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## New answers tagged error-propagation

1

By Newton's binomial theorem: \begin{align} &1+\theta_n'=\prod_{j=1}^{n} (1 + \epsilon_j)^{-1} \leq (1 -|u|)^{-n} = 1+ \sum_{k=1}^\infty {-n \choose k} (-1)^k |u|^k \\ = & 1+ \sum_{k=1}^\infty {n+k-1 \choose k} |u|^k \\ \leq & 1+ \sum_{k=1}^\infty n^k |u|^k = 1+\frac{n|u|}{1-n|u|}, \end{align} because \begin{align} &{n+k-1 \choose k} \leq ...

0

That might help: From $$\prod_{j=1}^{n} (1 + \epsilon_j)^{-1} = 1 + \theta_n',$$ you get $$\prod_{j=1}^{n} (1 + \epsilon_j) = \frac{1}{1 + \theta_n'}$$ so from the previous part $$\frac{1}{1 + \theta_n'}-1\leq \frac{n|u|}{1-n|u|}$$

1

(b) We have $$\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$$ from which it follows that $$\frac{1-\cos(x)}{x^2} = \frac{1}{2} - \frac{x^2}{24} + O(x^4).$$ Therefore $$f(x) - \frac{1}{2} = O(x^2).$$ (c) When $x \rightarrow 0$, $x \not = 0$ the expression \$1 - ...

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