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Parts:
- Negating: \begin{align*}\begin{align*} \neg (f \rightrightarrows f) &\iff \neg (\forall\varepsilon, \exists N(\varepsilon) {~\mathrel{\Big\vert}~}\forall x\in X, \quad & n > N \implies {\left\lvert {f_n(x) - f(x)} \right\rvert} < \varepsilon)\\ &\iff \neg (\forall\varepsilon, \exists N(\varepsilon) {~\mathrel{\Big\vert}~}\forall x\in X, \quad &{\left\lvert {f_n(x) - f(x)} \right\rvert} < \varepsilon \text{ or } n \leq N)\\ &\iff \exists \varepsilon {~\mathrel{\Big\vert}~}\forall N, \exists x\in X {~\mathrel{\Big\vert}~}\quad &n \geq N ~\&~ {\left\lvert {f_n(x) - f(x)} \right\rvert} \geq \varepsilon \end{align*}\end{align*} In words: there is some \(\varepsilon\) such that no matter what \(N\) you choose, there is at least one point \(x\) where \(f_n(x)\) is not \(\varepsilon{\hbox{-}}\)close to \(f(x)\) for every \(n \geq N\).
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Can use any function that converges pointwise but not uniformly. Example on \([0, 1]\):
\begin{align*}
f_n(x) = x^n \implies f_n(x) \to \begin{cases}0 & x\in [0, 1) \\ 1 & x=1 \end{cases} \coloneqq f(x).
\end{align*}
Proof that \(f_n\) converges to \(f\) pointwise:
- Note that \(f_n(1) = 1, f_n(0) = 0\), so no issues there.
- For \(x\in (0, 1)\), need to show \begin{align*} f_n \to f \iff \forall \varepsilon, \forall x\in X, \exists N(\varepsilon, x) {~\mathrel{\Big\vert}~}n > N \implies {\left\lvert {f_n(x) - f(x)} \right\rvert} < \varepsilon \end{align*} In this case, \(f(x) = 0\), so just need to show \({\left\lvert {f_n(x)} \right\rvert} < \varepsilon\). To get \(x^n < \varepsilon\), just take \(n > \frac{\ln\varepsilon}{\ln x}\). Proof that \(f_n\) does not converge uniformly:
- Let \(\varepsilon = \frac {1}{10}\) and \(N\) be arbitrary, so \(f_n(x) = x^N\). Then consider \(x = \frac{9}{10}^{\frac 1 N}\), so \(f_n(x) = \frac{9}{10}\), and we have \({\left\lvert {f_n(x)} \right\rvert} = \frac 9 {10} \geq \frac 1 {10} = \varepsilon\).
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Apply the spectral theorem: \(D\) will have the eigenvalues on the diagonal, and \(A\) will be the column of eigenvectors. Thus \begin{align*} A = \left[ \begin{array} { l l } { 2 } & { 1 } \\ { 1 } & { 1 } \end{array} \right], \quad D = \left[\begin{array}{rr} 1 & 0 \\ 0 & -2 \end{array}\right]. \end{align*}
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We want some power series centered at 4 with convergence radius 1, so something of the form \(\sum_n f(n)(x-4)^n\) with \(L = \lim \frac{a_{n+1}}{a_n} = 1\).
Then, by plugging in \(5\) and \(3\), we find that we want \(\sum (-1)^n f(n)\) to converge but \(\sum f(n)\) to diverge. The canonical example of this is the harmonic series with \(f(n) = \frac 1 n\), so we can just take \begin{align*}f(x) = \sum_n \frac{(x-4)^n}{n}. \hfill\blacksquare\end{align*}
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We need to show \begin{align*} \forall \varepsilon \exists \delta(\varepsilon) {~\mathrel{\Big\vert}~}\quad {\left\lvert {x-2} \right\rvert} < \delta \implies {\left\lvert {\frac 1 {3+x} - \frac 1 5} \right\rvert} < \varepsilon \end{align*}
Choose \(\delta = 20 \varepsilon\). Then note that we can take \(1 < x < 3\), and so \({\left\lvert {5(3+x)} \right\rvert} < 20\). Then \begin{align*} {\left\lvert {\frac 1 {3+x} - \frac 1 5} \right\rvert} = {\left\lvert {\frac{x-2}{5(3+x)}} \right\rvert} < \frac 1 {20} {\left\lvert {x-2} \right\rvert} < \frac 1 {20} (20 \varepsilon) < \varepsilon. \hfill\blacksquare \end{align*}
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Parts
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We have \begin{align*} \int_C F_1 dx + F_2 dy + \cdots = \int _ { C } \vec { F } \cdot d \vec { r } = \int _ { a } ^ { b } \vec { F } ( \vec { r } ( t ) ) \cdot \vec { r } ^ { \prime } ( t ) d t \end{align*} and we can paramaterize a line segment \(r(t) = t[2, 1] + (1-t)[3, 5] = [3-t, 5-4t]\), so just plug everything in to get \begin{align*} \int_0^1 22t-31~dt = -20 \end{align*}
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\(F = [P, Q] = [2x+y, y]\) is conservative iff there is a potential function \(\phi\) satisfying \(\nabla \phi = F\) iff \(P_y = Q_x\). Here we see that \((2x+y)_y = 1\) but \((y)_x = 0\), so the integral is path-dependent.
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Use the fact that \(\sum_{i=1}^n i = \frac 1 2 n(n+1)\), so \begin{align*}\begin{align*} \sum_{i=1}^{n+1} i^3 &= \sum_{i=1}^{n} i^3 + (n+1)^3 \\ &= \left( \sum_{i=1}^{n} i \right)^2 + (n+1)^3 \\ &= \left( \frac 1 2 n(n+1) \right)^2 + (n+1)^3 \\ &= \frac 1 4 n^2(n+1)^2 + (n+1)(n+1)^2 \\ &= \frac 1 4 (n+1)^2 (n^2 + 4(n+1)) \\ &= \frac 1 4 (n+1)^2(n+2)^2 \\ &= \left( \frac 1 2 (n+1)(n+2)\right)^2 \\ &= \left( \sum_{i=1}^{n+1} i \right)^2. \hfill\blacksquare \end{align*}\end{align*}
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Claim: this holds iff \(f\) is injective iff \(f\) has a left inverse.
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\(\implies\): Suppose this holds, and let \(x_1,x_2 \in X\). Then suppose \(f(a) = f(b)\). We have \(f^{-1}(f(\left\{{a}\right\})) = \left\{{a}\right\}\) and \(f^{-1}(f(\left\{{b}\right\})) = \left\{{b}\right\}\) by assumption, so combining these we have \begin{align*} a = \left\{{a}\right\} = f^{-1}f(\left\{{a}\right\}) = f^{-1}f(a) = f^{-1}f(b) f^{-1}f(\left\{{b}\right\})= \left\{{b}\right\} = b, \end{align*} so \(f\) is injective.
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\(\neg \implies\) Suppose that this does not hold, then there is some \(S \subset X\) such that \(f^{-1}f(S) \neq S\). Then \(S - f^{-1}f(S) \neq \emptyset\), so pick an element \(x\) from it.
Then \(f(x) \coloneqq y \in f(S)\) since \(x\in S\), but \(f^{-1}(y) \in f^{-1}f(S)\) and so there is some \(x'\) such that \(f^{-1}(y) = x'\), where \(x' \in f^{-1}(f(S))\). Since \(x \in S-f^{-1}f(S)\) but \(x' \in f^{-1}f (S)\), we have \(x\neq x'\), and by definition we have \(f(x) = y = f(x')\), we must conclude that \(f\) is not injective.
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The gradient \(\nabla f = [y, x]\) will be orthogonal to level curves:
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\(A = \left[ \begin{array} { r r r } { 1 } & { - 2 } & { 1 } \\ { 0 } & { 5 } & { - 3 } \\ { 0 } & { 0 } & { 0 } \end{array} \right]\); a straightforward computation of \(\mathrm{nullspace}(A)\) shows that \(\ker A = \mathrm{span}({\left[ {1,3,5} \right]})\).