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_ …

At first I thought this was the same as taking a Cartesian product, but he used the usual $\prod$ symbol for that further down the page, so I am inclined to believe there is some difference. Does anyone …

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 $0$ …

Nov 26, 2025 · Compute $$\lim_ {n \to \infty} \sqrt [n] {\prod_ {k=1}^n \left (1+ \frac {k} {n}\right)}$$ I've tried to solve it using limits of Riemann sums of the logarithm of the expression:

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

May 8, 2014 · 29 I've been looking at proofs of Euler's Sine Expansion, that is $$ \frac {\sin\left (x\right)} {x} = \prod_ {k = 1}^ {\infty} \left (1-\frac {x^ {2}} {k^ {2}\pi^ {2}}\right) $$ All the proofs seem to rely on …

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

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.

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, …