Homework 4

Q1

If \(f\) has no zeroes on \([a,+\infty)\), by IVT for all \(x > a\), we have \(f(x) > 0\). Choose \(y\) large enough (specifically \(y > \frac{-f(a)}{f'(a)} + a\)) such that by mean value theorem, there exists \(c\in (a,y)\) \[ \frac{f(y) - f(a)}{y - a} = f'(c) > \frac{-f(a)}{y-a} > f'(a) \] then by mean value theorem again there exists \(d\in (a,c)\) such that \[ f''(d) = \frac{f'(c) - f'(a)}{c - a} > 0 \] contradicting hypothesis.

If \(f\) has two zeroes \(x_1, x_2\in (a,+\infty)\), then by mean value theorem there exists \(c\in (x_1, x_2)\) with \(f'(c) = 0\), then by mean value theorem there exists \(d\in (a,c)\) with \[f''(d) = \frac{f'(c) - f'(a)}{c - a} > 0\] which contradicts hypothesis again.

Q2

By induction on \(n\) we show for all \(n\in\mathbb{N}\), for all \(f, x_0\) such that \(f\) is \(n\)-th differentiable at \(x_0\), we have \[ f^{(n)}(x_0) = \lim_{h\to 0} \frac1{h^n} \sum_{k=0}^n (-1)^{n-k}\binom{n}{k} f(x_0 + kh) \] if the limit exists.

When \(n = 0\) result trivially holds. Suppose result holds for \(n\), and \(f\) is \(n+1\)-th differentiable at \(x_0\), we want to show that \[f^{(n+1)}(x_0) = \lim_{h\to 0} \frac1{h^{n+1}} \sum_{k=0}^{n+1} (-1)^{n+1-k}\binom{n+1}{k} f(x_0 + kh) \] provided that the limit exists. First we rewrite some terms \[\begin{align*} &\quad\ \sum_{k=1}^{n} (-1)^{n+1-k}\binom{n+1}{k} f(x_0 + kh) \\ &= \sum_{k=1}^n (-1)^{n+1-k}\binom{n}{k} f(x_0 + kh) + \sum_{k=1}^n (-1)^{n-(k-1)} \binom{n}{k-1}f(x_0 + (k-1)h + h) \\ &= -\sum_{k=1}^n (-1)^{n-k}\binom{n}{k} f(x_0 + kh) + \sum_{k=0}^{n-1} (-1)^{n-k} \binom{n}{k}f(x_0 + kh + h) \end{align*}\] so that \[\begin{align*} \sum_{k=0}^{n+1} (-1)^{n+1-k}\binom{n+1}{k} f(x_0 + kh) &= \sum_{k=0}^{n-1} (-1)^{n-k} \binom{n}{k}f(x_0 + kh + h) + f(x_0 + kh + h) \\ &\qquad -(-1)^{n} f(x_0) -\sum_{k=1}^n (-1)^{n-k}\binom{n}{k} f(x_0 + kh) \\ &= \sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k}f(x_0 + kh + h) - \sum_{k=0}^n (-1)^{n-k}\binom{n}{k} f(x_0 + kh) \end{align*}\] armed with this we can examine the limit (which is assumed to exist) \[\begin{align*} &\quad\ \lim_{h\to 0} \frac1{h^{n+1}} \sum_{k=0}^{n+1} (-1)^{n+1-k}\binom{n+1}{k} f(x_0 + kh) \\ &= \lim_{h\to 0} \frac1h \frac1{h^n} \left[ \sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k}f(x_0 + kh + h) - \sum_{k=0}^n (-1)^{n-k}\binom{n}{k} f(x_0 + kh) \right] \\ &= \lim_{(h_1,h_2)\to(0,0)} \frac1{h_1} \left[ \frac1{h_2^n} \sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k}f(x_0 + h_1 + kh_2) - \frac1{h_2^n}\sum_{k=0}^n (-1)^{n-k}\binom{n}{k} f(x_0 + kh_2) \right] \\ &= \lim_{h_1\to 0} \frac1{h_1} \left[ \lim_{h_2\to 0} \frac1{h_2^n} \sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k}f(x_0 + h_1 + kh_2) - \lim_{h_2\to 0} \frac1{h_2^n}\sum_{k=0}^n (-1)^{n-k}\binom{n}{k} f(x_0 + kh_2) \right] \\ &= \lim_{h_1\to 0} \frac1{h_1} \left( f^{(n)}(x_0+h_1) - f^{(n)} (x_0) \right) = f^{(n+1)}(x_0) \end{align*}\]

Q3

Let \(h\) be small and without loss of generality positive such that \(f\) is continuous at \([x_0-h, x_0+h]\) and twice differentiable at \((x_0-h, x_0+h)\), define \(g(t) = f(x_0 + t) + f(x_0 - t)\) and it suffices to show that \[ \frac{g(h) - g(0)}{h^2} = \frac{g''(c)}2 \] for some \(c\in (-h,h)\). By Taylor’s theorem on \([0,h]\) there exists \(c\in(0,h)\) such that \[g(h) = g(0) + g'(0)h + \frac{g''(c)}2 h^2 \] and since \(g'(t) = f'(x_0+t) - f'(x_0 -t)\) is an odd function \(g'(0) = 0\), then the result follows.

Homework 5

Q1

By Taylor expansion on \(\ln(1+x)\) for any \(x\) there exists \(c\in(0,x)\) such that \[ \ln(1+x) = x - \frac{x^2}{2(1+c)^2}. \] Then for any \(n\), there exists \(c_1\in(0,\frac1{n+1}), c_2\in(0,\frac1n)\) such that \[\begin{align*} n^s\left( \left( 1 + \frac1{n+1}\right)^{n+1} - \left( 1 + \frac1{n}\right)^{n} \right) &= n^s\left( e^{(n+1)\ln(1+\frac1{n+1})} - e^{n\ln(1+\frac1{n})} \right) \\ &= n^s e \left( \exp\left(- \frac{1}{2(1+c_1)^2(n+1)}\right)- \exp\left(- \frac{1}{2(1+c_1)^2n}\right) \right) \\ \end{align*}\] then there exists \(d_1\in \left(- \frac{1}{2(1+c_1)^2(n+1)},0\right), d_2\in \left(- \frac{1}{2(1+c_1)^2n)},0\right)\) such that \[\begin{align*} n^s\left( \left( 1 + \frac1{n+1}\right)^{n+1} - \left( 1 + \frac1{n}\right)^{n} \right) &= n^s e \left( 1 - \frac{e^{d_1}}{2(1+c_1)^2(n+1)} - 1 + \frac{e^{d_2}}{2(1+c_2)^2n} \right) \\ &= n^s \frac{e}2 \left( \frac{e^{d_2}}{n(1+c_2)^2} - \frac{e^{d_1}}{(n+1)(1+c_1)^2} \right) \\ &= \frac{n^s}{n(n+1)} \frac{e}2 \left( \frac{e^{d_2}(n+1)}{(1+c_2)^2} - \frac{e^{d_1}n}{(1+c_1)^2} \right) \\ &= \frac{n^s}{n(n+1)} \frac{e}2 \left( n\left(\frac{e^{d_2}}{(1+c_2)^2} - \frac{e^{d_1}}{(1+c_1)^2}\right) + \frac{e^{d_2}}{(1+c_2)^2} \right) \end{align*}\] as \(n\to\infty\), each \(c_i, d_i\to 0\) and \(\frac{e^{d_i}}{(1+c_i)^2} \to 1\), so \(f(2) = \frac{e}2\) and \(f(s) = 0\) for all \(s < 2\), while the limit diverges for all \(s > 2\), and the domain is \((-\infty, 2]\) and range \(\left\{0,\frac{e}2\right\}\).

Q2

First we note that for all \(n\), as long as \(x \geq 0\), \(P_n(x) > 0\). Also by Taylor’s theorem on \(e^{-x}\) we have for any \(n\), there exists \(c\in (0,x)\) such that \[ e^{-x} = P_n(-x) + \frac{(-1)^{n+1} e^{-c}}{(n+1)!}x^{n+1}. \]

(i)

If \(n\) is even, for any \(x > 0\) then there exists \(c\in (0,x)\) such that \[ P_n(-x) = e^{-x} + \frac{e^{-c}}{(n+1)!}x^{n+1} >0 \] which completes the proof.

(ii)

If \(n\) is odd, then we can evaluate the derivative of \(P_n(x)\), which turns out to be \(P_{n-1}(x)\) (case \(n=1\) is obvious). By our earlier result in part (i) \(P_{n-1} > 0\) so \(P_n(x)\) cannot have two or more zeroes. \(P_n\) has to have a root though, since it is an odd degree polynomial.

(iii)

Fix \(n\) and we compare \(P_{2n+1}\) with \(P_{2n-1}\), specifically \[ P_{2n+1}(x) = P_{2n-1}(x) + \frac{x^{2n}}{(2n)!} + \frac{x^{2n+1}}{(2n+1)!} \] it suffices to show that \(P_{2n+1}(x_{n-1}) > 0\), then by earlier result that \(P_{2n+1}\) has strictly positive derivative we can conclude that \(x_n < x_{n-1}\). To show that \(P_{2n+1}(x_{n-1}) > 0\) we only have to check that \[ x_{n-1}^{2n}(2n+1) + x_{n-1}^{2n+1} > 0 \] which can be reduced to just showing that \(-(2n+1) < x_{n-1} < 0\), and note that \(x_{n-1}\) being negative was shown earlier.

Suppose for a contradiction \(x_{n-1} \leq -(2n+1)\), then there exists \(c \in (x_{n-1}, 0)\) such that \[ P_{2n-1}(x_{n-1}) = e^{x_{n-1}} - \frac{e^c}{(2n)!} x_{n-1}^{2n} \] but \(e^c \geq e^{x_{n-1}}\) so \[ P_{2n-1}(x_{n-1}) \leq e^{x_{n-1}}\left[ 1 - \frac{x_{n-1}^{2n}}{(2n)!} \right]\] and \(x_{n-1}^{2n} \geq (2n+1)^{2n} \geq (2n)^{2n} > (2n)!\) for all \(n>0\), which means \(P_{2n-1}(x_{n-1}) < 0\).

Q3

By taking Taylor’s expansions at \(\frac{a+b}2\), there exists \(c\in (\frac{a+b}2, b), d\in (a,\frac{a+b}2)\) such that \[\begin{align*} f(b) &= f\left(\frac{a+b}2\right) + \frac{f''(c)}2 \cdot\left(\frac{b-a}2\right)^2 \\ f(a) &= f\left(\frac{a+b}2\right) + \frac{f''(d)}2 \cdot\left(\frac{b-a}2\right)^2 \end{align*}\] then \[\begin{align*} f(b) - f(a) &= \frac{f''(d) - f''(c)}8 (b-a)^2 \\ \left\lvert f''(d) - f''(c)\right\rvert &= \frac{8}{(b-a)^2}\left\lvert f(b)-f(a)\right\rvert \end{align*}\] and since \(\max\left(\left\lvert f''(c)\right\rvert,\left\lvert f''(d)\right\rvert\right) \geq \dfrac12\left\lvert f''(d)-f''(c)\right\rvert\), and by choosing \(\xi\in\left\{c,d\right\}\) such that \(\left\lvert f''(\xi)\right\rvert = \max\left(\left\lvert f''(c)\right\rvert,\left\lvert f''(d)\right\rvert\right)\) we complete the proof.

Homework 6

Q1

Let \(P\) be any partition \(a = x_0 < x_1 < \dots < x_{p-1} < x_p = b\), then for all \(j\), by mean value theorem there exists \(d_j \in [x_{j-1}, x_j]\) such that \[\begin{align*} x_j^{m+1} - x_{j-1}^{m+1} &= (m+1) d_j^m (x_j - x_{j-1}) \\ d_j^m &= \frac1{m+1}(x_j^{m+1} - x_{j-1}^{m+1}) \end{align*}\] and when we let \(\mathbf{d}\) denote \((d_1,d_2,\dots,d_p)\) this sum will telescope as follows \[\begin{align*} R(x^m, x, P, \mathbf{d}) &= \sum_{j=1}^p d_j^m (x_j - x_{j-1}) \\ &= \frac1{m+1} \sum_{j=1}^p (x_j^{m+1} - x_{j-1}^{m+1}) \\ &= \frac1{m+1} \left(b^{m+1} - a^{m+1}\right) \end{align*}\] This reduces our problem to showing that for any \(\varepsilon>0\), there exists \(\delta>0\) such that when \(\left\lvert P\right\rvert \leq \delta\) and \(\mathbf{t} = (t_1, t_2, \dots, t_p)\) such that each \(t_j \in [x_{j-1}, x_j]\) \[ \left\lvert R(x^m, x, P, \mathbf{t}) - R(x^m, x, P, \mathbf{d})\right\rvert < \varepsilon. \]

First observe that \(x^m\) is continuous on \([a,b]\), and is hence uniformly continuous. Fix \(\varepsilon> 0\) and use uniform continuity of \(x^m\) to get a \(\delta_1\) such that whenever \(a \leq x < y \leq b\), \[ \left\lvert x-y\right\rvert < \delta_1 \implies \left\lvert x^m - y^m\right\rvert < \frac{\varepsilon}{2(b-a)} \] choose \(\delta = \delta_1\) and consider any partition \(\left\lvert P\right\rvert \leq \delta\) that is uniform enough, in the sense that \(p < \frac{2(b-a)}{\delta}\). Let \(\mathbf{t}\) be any set of tags, then \[\begin{align*} \left\lvert R(x^m,x,P,\mathbf{t}) - R(x^m,x,P,\mathbf{d})\right\rvert &= \left\lvert\sum_{j=1}^p\left(t_j^m - d_j^m\right)(x_j-x_{j-1})\right\rvert \\ &\leq p\delta\frac{\varepsilon}{2(b-a)} < \varepsilon \end{align*}\] then reusing the fact proven in Proposition 3.3 (regarding refinements) we get the general case for arbitrary partitions. In greater detail, when \(P^* \supset P\) is a refinement and \(\mathbf{t}^*\) is a set of tags for \(P^*\), we can check that there are tag assignments \(\mathbf{t}_1, \mathbf{t}_2\) satisfying \[ R(x^m,x,P,\mathbf{t}_1) \leq R(x^m,x,P^*,\mathbf{t}^*) \leq R(x^m,x,P,\mathbf{t}_2). \]

Q2

Suppose \(f\) is continuous at \(x\) but \(f(x) \ne 0\), then by continuity there exists \(\delta_0>0\) such that for all \(y \in (x-\delta_0, x+\delta_0)\), we have \(f(y) \in (\frac{f(x)}2, \frac{3f(x)}2)\), so in particular \(f(y) > \frac{f(x)}2\).

We now show that the integral \(\int_a^b f\ dx\) cannot be \(0\). Let \(\varepsilon= \frac{\delta_0 f(x)}2\) and for any \(\delta>0\), we can produce a mesh \(P\) with \(\left\lvert P\right\rvert < \delta\) such that \(\left\lvert R(f,x,P)\right\rvert > \varepsilon\). Specifically let \(P\) be \(a = x_0 < x_1 < \dots < x_i = x - \delta_0 < \dots < x_j = x + \delta_0 < \dots < x_p = b\) and \(P\) just have to be fine enough, then \[\begin{align*} \left\lvert R(f,x,P)\right\rvert &= \sum_{k=1}^p f(t_k)(x_k-x_{k-1}) \\ &\geq \sum_{k=i}^j f(t_k)(x_k-x_{k-1}) \\ &\geq \frac{f(x)}2 \sum_{k=i}^j (x_k-x_{k-1}) \\ &\geq \frac{f(x)}2 (x_j - x_i) > \varepsilon \end{align*}\]

Q3

Let \(c_1, c_2, \dots, c_r \in [a,b]\) enumerate all points which \(f\) and \(g\) differ and let \(d = \max\left\lvert g(c_i) - f(c_i)\right\rvert\).

Suppose \(f\) is Riemann integrable on \([a,b]\) and let \(\varepsilon> 0\) let \[ L = \int_a^b f(x)\ dx \] then by integrability of \(f\) there exists \(\delta_0>0\) such that for all partitions \(P\) with \(\left\lvert P\right\rvert < \delta_0\) \[ \left\lvert R(f,x,P) - L\right\rvert < \frac\varepsilon 2 \] choose \(\delta = \min(\delta_0, \frac{\varepsilon}{2rd})\) and let \(P\) be a partition such that \(\left\lvert P\right\rvert < \delta\), let \(\mathbf{t} = \left\{t_1, t_2, \dots, t_p\right\}\) be any tag set \[\begin{align*} R(g,x,P,\mathbf{t}) &= \sum_{k=1}^p g(t_k) (x_k - x_{k-1}) \\ &= \sum_{k=1}^p \left(f(t_k) + g(t_k) - f(t_k) \right) (x_k - x_{k-1}) \\ &= R(f,x,P,\mathbf{t}) + \sum_{k=1}^p \left(g(t_k) - f(t_k) \right) (x_k - x_{k-1}) \\ \left\lvert R(g,x,P,\mathbf{t}) - L\right\rvert &\leq \left\lvert R(f,x,P,\mathbf{t}) - L\right\rvert + \sum_{k=1}^p \left\lvert g(t_k) - f(t_k)\right\rvert (x_k - x_{k-1}) \\ \end{align*}\] note that in the worst case \(\{t_k\}_{k\in\left\{1,\dots,p\right\}}\) covers \(\{c_l\}_{l\in\left\{1,\dots,r\right\}}\), but we still have the bound \[ \sum_{k=1}^p \left\lvert g(t_k) - f(t_k)\right\rvert(x_k-x_{k-1}) \leq \delta \sum_{l=1}^r \left\lvert g(c_l) - f(c_l)\right\rvert \leq \delta r d \] which substituted back gives \[ \left\lvert R(g,x,P,\mathbf{t}) -L\right\rvert < \frac{\varepsilon}2 + \delta rd \leq \varepsilon. \]