Search results

I can't make sense of the question, but if it's just asking you solve for x then we have: \begin{align*}\frac{\sqrt{2}(\cos(x)-i\sin(x))}{{2+i}} = \frac{1-3i}{5} &\Leftrightarrow \cos(-x)+i\sin(-x)) = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} \\ &\Leftrightarrow x = \frac{\pi}{4} + 2k\pi, \ k \in...

Yep!

I borrowed a book from the library today that was written in the late 19th century I believe, and it's on it's fourth edition (from the 60's I believe), and still widely referred to. If Terry Lee truly needed 7 editions to get his book right then he's either incompetent or just damn greedy. In...

\text{We have } x = \cos^2(\theta) \Rightarrow dx = -2\sin(\theta)\cos(\theta) \, d\theta \ \text{ and } \ 1 - x = \sin^2(\theta). Now, \begin{align*}x = 0 &\Rightarrow \cos^2(\theta) = 0 \\ &\Rightarrow \theta = \frac{\pi}{2}, \ \text{ and} \\ x = \frac{1}{2} &\Rightarrow \cos(\theta) = \pm...

Damn, I think that question has been made this forums bitch :)

Muchos gracias senior.

Letting y = 1 + 2cos(x) we have -dy/2 = sin(x) dx and: \int \frac{\sin(x)}{\sqrt{1 + 2\cos(x)} }\, dx = - \int \frac{dy}{2\sqrt{y}} = -\sqrt{y} + C = -\sqrt{1 + 2\cos(x)} + C. So \int_{0}^{\frac{\pi}{2}} \frac{\sin(x)}{\sqrt{1 + 2\cos(x)}} \, dx = -\sqrt{1 + 2\cos(x)} \...

Let me clearer: You want to show that the range of the one-to-one function f is positive. You start by considering the inverse of f, whose domain is positive. But this is equivalent to what you want to show, that is, you start off by assuming what you want to show. The comments about n at the...

I have some problems with this. In using the log function, which is the inverse of the exponential function (a fact that you use later on in your proof) and only defined for positive y, you're implicitly assuming the exponential function is positive before you start. But that is the fact you're...

A differentiable function En(x) is greater than 0 for all x if all the turning points occur "above the x-axis" and \lim_{x \to \infty} E_n(x) > 0 \text{ and } \lim_{x \to -\infty} E_n(x) > 0 Since En(x) is a polynomial (and hence differentiable everywhere) and n is even, both limits tend to...

The second derivative test can be a conclusive wrt to mins or maxs, but often isn't. The classic example y = x4 that jetblack gave doesn't play nice with that test.

\fbox{\text{Here is some text and some} \mu \alpha \tau \eta \sigma}

Well 0 isn't a positive number, I don't know what else to tell you. You can say the absolute value function is "non-negative" though. That identity you gave only works for real numbers. In general: |x + iy| = \sqrt{x^2 + y^2}, which with a bit of Pythagoras you can see is indeed the...

I do know what you mean, but is wrong.

Hahaha. Ever notice it's always the simplest things? :( \begin{align*}6x &\neq 2x + 3x & \text{for } x \neq 0 \\ &= 3x + 2x & \text{(since } \mathbb{R} \text{ is commutative)} \\ &\neq 6x & \text{for } x \neq 0 \end{align*}

What? |0| = 0 which isn't positive.

Nice. Maybe mention the align* and cases environment too? \begin{align*}6x &= 2x + 3x \\ &= 3x + 2x & \text{(since } \mathbb{R} \text{ is commutative)} \\ &= 6x \end{align*} and |x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}

Whoops, I didn't put too much thought into what words I used to give the pronunciation. I had the word technology in mind when I wrote that, so lay-tech as in technology. Lay-teck might be a little less ambiguous though :)

The legendary ... Donald Knuth

A'la French? :)