Paul Pichaureau
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 Feb 1 comment Counterexample - Increasing function If the derivative is continuous in $c$ then the conclusion is true. Nov 11 awarded Yearling Nov 3 answered Why is intersection of two independent set probability a multiplication process? Feb 5 answered How does $\exp(x+y) = \exp(x)\exp(y)$ imply $\exp(x) = [\exp(1)]^x$? Feb 5 comment limit of a series There is an error : $(3k-2)^3 = 3^3k^3 - 3*3^2*2*k^2+3*3*2^2*k - 2^3 = 27k^3-54*k+36*k-8$ Feb 5 comment limit of a series The numerator and the denominator can be expressed by a closed form. For example : $1^3 + 4^3 + ... + (3n-2)^3 = \sum_{k=1}^{n} (3k-2)^3 = 27 \sum k^3 -54 \sum k^2 + 36 \sum k + 8n$... Feb 5 awarded Student Feb 5 comment Preimage of Intersection of Two Sets = Intersection of Preimage of Each Set : $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$ You should start by taking $y \in f^{-1} (A \cap B)$. In this case, there is some $x \in A \cap B$ such that $y=f(x)$. Now you want to show that $y \in f^{-1}(A) \cap f^{-1}(B)$... Don't forget that $f^{-1}(A) = \{ x : f(x) \in A \}$. Feb 5 comment Why is this probability measure countably additive? I think we should read $\int_{\bigcup_{n=1}^N B_n} e^{-x} dx = ...$ Feb 5 answered Compute $\lim\limits_{a \to 0^+} \left(a \int_1^{\infty} e^{-ax}\cos \left(\frac{2\pi}{1+x^{2}} \right)\,\mathrm dx\right)$ Feb 5 awarded Scholar Feb 5 accepted Probability over a given $\sigma$-algebra Feb 5 comment Probability over a given $\sigma$-algebra No, it's ok ! I was looking for something much more complicated. Thank you. Feb 5 asked Probability over a given $\sigma$-algebra Nov 18 comment Why $\lim \limits_ {n\to \infty}\left (\frac{n+3}{n+4}\right)^n \neq 1$? I do agree : $1^\infty$ is tricky. You can rewrite it as $\exp(\infty \times \ln(1)) = \exp(\infty \times 0)$ to make the trick even more visible. Nov 18 answered How to show every subgroup of a cyclic group is cyclic? Nov 15 awarded Supporter Nov 15 awarded Critic Nov 15 answered Show $\sum_{k=1}^n (p_k + \frac{1}{p_k})^2\geq n^3 + 2n + \frac{1}{n}$ ; $p_k\geq 0 \forall k$ and $\sum_kp_k=1$ Nov 14 answered Is this a bad way of approaching math problems?