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What you have done is correct (assuming the $x$s in your final answer should be $\lambda$s). But two points. I would say there is no need to avoid negatives in describing $F_7$, so your answer can be simplified to $-\lambda^3-\lambda^2+3\lambda-1$. (However your instructor may disagree - better check.) IMHO it would actually be easier if you used $F_7$ ...


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\We can also say $f_{n_1 \cdots n_r}(x) = \prod_{i=1}^r f_{n_i}(x) = \odot_{i=1}^r f_{n_i}(x)$. Also for boolean or we have $f_{n_1}(x) \oplus \dots \oplus f_{n_r}(x) = \lfloor \max\{g_{n_1}(x), \dots, g_{n_r}(x)\}\rfloor$


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I found an answer to this problem here: http://www.les-mathematiques.net/phorum/read.php?12,488850,489042 As French language may be a problem, here's a translation of this solution: Let $t\in\left[-\frac{1}{T},\frac{1}{T}\right]$. If $|\phi(t)|=1$, then there exists $\theta=\theta(t)$ such that $\phi(t)=e^{i\theta}$. So: $$E[1-e^{i(tX-\theta)}]=0$$ This ...


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First fix $t$. From $$(\mathbb{E}(\cos tX))^2+(\mathbb{E}(\sin tX))^2 = \mathbb{E}(\cos^2 tX+\sin^2 tX)$$ you get $$\textrm{var}{(\cos{tX})} + \textrm{var}{(\sin{tX})} = 0$$ Then, almost surely $$\cos{tX} = c(t) \qquad\text{and}\qquad \sin{tX} = s(t)$$ for some constants $c(t)$ and $s(t)$ satisfying $c(t)^2+s(t)^2=1$. Then you could say $X = ...


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Let $g=\chi_E$. We have $0\le g \le 1$ and $g$ is integrable. Thus, for any $\epsilon > 0$ there exists a continuous function $g_\epsilon$ with $0\le g_\epsilon \le 1$ and $\| g - g_\epsilon \|_1 < \epsilon$. Let $f_\epsilon = g_\epsilon * g_\epsilon$ (convolution). Notice that $f_\epsilon$ is continuous and for each $x\in\mathbb R$ we have ...



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