DevOps engineers are those who are good at debugging, troubleshooting, analyzing prod issues and providing solutions. Who have good hands on technologies like unix shell scripting, perl, SQL etc.

Dec 28, 2013 · 21 The symbol $\Pi$ is the pi-product. It is like the summation symbol $\sum$ but rather than addition its operation is multiplication. For example, $$ \prod_ …

Sep 10, 2024 · By L'Hospital: The derivative of the denominator is (by pulling one cosine at a time from the product) $$\sum_ {i=1}^n\frac {i\sin (ix)} {\cos (ix)}\prod_ {i=1}^n\cos (ix).$$ This still tends to $0$ …

Aug 20, 2025 · Please edit to include your efforts. If, as you suggest, you know a proof of the statement involving $1$, why not include it? Presumably it sheds some light on how to proceed more generally.

Nov 1, 2024 · There are simple reasons for the others - it is that $1$ and $4$ are squares of integers.

Sep 5, 2025 · How can I prove that if $X_\alpha$ are locally compact and all but a finite number of factors are compact then $\prod_\alpha X_\alpha$ is locally compact? What needs to be proved is

Sep 13, 2016 · Compute: $$\prod_ {n=1}^ {\infty}\left (1+\frac {1} {2^n}\right)$$ I and my friend came across this product. Is the product till infinity equal to $1$? If no, what is the answer?

Jan 17, 2025 · $$ \prod_ {p \leq x} p \geq\prod_ {\sqrt {x} < p \leq x} p \geq \left (\sqrt {x}\right)^ {\pi (x) - \pi (\sqrt {x})} \ge e^ {\frac1 {2} (\frac {1} {2} x - 4 \sqrt {x})}, $$ But I have no idea what to do with this, …

Aug 4, 2024 · $$\displaystyle\prod\limits_ {i=1}^ {n} \left (1+a_i\right) \,\, = \,\, \displaystyle\sum_ {S \,\subseteq \, \ {1,\, 2,\, 3,\, \dots\,,\, n\}} \,\,\,\left (\,\prod ...

Thus, if we apply Kirchhoff's theorem, we get $$\prod_ {m=1}^ {n-1} 4\sin^2 (\frac {m\pi} {n}) = n^2.$$ By taking square root and dividing both sides by $2^ {n-1}$, we get the desired formula.