zeraoulia rafik
• Member for 6 years, 9 months
• Last seen more than 5 years ago
• Batna, Algeria

Let $a_n = 1 + \dfrac {1}{\sqrt{2}} + \cdots + \dfrac {1}{\sqrt{n}}$ and $b_n = \sqrt{n}$ for all $n\geq 1$. We have $$\lim_{n\to \infty} \dfrac {a_{n+1} - a_n} {b_{n+1} - b_n} = \lim_{n\to \infty} \... View answer Accepted answer 13 votes for any 0 \le x \le 1, we have$$1 - \prod_{i=1}^n (1 - x^{a_i}) \ge 0$$and hence$$f(x) = \sum_{i=1}^n x^{a_i} - \sum_{i < j} x^{a_i + a_j} + \cdots + (-1)^{n-1} x^{a_1 + \cdots + a_n} \ge 0.$$... View answer Accepted answer 8 votes In the First : \sinh x = \frac{e^x - e^{-x} }{2} \Longrightarrow \sinh^{-1}x = \ln(x+\sqrt{x^2+1}). to get this just solve the equation y=\sinh x to get the above inverse function (notice that ... View answer Accepted answer 5 votes According to an answer by rae306 on Art of Problem Solving: Using integration by parts:$$\int_0^\infty \ln\left(1+\frac{a^2}{x^2}\right)\,dx\,\,\begin{bmatrix}u=\ln\left(1+\frac{a^2}{x^2}\right)&...

\int\frac {\mathrm{dx}}{1+\sin x -\cos x}= \int\frac {\mathrm{dx}}{2\sin^2 \frac{x}{2}+2\sin \frac{x}{2} \cos \frac{x}{2}}\\= \frac{1}{2}\int\frac {\mathrm{dx}}{\sin^2 \frac{x}{2}+\sin \frac{x}{2} \... View answer Accepted answer 5 votes Hint :x+3\sqrt[3]{xy^2}=x+\sqrt[3]{xy^2}+\sqrt[3]{xy^2}+\sqrt[3]{xy^2}\ge 4\sqrt[4]{x\cdot\sqrt[3]{xy^2}\cdot\sqrt[3]{xy^2}\cdot\sqrt[3]{xy^2}}=4\sqrt{xy} View answer Accepted answer 4 votes Let p^4-20p^2+19 = (p^2-1)(p^2-19), and 180 = 6^2\cdot 5. But p\equiv \pm 1\pmod{6}, so p^2\equiv 1\pmod{6}, thus p^2-1 \equiv 1-1 = 0 \pmod{6} and p^2-19 \equiv 1-19 = -18 \equiv 0 \pmod{6}... View answer Accepted answer 3 votes \begin{gathered} \int {\frac{{\sqrt { - {x^2} - x + 2} }}{{{x^2}}}} \,dx \hfill \\ = - \frac{{\sqrt { - {x^2} - x + 2} }}{x} - \int {\frac{{\left( { - 2x - 1} \right)}}{{2\sqrt { - {x^2} - x + ... View answer Accepted answer 3 votes First of all, it is obvious that the minimum is when 1>t>0, (since e^{-x^2}<1 on the interval of integration). \begin{align*}f(t)=\int\limits_0^1 x |e^{-x^2} - t|\ \mathrm{d}x &= \int\... View answer 3 votes Try to take mod 3 \rightarrow y^3 \equiv 1 \pmod {3} \rightarrow y \equiv 1 \pmod {3} \implies y^3 \equiv 1 \pmod {9} Therefore 3x^2 + 3x + 7 \equiv 1 \pmod {9} \rightarrow 3x^2 + 3x \equiv 3 \... View answer Accepted answer 3 votes Use the substitution u^2 = x + b. Let I = \int\frac{1}{(x+ a)\sqrt{x + b}}\,dx  Let \begin{align*} &u^2 = x + b\implies x + a = u^2 + a - b\\ &2u\,du = dx \end{align*} Then \begin{align*... View answer Accepted answer 2 votes Maybe this answer from AOPS is helpful to you: I =\int_0 ^\frac{\pi}{2}\frac{\sec^{2}x}{(\sec x+\tan x)^{n}}\ dx = \int_0 ^\frac{\pi}{2}\frac{1}{(\sec x+\tan x)^{n}} \cdot \frac{1}{\cos^{2}x}\ dx ... View answer Accepted answer 2 votes We begin with the finite geometric series f_n(z) = \frac{z^{n+1} - 1}{z-1} = \sum_{k=0}^n z^k.  Integration term-by-term with respect to z then gives  \int f_n(z) \, dz = \sum_{k=0}^n \frac{z^{k+... View answer 2 votes Hinit: What you meant is called halley method such that the Algorithm can be written as follow : x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}[1-\frac{f(x_{n})}{f'(x_{n})}\frac{f''(x_{n})}{2f'(x_{n})}...

look the integral that you meant is :$$\int_{e}^{e^2}(\frac{1}{\log{x}}-(\frac{1}{\log x})^2).\mathrm{d}x$$ Hint : your integral doesn't converge ,you have a singularity at x=1 and should be begin ...

Hint :$\sum_{n=1}^\infty 2^{-n}$ is a power of geometric series is already convergent and $\{ x_n\} \mapsto x_k$ is continuous for every $k$ ( Uniform convergence). pleas look ...

Suppose that : $\lim_{x\to\infty}f '(x)^2 + f(x)^3 = 0.$ Show $\limsup_{x\to\infty} f(x) = 0:$ Showing $\,\limsup_{x\to\infty} f(x) \le 0$ is easy. If $\limsup_{x\to\infty} f(x) < 0,$ then $f(x) &... View answer 1 votes Hint :look Lambert W function. solutions represented as: $$x=\frac{\ln(y)}{W(\ln y)}$$ View answer 0 votes Hint :if you seek for a proof just to look the following answer by "Mark Bennet" for related question Answer: Suppose we want to find the antiderivative of$\cfrac 1x$- it doesn't come in the ... View answer 0 votes I cannot say that this is an elegant solution but oh well, i will use$s$for$sin$,$c$for$cos$and$t$for tan we know that$\frac{s}{c} = t$so$\frac{16}{c^{6}*t^{6}} + \frac{81}{c^{6}} = 625$... View answer -1 votes To solve:$3\sin^{3}x+2\cos^{3}x=2\sin x+\cos x$. Use the substitution$\sin x = \frac{2\tan \frac{x}{2}}{1+\tan^{2}\frac{x}{2}}=\frac{2t}{1+t^{2}}$and$\cos x =\frac{1-t^{2}}{1+t^{2}}$. We become$t^...