Recent content by Ferramentarius

As to the absolute value it is a generalization which works in the case of negative numbers for which the real logarithm ln(x) is undefined. That it applies can be seen by forming the derivative in the separate cases when x > 0 and x < 0 respectively. One should still watch for intervals...

z = re^{i\phi} and Re(z) = \frac{1}{2}(z+\bar{z}) .

The described implication does not hold exclusively because cos(x) = cos(-x) = cos(x-2x) for any x. As cos(z) = cos(w) only when z and w are at the same x-coordinate of the unit circle we have z = \pm w + 2\pi k. In the quoted case it is x = \pm (x+y) \rightarrow y = 2\pi k \vee y =...

In combinatorics {{n}\choose{n}} = {{n}\choose{0}} = 1, while \varphi(1) = \varphi(2) = 1 where \varphi is Euler's totient function, also F_1=F_2=1 for the Fibonacci sequence and famously e^{i\pi}+1 = 0 known as Euler's identity.

(a-b)(a+b) = a^2-b^2 and a = \sqrt{a} \sqrt{a} .

Indeed already the Greek mathematician Euclid of Alexandria (fl. 300 BC) proved in his Elements, Book IX, Proposition 20 that the number of primes is infinite wherefore there cannot be the largest one.

While there are several approaches to differentials Wikipedia has a way of explaining this through linear mapping.

By the matrix definition of the cross product we have \vec{a}\times \vec{b} \cdot \vec{c} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ a_i & a_j & a_k \\ b_i & b_j & b_k \end{vmatrix} \cdot \vec{c} = (\vec{i} \begin{vmatrix} a_j & a_k \\ b_j & b_k \end{vmatrix} -\vec{j}...

x → cx, where c is a constant, is a linear map.

Consider the derivatives of trigonometric functions.

Wikipedia explains the basic properties of vectors such as addition, scalar multiplication, length as well as dot and cross products.

As the derivative is graphically the slope of the tangent line at a given point on the function, at the point (t, y(t)) the direction field should match the value of y'.

The key is that the numbers have a finite length, and can be separated far enough from each other by multiplication for further examination.

One more way to solve it is integration by parts. \int u(x)v'(x)dx = u(x)v(x) - \int u'(x)v(x)dx Substituting u(x) = 2x and v(x) = - \frac{1}{(x-1)} : \int \frac{2x}{(x-1)^2}dx = - \frac{2x}{x-1} + \int \frac{2}{x-1}dx = - \frac{2x}{x-1} + 2ln(x-1).

There are matrix exponentials for a given matrix X of nxn dimensions defined similarly to the ordinary exponential of a number. eX = \sum^{∞}_{k=0} \frac{1}{k!} Xk