Fall 2017

The first negation is: \[ \text{There is an }x\in\mathbb R\text{ such that }|x-y|<2017 \text{ for every }y\in\mathbb R. \]

Using the pointwise definition of continuity on \(\mathbb R\), the second negation is: \[ \exists x\in\mathbb R\ \exists\varepsilon>0\ \forall\delta>0\ \exists y\in\mathbb R: |x-y|<\delta\ \text{ and }\ |f(x)-f(y)|\geq\varepsilon. \]

Let \(\varphi\colon\mathbb R^3\to\mathbb R\) be \(\varphi(x,y,z)=3x+4y+5z\), so that \(V=\ker\varphi\).

<1>1. \(V\) is a linear subspace of \(\mathbb R^3\) of dimension \(2\), with basis \(v_1=(4,-3,0)\), \(v_2=(5,0,-3)\).

<1>2. Some linear map \(S\colon\mathbb R^3\to\mathbb R^3\) has kernel \(V\).

<1>3. Some linear map \(T\colon\mathbb R^3\to\mathbb R^3\) has image \(V\).

<1>4. No linear map \(U\colon\mathbb R^3\to\mathbb R^3\) has \(\ker U=\operatorname{im}U=V\).

For \(n=0\), \[ \int_0^\infty e^{-x}\,dx=1=0!. \]

For the inductive step, integration by parts gives \[ \int_0^\infty x^ne^{-x} = -x^ne^{-x}\bigg\rvert_0^\infty - \int_0^\infty -nx^{n-1}e^{-x} = -x^ne^{-x}\bigg\rvert_0^\infty + n \int_0^\infty x^{n-1}e^{-x} \\ \equalsbecause{I.H.} -x^ne^{-x}\bigg\rvert_0^\infty + n (n-1)! \\ \equalsbecause{L.H.$\times n$}0 + n(n-1)! = n!. \qed \]

1. Since \(A\) is \(2\times 2\) and has 2 eigenvalues, noting that \(\deg \chi_A(x) = 2\), we have \(\chi_A(x) = (x-1)(x+1) = x^2 -1\). The minimal polynomial of \(A\) divides \(\chi_A(x)\), so we have \(\chi_A(A) = 0\) and thus \(A^2 - I_2 = 0 \implies A^2 = I_2\).

2. This will happen when \(x^2-1 = (x+1)(x-1)\) is not the minimal polynomial, and we can force the minimal polynomial to be degree 3 by inserting a nontrivial Jordan block to a diagonal matrix containing just the eigenvalues \(\pm 1\). An example that works: \[ \left(\begin{array}{rr|r} 1 & 1 & 0 \\ 0 & 1 & 0 \\ \hline 0 & 0 & -1 \end{array}\right),~ A^2 = \left(\begin{array}{rrr} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right), \quad \spec(A) = [-1, 1, 1] \]

3. Every symmetric matrix with \(A\) real spectrum admits a real eigendecomposition \(\Lambda D \Lambda^T\), where \(D\) is diagonal with entries the eigenvalues of \(A\) and \(\Lambda\) are orthogonal (which are also invertible). Here, we only need the fact that \(A\) is diagonalizable by invertible matrices. In our case we have \([D^2]_{ii} = (\pm 1)^2 = 1\) so \(D^2 = I_n\). Thus we have \[A^2 = (\Lambda D \inverseof{\Lambda})^2 = \Lambda D^2 \inverseof{\Lambda} = \Lambda I_n \inverseof{\Lambda} = I_n. \qed\]

Uniform convergence to zero gives \[ \|f_n\|_{\infty}=\sup_{x\in[0,1]}|f_n(x)|\longrightarrow0. \] Therefore \[ \left|\int_0^1f_n(x)\,dx\right| \leq\int_0^1|f_n(x)|\,dx \leq\|f_n\|_{\infty}\longrightarrow0. \]

Let \(\varepsilon>0\) and choose \[ \delta=\min\left\{\frac12,\sqrt{\frac\varepsilon2}\right\}. \] If \(|x-1|<\delta\), then \(|x|>1/2\). Hence \[ \left|\frac{x^2+1}{x}-2\right| =\frac{|x-1|^2}{|x|} <2|x-1|^2 <2\delta^2 \leq\varepsilon. \]

1. Definitions:

• \(f: \RR^2 \to \RR, \quad (x,y) \mapsto z = f(x,y)\)

• Level curves are given by \(f(x, y) = c\);

• \(\nabla f: \RR^2 \to \RR^2, \quad \nabla f(\vector p) = \thevector{f_x(\vector p), f_y(\vector p)}\)

• \(\gamma(t): \RR \to \RR^2, \quad \gamma(t) = \thevector{x(t), y(t)}, \quad \gamma_t(t) = \thevector{x_t(t), y_t(t)}\)

• \(g(t) = f \circ \gamma:\RR \to \RR, \quad (f \circ \gamma)(t) = f(x(t), y(t)), \quad g_t(t) = f_t(\gamma(t))\cdot \gamma_t(t)\)

• A vector \(v\) is perpendicular to a surface at a point \(p\) iff \(v\) is perpendicular to the tangent vector of every curve passing through \(p\).

• Proof of actual statement

  • Let \(\vector p = \thevector{x_0, y_0}\) be a point on the level surface, so \(f(\vector p) = c\) for some constant.

  • Let \(\gamma(t) = \thevector{x(t), y(t)}\) be a curve on the level surface, so \(\gamma(t_0) = \vector p\) for some \(t_0\). Let \(\gamma'(t)\) be its tangent vector.

  • Let \(g(t) = f(x(t), y(t)) = (f \circ \gamma)(t)\), and note that for some \(\vector x\) on the level surface, and so we have \(g(t_0) = f(\vector x) = c\) and thus \(\dd{g}{t}(t) = 0\).

  • By the chain rule, compute \[\begin{align*} \dd{g}{t}(t) &= \left(\dd{f}{x}\dd{x}{t} + \dd{f}{y}\dd{y}{t}\right)(t) \\ &= \inner{ \thevector{\dd{f}{x}(x(t), y(t)), \dd{f}{y}(x(t), y(t))} } { \thevector{\dd{x}{t}(t), \dd{y}{t}(t)} } \\ &= \inner{\nabla f(\gamma(t))}{\dd{\gamma}{t} (t)}. \end{align*}\]

  From above, we know that this is zero. Now note that we have \[ \dd{g}{t}(t_0) = \inner{\nabla f(\gamma(t_0))}{\dd{\gamma}{t}(t_0)} = \inner{\nabla f(\vector p)}{\gamma'(t_0)} \] but by the previous statement, \(\dd{g}{t}(t_0) = 0\), which exacty says that the gradient of \(f\) is orthogonal to \(\gamma\) at \(\vector p\). But \(\vector p\) was an arbitrary point on the level surface, and \(\gamma\) was an arbitrary curve through it. So \(\nabla f\) is orthogonal to every level curve through \(\vector p\), and this orthogonal to the tangent plane at \(\vector p\), and thus normal to the surface at \(\vector p\). Since \(\vector p\) was an arbitrary point on the level curve, this holds everywhere on the level curve, and for arbitrary level curves. Thus \(\nabla f\) is orthogonal to every level curve. \(\qed\)

Goal: Prove that for a continuously differentiable function \(f: \mathbb{R}^2 \to \mathbb{R}\), the gradient vector \(\nabla f(\mathbf{p})\) at any point \(\mathbf{p}\) on a level curve \(C = \{(x,y) \in \mathbb{R}^2 \mid f(x,y) = c\}\) is orthogonal to the tangent vector of every smooth curve in \(C\) passing through \(\mathbf{p}\).

<1>1. Let \(\mathbf{p} = (x_0, y_0) \in C\) such that \(f(\mathbf{p}) = c\). Let \(\gamma: (-\varepsilon, \varepsilon) \to \mathbb{R}^2\) be a differentiable parametrized curve such that \(\gamma(0) = \mathbf{p}\) and \(\gamma(t) \in C\) for all \(t \in (-\varepsilon, \varepsilon)\).

<1>2. The composite function \(g(t) = (f \circ \gamma)(t) = f(\gamma(t))\) is constant on \((-\varepsilon, \varepsilon)\) with value \(c\).

<1>3. \(g'(0) = 0\).

<1>4. By the multivariable Chain Rule, \(g'(0) = \nabla f(\gamma(0)) \cdot \gamma'(0) = \nabla f(\mathbf{p}) \cdot \gamma'(0)\).

<1>5. \(\nabla f(\mathbf{p}) \cdot \gamma'(0) = 0\).

<1>6. Conclusion: \(\nabla f(\mathbf{p})\) is orthogonal to the tangent vector \(\gamma'(0)\) of any curve lying on the level set through \(\mathbf{p}\).

If \(f(x_1)=f(x_2)\), then applying \(g\) gives \[ x_1=g(f(x_1))=g(f(x_2))=x_2. \] Thus \(f\) is injective.

For each \(x\in X\), the element \(f(x)\in Y\) satisfies \(g(f(x))=x\). Thus \(g\) is surjective.

Let \(X=\{1\}\), \(Y=\{1,2\}\), \(f(1)=1\), and \(g(1)=g(2)=1\). Then \(g\circ f=\operatorname{id}_X\), while \(f\) is not surjective and \(g\) is not injective.