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Hints: (1) $f$ extends to a continuous function $[0,\infty)\to\mathbb{R}$. (2) $\lim_{x\to\infty}f(x)=0$.

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I don't see an elegant way to do this so we will apply brute force starting from this formula $$w_k=\frac{a_n}{a_{n-1}}\frac{\int_a^bw(x)p_{n-1 }(x)^2dx}{p_n^{\prime}(x_k)p_{n-1}(x_k)}$$ In our case we write it as $$v_k=\frac{b_m}{b_{m-1}}\frac{\int_0^1\sqrt y\,q_{m-1}(y)^2dy}{q_m^{\prime}(y_k)q_{m-1}(y_k)}$$ Now $b_m$ is the leading coefficient of $q_m(y)=\... 4 You can also directly apply Dirichlet's test (the integral of the numerator is zero when taken over a period, therefore uniformly bounded on all finite intervals). We have $$\cos^{2 k} u = 2^{-2 k} \binom {2 k} k + 2^{1 - 2 k} \sum_{1 \leq j \leq k} \binom {2 k} {k - j} \cos 2 j u, \\ \sum_{k \geq 0} \frac {(i \alpha \cos u)^{2 k}} {(2 k)!} = J_0(\alpha) + ... 1 Disclaimer: I don't have a full answer. Just some ideas. You have already managed to get \alpha_n, \gamma_n and \beta_{n-1}. So it is safe to count them as known parameters. Now let \lambda=0, then$$\prod_{i=1}^{2n}\lambda_i+\gamma_n\prod_{i=1}^{2n-2}\mu_i=\delta_{n-1}\prod_{i=2}^{n-1}\delta_i=\delta_{n-1}^2\prod_{i=2}^{n-2}\delta_i$$The left side ... 1 Let me propose one more method of proof. "Beginner's guide to mapping simplexes affinely", section "Lagrange interpolation", describes a determinant form of Lagrange polynomial that interpolates (a_0;b_0), \dots, (a_n;b_n)$$ P(x) = (-1) \frac{ \det \begin{pmatrix} 0 & b_0 & b_1 & \cdots & b_n \\ ... 1 Consider$F$as a function of$t$and$x$: then we have $$\begin{split} x''(t)&=\lim_{h\to 0}\frac{x'(t+h)-x'(t)}{h}=\lim_{h\to 0} \frac{F(t+h,x(t+h))-F(t,x(t))}{h}\\ &=\lim_{h\to 0} \frac{F(t+h,x(t+h))- F(t,x(t+h))+F(t,x(t+h))-F(t,x(t))}{h}\\ &=\lim_{h\to 0} \frac{F(t+h,x(t+h))- F(t,x(t+h))}{h}+\lim_{h\to 0}\frac{F(t,x(t+h))-F(t,x(t))}{h}\\ &... 1 There are quite a few ways to derive this formula, but thinking of the determinant as a volume has a straightforward geometric appeal. Place the triangle \triangle{ABC} on the plane x=1 in \mathbb R^3 and consider the pyramid with this triangle as its base and the origin for its apex: This pyramid’s altitude is 1, so its volume is equal to the ... 1 The last significant bit has a weight 2^{-23}, which is 0.000000119209\cdots or 1.19209\cdots\times10^{-7}. This is maximum relative error on any number. You can obtain this result by solving$$10^{-d}=2^{-23},$$or$$d=\log_{10}2^{23}=6.924\cdots$$and rounding. 1 Floating point numbers are encoded as \pm 2^{e}\cdot(1.m_1...m_{23})_2. Representable numbers between 1 and 2 have a distance of 2^{-23}=(8⋅1024^2)^{-1} in their sequence. The distance from any number in [1,2] to a representable number is half that distance, (16⋅1024^2)^{-1}. This justifies to say that the accuracy is about 7 correct decimal ... 1 All values and derivatives are Taylor expanded at t so they can be compared. You could just as well start with a polynomial or power series, y(t+s)=\sum a_ks^k, y'(t+s)=\sum ka_ks^{k-1} and then start comparing coefficients starting at the lowest degree. Note that the resulting method is extremely unstable, as local errors, truncation or floating ... 1 For the Gauss - Seidel Method to work, the matrix must be in diagonally dominant form and your current matrix is not, so we expect it to fail. Sometimes, we cannot easily see a way to put the matrix in such a form without playing around with it. We have$$ \left[ \begin{array}{cccc|c} 2 & 3 & -4 & 1 &3\\ 1 & -2 & -5 & 1 &... 1$\lim_{x\to 0^+}f(x)=0$so we can extend$f$to$[0,\infty)$by letting$f(0)=0$. Then show that$f$is uniformly continuous on$[0,\infty)$and hence uniformly continuous on the subset$(0,\infty).$(i).$f$is continuous on$[0,\infty)$so f is uniformly continuous on the closed bounded interval$[0,1].$(ii). If$x\ge 1$then$|f'(x)|=|\sin (x^{-2})+2x^...

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Notice that $$-x<x\sin(\frac{1}{x^2}<x$$ so by sandwich theorem $$\lim_{x\rightarrow 0} x\sin \left(\frac{1}{x^2}\right)=0.$$

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