) = M\)

Q: Theorem. Let \(f:[a,b] \to \mathbb{R}\) be continous and stricly increasing for \(x \in [a,b]\). Let \(c = f(a), d = f(b)\). Then \(f:[a,b] \to [c,d]\) is bijective, and the inverse \(g := f^{-1}[c,d] \to [a,b]\) is continuous and strictly increasing.
A: Let \(c < k <d\). From IVT, there exists h such that \(f(h) = k\). This h must be unique since the function is strictly increasing.
Then we can define \(g(k)=h\), giving us an inverse for f.
First note that g is strictly increasing. Indeed for \(y_1 < y_2\), then \(y_1 = f(x_1), y_2 = f(x_2)\). This means if \(x_2 \leq x_1\) then since f increasing \(y_2 \leq y_1\) which is a contradiction.
Now, note that g is continuous. Given \(\epsilon > 0\), let \(k_1 = f(h-\epsilon)\) and \(k_2 = f(h+\epsilon)\). If f is strictly increasing then \(k_1 < k < k_2\). Then \(h-\epsilon < g(y) < h + \epsilon\). So let \(\delta = \min (k_2 - k, k - k_1)\) establisihing continuity as claimed.

Q: Defintion. Let \(x \in E\) be a point such that there exists a sequence \(x_n \in E\) with \(x_n \neq x\) but \(x_n \to x\), i.e. x is a limit point. \(f\) is said to be differentiable at x with derivative \(f'(x)\) if
A: \[\lim_{y \to x} \frac{f(y) - f(x)}{y - x} = f'(x)\]we could rewrite the definition as the following \[\epsilon(h) := f(x+h) - f(x) - hf'(x); \; \; \lim_{h\to 0} \frac{\epsilon(h)}{h} = 0\]with f being differentiable at x if there exists A and \(\epsilon\) such that \[f(x+h) = f(x) + hA + \epsilon(h)\]
so then \[f'(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h}\]

Q: Theorem. Let \(f: U \to \mathbb{C}\) be such that \(f(x) \in V\) for all \(x \in U\). If f is differentiable at \(a \in U\), and \(g: V \to \mathbb{C}\) is differentiable at \(f(a) \in V\), then \(g \circ f\) is differentiable at a with \[(gf)'(a) = f'(a)g'(f(a))\]
A: Since f,g differentiable we can write \[f(x) = f(a) + (x-a)f'(a) + \epsilon_f(x)(x-a) \; \; \; \lim_{x \to a} \epsilon_f(x) = 0\]\[g(y) = g(b) + (y-b)g'(b) + \epsilon_g(y)(y-b) \; \; \; \lim_{y \to b} \epsilon_g(y) = 0 \;\;\;\; f(a) = b\]Now compose function so \(y = f(x)\), we get \[g(f(x)) = g(f(a)) + (x-a)f'(a)g'(b) + (x-a)\sigma(x)\]And you can show that \(\lim_{x \to a} \sigma(x) = 0\) so we get the result as required.

Q: Theorem. Rolle's Theorem. Let \(f: [a,b] \to \mathbb{R}\) be a continuous function on \([a,b]\) and differentiable on \((a,b)\). If \(f(a) = f(b)\), then there exists \(c \in (a,b)\) such that \(f'(c) = 0\)
A: Let M be the maximum point and m be the minimum point of the function. We know that any function achieves its bounds. Let \(k = f(a)\). If M=m=k, then f must be constant, and clearly \(f'(c) = 0\).
Otherwise, either \(M > k\) or \(m < k\) . Suppose \(M > k\).
Then there exists some value \(c \in (a,b)\) such that \(f(c) = M\). We would like to show that \(f'(c) = 0\), so suppose instead that \(f'(c) > 0\), then there are values \(d > c\) where \(f(d) > f(c)\). Indeed \[f(h+c) - f(c) = h[f'(c) + \epsilon(h)]\]
This is positive if h is small, positive. This contradicts that M is the maximum. Similarly if \(f'(c) < 0\) there are values \(d < c\) with \(f(d) > f(c)\). Hence \(f'(c) = 0\)

Q: Theorem. Mean value theorem. Let \(f:[a,b] \to \mathbb{R}\) be a continuous function on \([a,b]\) and differentiable on \((a,b)\). Then there exists \(c \in (a,b)\) such that \[f(b) - f(a) = f'(c)(b-a)\]
A: Let \(\phi\) be a function defined by \(\phi(x) = f(x) - kx\), choosing a k such that \(\phi(a) = \phi(b)\). We can find that \[f(b) - bk = f(a) -ak \implies k = \frac{f(b) - f(a)}{b-a}\]
By Rolle's theorem, there exists \(c \in (a,b)\) such that \(\phi '(c) = 0\). Now note that \(f'(x) = \phi '(x) + k\), hence there exists c such that \(f'(c) = k\)
We can rewrite this as the following \[f(a+h) = f(a) + hf'(a + \theta h)\] where \(\theta \in (0,1)\).

Q: Theorem. Inverse function theorem. Let \(f: [a,b] \to \mathbb{R}\) be a continuous function on \([a,b]\) and differentiable on \((a,b)\), with \(f'(x) > 0\) everywhere on \((a,b)\). Let \(f(a) = c, f(b) = d\). Then the function \(f: [a,b] \to [c,d]\) is bijective, and \(f^{-1}:[c,d] \to [a,b]\) is differentiable on \((c,d)\) with \[(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}\]
A: We know that there exists a strictly increasing inverse \(g: [c,d] \to [a,b]\).
We must now show that g is differentiable and its derivative has the required form.
Let \(y = f(x)\), and let h be given by \[y + k = f(x+h)\] so \[x+h = g(y+k)\]Since g is continuous, if \(k \to 0\) then \(h \to 0\). Then \[\lim_{k\to 0} \frac{g(y+k) - g(y)}{k} = \lim_{h=0} \frac{h}{f(x+h) - f(x)}\] and the result follows.

Q: Theorem. Cauchy's Mean Value Theorem. If \(f,g: [a,b] \to \mathbb{R}\) are continuous, and differentiable on \((a,b)\), there exists \(t \in (a,b)\) such that \[(f(b) - f(a))g'(t) = f'(t)(g(b) - g(a)) \implies \frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(t)}{g'(t)}\]
A: Take \[\phi(x) = \left | \begin{matrix} 1 & 1 & 1 \\ f(a) & f(x) & f(b) \\ g(a) & g(x) & g(b)\end{matrix}\right |\]
It is continuous on \([a,b]\) and differentiable on \((a,b)\) and \(\phi(a) = \phi(b) = 0\) by observing linear dependence of columns.
By Rolle's Thoerem, there exists \(t \in (a,b)\) such that \(\phi ' (t) = 0\). We can expand this and show the required result.
\[\phi'(x) = f'(x)g(b) - g'(x)f(b) + f(a)g'(x) - g(a)f'(x) = f'(x)[g(b) - g(a)] + g'(x)[f(a) - f(b)]\]

Q: Theorem. Taylor's Theorem with Lagrange's Remainder. Suppose f and its derivatives up to order n-1 are continuous in \([a, a+h]\), and \(f^{(n)}\) exists for \(x \in (a, a+h)\). Then \[f(a+h) = f(a) + hf'(a) + \frac{h^2}{2!}f''(a) + \dots + \frac{h^{n-1}}{(n-1)!}f^{(n-1)}(a) + \frac{h^n}{n!}f^{(n)}(a+\theta h)\]
where \(\theta \in (0,1)\)
A: For \(0 \leq t \leq h\), we define \[\phi(t) = f(a+t) - f(a) - tf'(a) - \dots - \frac{t^{n-1}}{(n-1)!} f^{(n-1)}(a) - \frac{t^n}{n!}B\] where we choose B suitably such that \(\phi(h) = 0\).
Note that \[\phi(0) = \phi'(0) = \dots = \phi^{(n-1)}(0) = 0\]Now we can use Rolle's Theorem n times. Since \(\phi(0) = \phi(h) = 0\), there is a point \(0 < h_1 < h\) such that \(\phi'(h_1)=0\). Then since \(\phi'(0) = \phi'(h_1) = 0\), there is a point \(0 < h_2 < h_1\) such that \(\phi''(h_2) = 0\).
This continues until we find a point \(0 < h_n < h\) such that \(\phi^{(n)}(h_n) = 0\). Hence \(h_n = \theta h\) for some \(0 < \theta < 1\). Now \(\phi^{(n)}(t) = f^{(n)}(a+t) - B\). We can now see that \(B = f^{(n)}(a + \theta h)\), which gives the required result.

Q: Theorem. Taylor's Theorem with Cauchy's Remainder. Suppose f and its derivatives up to n-1 are continuous in \([a,a+h]\), and \(f^{(n)}\) exists for \(x \in (a, a+h)\). Then \[f(a+h) = f(a) + hf'(a) + \frac{h^2}{2!}f''(a) + \dots + \frac{h^{n-1}}{(n-1)!}f^{(n-1)}(a) + R_n\]where \[R_n = \frac{(1-\theta)^{n-1} h^n f^{(n)}(a+\theta h)}{(n-1)!}\] for \(\theta \in (0,1)\)
A: For simplicity let a=0. Define \[F(t) = f(h) - f(t) - (h-t)f'(t) - \dots - \frac{(h-t)^{n-1}f^{(n-1)}(t)}{(n-1)!}\]for \(t \in [0,h]\). Then \[F'(t) = -\frac{(h-t)^{n-1}}{(n-1)!} f^{(n)}(t)\]Let
\[\phi(t) = F(t) - \left [ \frac{h-t}{h}\right ]^p F(0)\]where \(p \in \mathbb{N}\) and \(1 \leq p \leq n\). Then \[\phi(0) = \phi(h) = 0\]
By Rolle's Theorem, there exists \(\theta \in (0,1)\) such that \[\phi '(\theta h) = 0\]We can compute \(\phi '\) to find \[\phi ' (\theta h) = F'(\theta h) + \frac{p(1-\theta)^{p-1}}{h}F(0) = 0\]
Substituting everything back into F gives \[ 0 = \frac{-h^{n-1}(1-\theta)^{n-1}}{(n-1)!}f^{(n)}(\theta h) + \frac{p(1 - \theta)^{p-1}}{h}\left [ f(h) - f(0) - h'(0) - \dots - \frac{H^{n-1}}{(n-1)!} f^{(n-1)}(0)\right ]\]Hence \[f(h) = f(0) + hf'(0) + \frac{h^2}{2!}f''(0) + \dots + \frac{h^{n-1}}{(n-1)!}f^{(n-1)}(0) + \frac{(1-\theta)^{n-1} h^n f^{(n)}(\theta h)}{(n-1)!}\]By letting p=n, we get Lagrange's Remainder. If p=1 we get Cauchy's Remainder.

Q: Definition. Power Series.
A: \[\sum_{n=0}^{\infty} a_nz^n\]
where \(z \in \mathbb{C}\) and the \(a_n\) is a given sequence of complex numbers

Q: Lemma. If the series \[\sum_{n=0}^{\infty} a_n z_1^n\]
converges for some point \(z_1\), and \(|z| < |z_1|\), then the series \[\sum_{n=0}^{\infty} a_n z^n\] also converges absolutely.
A: Since \(\sum_{n=0}^{\infty}a_n z_1^n\) converges, \(a_n z_1^n \to 0\). Thus the sequences \(a_n z_1^n\) is bounded by some k>0. Then \[|a_n z^n| \leq k \left | \frac{z}{z_1}\right |^n\]
Since the geometric series \(\sum_{n=0}^{\infty} \left | \frac{z}{z_1} \right |^n \) converges, the lemma follows by comparison.

Q: Theorem. Any power series either
i)converges absolutely for all z
ii)converges absolutely for all z where \(|z|<R\) and diverges for all z where \(|z|>R\)
iii)converges for z = 0 only
A: Let \[S = \left \{ x \in \mathbb{R} : x \geq 0, \sum_0^{\infty} a_nx^n \text{ converges }\right \}\]Clearly, \(0 \in S\). If \(x_1 \in S, \text{ then } [0,x_1] \subseteq S\). If \(S = [0, \infty)\), then we have case i).
If \(S \neq [0, \infty)\), there exists a supremum \(0 \leq R = \sup S < \infty\).
For all \(z_1\) with \(|z_1| < R\) there exists \(R_0\) such that \(|z_1| < R_0 < R\), and abolute convergence follows from our lemma.
If \(|z_1| > R\), there exists \(R_0\) such that \(|z_1| > R_0 > R\). If the series with \(z_1\) converges, then by the lemma the same would be true for \(R_0\). But \(R_0\) does not converge, so this is a contradiction.

Q: Lemma. If \[\left | \frac{a_{n+1}}{a_n}\right | \to l\] as \(n \to \infty\), then \(R = \frac{1}{l}\)
A: By the ratio test we have absolute convergence if \[\left | \frac{a_{n+1}}{a_n} \frac{z^{n+1}}{z^n}\right | < 1\]
So we have absolute convergence if \(|z| < \frac{1}{l}\) and divergence if \(|z| > \frac{1}{l}\) as required.

Q: Lemma. If \[|a_n^{1/n}| \to l\] as \(n \to \infty\), then \(R = \frac{1}{l}\)
A: By the root test we have abolsute convergence if
\[\left| a_n^{1/n}z \right | < 1\]So we have radius of convergence of \(\frac{1}{l}\) as required.

Q: Theorem. Let \(f(z) = \sum_0^{\infty}a_nz^n\) have a radius of convergence R. Then f is complex differentiable at all points with \(|z| < R\), with \[[...]\] with the same radius of convergence as the original series
A: f'(z) = \sum_1^{\infty} na_n z^{n-1}

Q: Theorem. Take the exponential function \(e: \mathbb{R} \to \mathbb{R}\). Then
iii) \( e(x) > 0 \)
iv) e is strictly increasing
v) \(e(x) \to \infty\) as \(x \to \infty\), and \(e(x) \to 0\) as \(x \to -\infty\)
vi) \(e:\mathbb{R} \to (0,\infty)\) is a bijection
A: iii) Clearly e(x) > 0 for all \(x \geq 0\) by considering the power series and the fact that \(e(0) = 1\). Also \(e(0) = e(x-x) = e(x)e(-x)\), hence e(x) is positive for all negative x also.
iv)Since \(e'(x) = e(x)\) which is always positive e is strictly increasing.
v) By considering partial sums, if x>0, we have \(e(x) > 1 + x\), so if \(x \to \infty, e(x) \to \infty\). Using the fact that \(e(-x) = \frac{1}{e(x)}\) we get the other part.
vi) We get injectivity from being strictly increasing. We get surjectivity from the intermediate value theorem and the fact that e spans all the reals as shown in v).

Q: Theorem. Let l be the inverse function of e(exponential) as previously defined.
i) \(l : (0,\infty) \to \mathbb{R}\) is a bijection, and \(l(e(x))=x\) for all \(x \in \mathbb{R}\), and \(e(l(x)) = x\) for all \(x \in (0,\infty)\).
ii) l is differentiable and its derivative is \(l'(t) = \frac{1}{t}\).
iii) \(l(xy) = l(x) + l(y)\)
A: i) Comes from definition
ii)By the inverse function theorem l is differentiable everywhere with derivative \(\frac{1}{t}\) as required.
iii)From IA Groups, if e is an isomorphism, so is its inverse.

Q: Theorem. Suppose \(x,y > 0\) and \(\alpha, \beta \in \mathbb{R}\) . Let \(r_{\alpha}(x) = e(\alpha l(x))\). This is like raising x to the power \(\alpha\)
i)\(r_{\alpha}(xy) = r_{\alpha}(x)r_{\alpha}(y)\)
ii)\(r_{\alpha + \beta}(x) = r_{\alpha}(x)r_{\beta}(y)\)
iii)\(r_{\alpha}(r_{\beta}(x)) = r_{\alpha \beta}(x)\)
iv) \(r_1(x) = x\), and \(r_0(x) = 1\)
A: I will only prove i)
\(r_{\alpha}(xy) = e(\alpha l(xy)) = e(\alpha l(x) + \alpha l(y)) = e(\alpha l(x)) e(\alpha l(y)) = r_{\alpha}(x) r_{\alpha}(y)\)

Q: Lemma. For any \(r>0\)
i)\(x^r e^{-x} \to 0\) as \(x \to \infty\)
ii) \(x^{-r} \log x \to 0\) as \(x \to \infty\)
iii)\(x^r \log x \to 0\) as \(x \to 0^+\)
A: i)For \(x > 0\), \(e^{x} = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} + \dots>\frac{x^n}{n!}\)
For r>0, pick n such that \(n-r \geq 1\). Then \(0 \leq \frac{x^r}{e^x} \leq \frac{n!}{x^{n-r}} \leq \frac{n!}{x}\)
So \(x^r e^{-x} \to 0\) as \(x \to \infty\)
ii) Pick \(m \in \mathbb{N}\) such that \(m \geq \frac{2}{r}\).
Then \(0 \leq \frac{x^{1/r}}{e^x} \leq \frac{m!}{x^{m-1/r}} \leq \frac{m!}{x^{1/r}}\) as \(x > 1\)
For \(y > e\) let \(x = \log y\)
\(0 \leq \frac{(\log y)^{1/r}}{y} \leq \frac{m!}{(\log y)^{1/r}} \implies 0 \leq \frac{\log y}{y^r} \leq \frac{(m!)^r}{\log y}\)
But \(\log y \to \infty\) as \(y \to \infty\) so \(\frac{\log y}{y^r} \to 0\) as \(y \to \infty\)
iii) Set \(\zeta = \frac{1}{y}\) as above.
\(0 \leq - \zeta^r \log \zeta \leq \frac{(m!)^r}{- \log \zeta}\).
As \(\zeta \to 0^+, \log \zeta \to \infty\), so \(\zeta^r \log z \to 0\)

Q: Proposition. There is a smallest positive number \(\pi\) such that \[\cos \frac{\pi}{2} = 0\] and we have \(\sqrt{2} < \frac{\pi}{2} < \sqrt{3}\)
A: If 0<x<2, \[\sin x = \left(x - \frac{x^3}{3!} \right ) + \left ( \frac{x^5}{5!} - \frac{x^7}{7!}\right ) + \cdots\]Each parenthesised block is positive, so \(\sin x > 0\). So in this range, \[\frac{d}{dx} \cos x < 0\]So cos is strictly decreasing on this interval. Then sub in \(\sqrt{2}\) and \(\sqrt{3}\) into the power series for cos and conclude by the intermediate value theorem that such a \(\pi\) exists.

Q: Definition. Dissection or Partition \(\boldsymbol{\mathcal{D}}\) of \([a,b]\) is [...]
A: a finite subset of \([a,b]\) containing the end points a and b. We write \[\mathcal{D} = \{x_0, x_1, \dots,x_n\}\]with \(a = x_0 < x_1 < \dots < x_{n-1} < x_n = b\)

Q: Definition. Upper and Lower sums. Of a bounded function f associated with a dissection or partition \(\mathcal{D}\) is defined by \[Upper: [...]\]\[Lower: [...]\]
A: S(f,\mathcal{D}) = \sum_{j=1}^n (x_j - x_{j-1}) \sup_{x\in [x_{j-1}, x_j]} f(x); s(f,\mathcal{D}) = \sum_{j=1}^n (x_j - x_{j-1}) \inf_{x\in [x_{j-1}, x_j]} f(x)

Q: Lemma. If \(\mathcal{D}\) and \(\mathcal{D}'\) are dissections with \(D \subseteq \mathcal{D}'\), then \[S(f,\mathcal{D}) \geq S(f, \mathcal{D}') \geq s(f, \mathcal{D}') \geq s(f,\mathcal{D})\]
A: Suppose \(\mathcal{D}'\) contains a single extra point y compared to \(\mathcal{D}\), where \(y \in (x_{r-1}, s_r)\). Clearly \[\sup_{x \in [x_{r-1}, y]} f(x), \sup_{x \in [y, x_r]} f(x) \leq \sup_{x \in [x_{r-1}, x_r]} f(x)\]Then \[(x_r - x_{r-1})\sup_{x \in [x_{r-1}, x_r]} f(x) \geq (y-x_{r-1}) \sup_{x \in [x_{r-1}, y]} f(x) + (x_r - y) \sup_{x\in[y,x_r]} f(x)\]Hence, \[S(f, \mathcal{D}) \geq S(f, \mathcal{D}')\]We can keep adding points in this way and the inequality holds.

Q: Lemma. If \(\mathcal{D}_1, \mathcal{D}_2\) are arbitrary dissections, then \[S(f, \mathcal{D_1}) \geq s(f, \mathcal{D_2})\]
A: Let \(\mathcal{D}' = \mathcal{D_1} \cup \mathcal{D_2}\), which is a refinement of both \(\mathcal{D_1}\) and \(\mathcal{D_2}\), apply the previous lemma with the chain of refinements and we get the result.

Q: Definition. The upper integral of f is \[[...]\]
A: I^* (f) = \inf_{\mathcal{D} } S(f,\mathcal{D})
Back Extra: The lower integral is \[I_*(f) = \sup_{\mathcal{D}} s(f,\mathcal{D})\]From the lemmas we've proven, the lower integral has to be less than or equal to the upper integral.

Q: Definition. A bounded function \(f:[a,b] \to \mathbb{R}\) is Riemann Integrable if [...].
A: \(I^*(f) = I_*(f)\)
Back Extra: If this equality holds, we write \[\int_{a}^b f(x)dx = I^*(f) = I_*(f) = \int_a^b f\]

Q: Theorem. A function \(f: [a,b] \to \mathbb{R}\) is Riemann integrable if and only if given \(\epsilon > 0\), there exists \(\mathcal{D}\) such that \[S(f, \mathcal{D}) - s(f, \mathcal{D}) < \epsilon\]
A: For every dissection \(\mathcal{D}\), we have that \(0 \leq I^*(f) - I_*(f) \leq S(f, \mathcal{D}) - s(f, \mathcal{D})< \epsilon\)  for all \(\epsilon > 0\). This immediately implies that f is Riemann integrable since the upper integral and lower integral match.
Conversely, if f is integrable, by the definition of the supremum and infimum, there are partitions \(\mathcal{D}_1\) and \(\mathcal{D}_2\) such that \[\int_a^b f - \frac{\epsilon}{2} = I_* (f) - \frac{\epsilon}{2} < s(f, \mathcal{D_1})\]
Also, \[\int_a^b f+ \frac{\epsilon}{2} + I^*(f) + \frac{\epsilon}{2} > S(f, \mathcal{D}_2)\].
Then we can use the fact that \(\mathcal{D_1} \cup \mathcal{D}_2\) is a refinement of both partitions to show that \[S(f, \mathcal{D}_1 \cup \mathcal{D}_2) - s(f, \mathcal{D}_1 \cup \mathcal{D}_2) \leq S(f, \mathcal{D}_2) - s(f, \mathcal{D}_1)\]
Now, \[S(f, \mathcal{D}_2) - s(f, \mathcal{D}_1) < \int_a^b f+\frac{\epsilon}{2} - \int_a^b f + \frac{\epsilon}{2} = \epsilon\]

Q: Theorem. Suppose a function \(f: [a,b] \to \mathbb{R}\) is monotonic. Then f is integrable.
A: Suppose f is increasing. Then \(\sup_{x \in [x_{j-1}, x_j]} f(x) = f(x_j)\), and similarly for the infimum. Thus, \[S(f, \mathcal{D}) - s(f, \mathcal{D}) = \sum_{j=1}^n (x_j - x_{j-1})[f(x_j) - f(x_{j-1})]\]
Let us choose the dissection \[\mathcal{D} = \left \{ a,a + \frac{b-a}{n}, a + 2\frac{b-a}{n} , \cdots, b\right \}\]
giving \[x_j = a+ j\frac{b-a}{n}\]
In this case, \[S(f, \mathcal{D}) - s(f, \mathcal{D}) = \frac{b-a}{n}\sum_{j=1}^n [f(x_j) - f(x_{j-1})] = \frac{b-a}{n}[f(b) - f(a)] \to 0\]

Q: Theorem. Suppose a function \(f: [a,b] \to \mathbb{R}\) is continuous. Then f is integrable.
A: We show the contrapositive, if f is not integrable, then f is not continuous.
Suppose f is not integrable. Then \(\exists \epsilon_0 > 0\) such that for all dissections \(\mathcal{D}\), \[S(f, \mathcal{D}) - s(f, \mathcal{D}) > \epsilon_0\]
Since \[S(f, \mathcal{D}) - s(f, \mathcal{D}) = \sum_{i=1}^n (x_i - x_{i-1})(\sup_{x\in [x_{i-1}, x_i]} f(x) - \inf_{x\in[x_{i-1}, x_i]} f(x))\]
It follows that for any dissection there is a j with \[\sup_{x \in [x_{j-1}, x_j]} f(x) - \inf_{x \in [x_{j-1}, x_j]}f(x) > \frac{\epsilon_0}{b-a}\]
In particular we can find \(y,z \in [x_{j-1}, x_j]\) with \[f(y) - f(z) > \frac{\epsilon_0}{b-a}\].
Now let \(\mathcal{D}_n\) be the uniform dissection into n equal intervals, by the argument above there must exists \(y_n, z_n\) in the same sub interval such that \(f(y_n) - f(z_n) > \frac{\epsilon_0}{b-a}\) and \(|y_n - z_n| < \frac{b-a}{n}\).
Now \((y_n)\) is a bounded sequence, so by Bolzano-Weierstrass we can take a convergent subsequence, \(y_{n_j} \to y \in [a,b]\), since \(|y_{n_j} - z_{n_j}| < \frac{1}{n_j}\) we have \(z_{n_j} \to y\).
But \(f(y_{n_j}) - f(z_{n_j}) > \frac{\epsilon_0}{b-a} > 0\) implies \(f(y_{n_j})\) and \(f(z_{n_j})\) cannot converge to the same limit. So f is not continuous at y.

Q: Proposition. If \(f \leq g\) on \([a,b]\), then \(\int f \leq \int g\). Both f,g bounded and integrable.
A: If \(f \leq g\), then \[\int f = I^*(f) \leq S(f, \mathcal{D}) \leq S(g, \mathcal D)\]Hence,
\[\int f = I^* (f) \leq I*(g) = \int g\]

Q: Proposition. \(f+g\) is integrable on \([a,b]\) and \(\int f+g = \int f + \int g\), given that f,g are bounded and integrable.
A: We have \[\sup_{[x_{j-1},x_j]}(f+g) \leq \sup_{[x_{j-1},x_j]}f + \sup_{[x_{j-1},x_j]}g\]Therefore, \[S(f+g, \mathcal{D}) \leq S(f, \mathcal{D}) + S(g, \mathcal{D})\]Now consider two dissections \(\mathcal{D}_1, \mathcal{D}_2\). Now, \[I^*(f+g) \leq S(f+g, \mathcal{D}_1 \cup \mathcal{D}_2) \leq S(f, \mathcal{D}_1 \cup \mathcal{D}_2) + S(g, \mathcal{D}_1 \cup \mathcal{D}_2) \leq S(f, \mathcal{D}_1) + S(g, \mathcal{D}_2)\]We can then fix \(\mathcal{D}_1\) and take the infimum over \(\mathcal{D}_2\) to get \[I^*(f+g) \leq S(f, \mathcal{D_1}) + I^*(g)\]Taking the infimum over \(\mathcal{D}_1\) gives \[I^*(f+g) \leq I^*(f) + I^*(g) = \int f + \int g\]A similar argument for the lower integral gives us the result.

Q: Proposition. \(|f|\) is integrable, and \(\left |\int f\right | \leq \int|f|\), given that f is bounded and integrable.
A: Consider first \(f_+ (x) = max(f(x) , 0)\), we want to show that this is integrable. We can check that \[\sup_{[x_{j-1}, x_j]} f_+ - \inf_{[x_{j-1}, x_j]} f_+ \leq \sup_{[x_{j-1}, x_j]}f - \inf_{[x_{j-1}, x_j]}f\]We know that given \(\epsilon > 0\), there exists \(\mathcal{D}\) such that \[S(f, \mathcal{D}) - s(f, \mathcal{D}) < \epsilon\].
Hence, \[S(f_+, \mathcal{D}) - s(f_+, \mathcal{D}) \leq S(f, \mathcal{D}) - s(f, \mathcal{D}) < \epsilon\]Therefore \(f_+\) is integrable. But \(|f| = 2f_+ - f\), hence \(|f|\) is integrable. Since \(-|f| \leq f \leq |f|\), the result follows due to monotonicity of integrals.

Q: Proposition. If f,g are bounded and integrable, fg is integrable.
A: Let f be integrable and positive. Then we can check that \[\sup_{[x_{j-1},x_j]} f^2 = (\sup_{[x_{j-1}, x_j]} f)^2 = M_j^2\]and similarly for the infimum. Then \[S(f^2, \mathcal{D}) - s(f^2, \mathcal{D}) = \sum_{j=1}^n(x_j - x_{j-1})(M_j ^2 - m_j ^2) = \sum_{j=1}(x_j - x_{j-1})(M_j - m_j)(M_j + m_j)\]Say f is bounded by some k. Then \[S(f^2, \mathcal{D}) - s(f^2, \mathcal{D}) \leq 2k(S(f, \mathcal{D}) - s(f, \mathcal{D}))\]So \(f^2\) is integrable. Now considering any f, \(|f|\) is a non negative integrable function. Since \(f^2 = |f^2|\), we deduce that \(f^2\) is integrable for any integrable f. Finally for fg, note that \[4fg = (f+g)^2 - (f-g)^2\]

Q: Theorem. Let \(F(x) = \int_a^x f(t) dt\), where f is integrable and bounded. Then F is continuous.
A: We know that \[F(x+h) - F(x) = \int_x^{x+h} f(t)dt\]
We want this quantity to vanish as \(h \to 0\). We find, given that f is bounded by k, \[|F(x+h) - F(x)| = \left | \int_x^{x+h} f(t) dt\right | \leq k|h|\]
So the result follows

Q: Theorem. If f is continuous at x, then F is differentiable, with \(F'(x) = f(x)\). When f is integrable and bounded with \(F(x) = \int_a^x f(t) dt\).
A: \[\left | \frac{F(x+h) - F(x)}{h} - f(x) \right | = \frac{1}{|h|} \left | \int_{x}^{x+h} f(t)dt - hf(x)\right | = \frac{1}{|h|} \left | \int_x^{x+h}[f(t) - f(x)]dt \right |\]Since f is continuous at x, given \(\epsilon > 0, \exists \delta > 0\) such that \(|t - x| < \delta \implies |f(t) - f(x)| < \epsilon\). If \(|h| < \delta\), then the integrand is bounded by \(\epsilon\). Hence, \[\left | \frac{F(x+h) -F(x)}{h} - f(x)\right| \leq \frac{1}{|h|} |h\epsilon| = \epsilon\]So the result follows.

Q: Corollary. If \(f = g'\) is a continuous function on \([a,b]\), then \[F(x) = \int_a^x f(t) \; \mathrm{d}t = g(x) - g(a)\]is a differentiable function on [a,b].
A: From the fundamental theorem of calculus, F - g has zero derivative in \([a,b]\), hence \(F-g\) is constant. Since \(F(a) = 0\), we get \(F(x) = g(x) - g(a)\).

Q: Theorem. Additivity of the integral. Let f be integrable on \([a,b]\). If \(a<c<b\), then f is integrable over \([a,c]\) and \([c,b]\), with \[\int_a^b f = \int_a^c f + \int_c^b f\]
A: If \(\mathcal{D}_1\) is a dissection of \([a,c]\) and \(\mathcal{D}_2\) is a dissection of \([c,b]\), then \(\mathcal{D} = \mathcal{D}_1 \cup \mathcal{D}_2\) is a dissection of \([a,b]\), and \[S(f, \mathcal{D}_1 \cup \mathcal{D}_2) = S(f|_{[a,c]}, \mathcal{D_1}) + S(f|_{[c,b]}, \mathcal{D}_2)\]Then also if \(\mathcal{D}\) is a dissection of \([a,b]\), then
\[ S(f, \mathcal{D}) \geq S(f, \mathcal{D} \cup \{c\}) = S(f|_{[a,c]}, \mathcal{D_1}) + S(f|_{[c,b]}, \mathcal{D}_2)\]By considering upper and lower integrals the result follows.

Q: Corollary. Integration by Parts. Suppose \(f'\) and \(g'\) exists are are continuous on \([a,b]\). Then \[\int_a^b f'g = f(b)g(b) - f(a)g(a) - \int_a^b fg'\]
A: \[\int_a^b f'g = \int_a^b [(fg)' - g'f] = \int_a^b (fg)' - \int_a^b g'f = f(b)g(b) - f(a)g(a) - \int_a^b g'f \text{ by FTC}\]

Q: Corollary. Integration by Substitution. Let \(g : [\alpha, \beta] \to [a,b]\) with \(g(\alpha) = a, g(\beta) = b\) and suppose \(g'\) exists and is continuous on \([\alpha, \beta]\). Let \(f: [a,b] \to \mathbb{R}\) be continuous. Then \[\int_a^b f(x) \; dx = \int_{\alpha}^{\beta} f(g(t)) g'(t) dt\]
A: Set \(F(x) = \int_a^x f(t) dt\). Let \(h(t) = F(g(t)) \; \; \; t \in [\alpha, \beta]\). Then \[\int_{\alpha}^{\beta} f(g(t))g'(t) dt = \int_{\alpha}^{\beta} F'(g(t)) g'(t) dt = \int_{\alpha}^{\beta} h'(t) dt = h(\beta) - h(\alpha) = F(b) - F(a) = \int_a^b f(x)dx\]Using FTC and the chain rule for the second equality.

Q: Theorem. Taylor's Theorem with Integral Remainder. Suppose f and its first n derivates exists and are continuous for \(x \in [0,h]\). Then \[f(h) = f(0) + hf'(0) + \dots + \frac{h^{n-1} f^{(n-1)}(0)}{(n-1)!} + R_n\]where \[R_n = \frac{h^n}{(n-1)!}\int_0^1 (1-t)^{n-1} f^{(n)}(th) dt\]
A: Substitute \(u = th\) into integral for \(R_n\).
\[R_n = \frac{1}{(n-1)!} \int_0^h (h-u)^{n-1} f^{(n)}(u)du\]Now integrating by parts over and over again gives you \[R_n = -\frac{h^{n-1}f^{(n-1)}(0)}{(n-1)!} - \dots - hf'(0) + \int_0^h f'(u)du\]And we get our result from rearranging and using FTC.

Q: Theorem. Mean Value Theorem for Integrals. Let \(f,g: [a,b] \to \mathbb{R}\) be continuous with \(g(x) \neq 0\) for all \(x \in (a,b)\). Then \[\exists c \in (a,b) s.t. \int_a^b f(x)g(x) \;dx = f(c) \int_a^b g(x) \;dx\]
A: We will use Cauchy's mean value theorem to get this result. Let \[F(x) = \int_a^x fg; \;\; G(x) = \int_a^x g\]Then there exists an intermediate point c such that \[(F(b) - F(a))G'(c) = F'(c)(G(b) - G(a))\]By the fundamental theorem of calculus, \[\left (\int_a^b fg \right )g(c) = f(c)g(c) \left (\int_a^b g \right)\]Now since \(g \neq 0\) everywhere, the result follows.

Q: Definition. We say the integral \(\int_a^{\infty} f\) exists, and its value is l, if [...]
A: \(f: [a, \infty) \to \mathbb{R}\) is integrable on every interval of the form \([a,R]\), and further, as \(R \to \infty\), we have \(\int_a^R f \to l\).
Back Extra: If the integral does not tend to a limit, we say the integral diverges.

Q: Lemma. Comparison test for integrals. If \(0 \leq f(x) \leq Kg(x), \forall x \geq a, K \text{ constant}\)then \[\int_a^{\infty} g(x) dx \text{ converges } \implies \int_a^{\infty} f(x) dx \text{ converges }\]and \[\int_a^{\infty} f \leq K\int_a^{\infty} g\]
A: Note that \(R \mapsto \int_a^R f\) is increasing so \[\int_a^R f \leq K \int_a^R g \leq K\int_a^{\infty} g\]so \(R \mapsto \int_a^R f\) is increasing, bounded above. Let \(l = \sup_{R \geq a} \int_a^R f\). Claim \(l = \lim_{R \to \infty} \int_a^R f \).
To see this fix \(\epsilon > 0\). Then \(\exists R_0\) s.t. \[\int_a^{R_0} f \geq l - \epsilon\].
So if \(R \geq R_0\), \[l - \epsilon \leq \int_a^{R_0} f \leq \int_a^R f \leq l\]
and therefore, \(\int_a^R f \to l\) as \(R \to \infty\)

Q: Theorem. Let f be a positive decreasing function for \(x \geq 1\). Then,
1)The integral \(\int_1^{\infty} f(x) dx\) and the series \(\sum_1^{\infty} f(x)\) both converge or diverge.
2) As \(n \to \infty\), \(\sum_{r=1}^n f(r) - \int_1^n f(x)dx\) tends to a limit l such that \(0 \leq l \leq f(1)\)
A: If \(n-1 \leq x \leq n\), then \[f(n-1) \geq f(x) \geq f(n)\]
Hence, \[f(n-1) \geq \int_{n-1}^n f(x) dx \geq f(n)\]
Adding up such integrals, we get \[\sum_1^{n-1} f(r) \geq \int_1^n f(x)dx \geq \sum_2^n f(r)\]
Then the first claim is obvious. For the second claim, let \[\phi(n) = \sum_1^n f(r) - \int_1^n f(x)dx\]
Then using the inequalities established above, \[\phi(n) - \phi(n-1) = f(n) - \int_{n-1}^n f(x) dx \leq 0\]
So \(\phi\) is a decreasing sequence. Further, \[0 \leq \phi(n) \leq f(1)\]
\(\phi\) is bounded, so it converges to some limit l.