Recent content by pasmith

It may be easier to express the first two as $$ (1 - sN_i)(1 - c_i \lambda) + sN_i = 1 - c_i\lambda + sc_iN_i \lambda = 0.$$ Newton's method is actually fairly straightforward here; we can even solve the resulting linear system at design time. If we order the variables as ##(N_1, \dots, N_5...

That appears to be the only method Excel provides for solving ##Ax = b##. The documentation does not explain what method Excel uses to calculate a matrix inverse, but assuming it is based on the standard LAPACK library it is using LU factorisation with partial pivoting.

I would rotate the sphere so that one point is at the north pole (0,0,1). Then ##\theta## is just the colatitude of the image of the second point.

If ##A## is invertible, then solving ##Ax = b## by Gaussian elimination where ##b## consists of zeros except for a 1 in the ##i##th row will give you the ##i##th column of ##A^{-1}##.

We don't solve linear systems numerically by finding an inverse matrix; that is incredibly inefficient. Instead we use other methods, like Gaussian elimination or LU decomposition.

Your working appears to be correct, and I obtained the same result by considering the discriminant of ##f##. What did you do to verify the text answer? According to the text, for ##k = -3##, ##f## should have more than one real root, but according to Wolfram Alpha there is only one real root...

Coordinates are independent, whether orthogonal or nonorthogonal. Introducing such a dependency restricts you to a curve or a surface rather than the whole space. Consider the following example. A point on the plane can be written as $$x(\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}}) + y(0,1) = xe_x...

Alternatively, we can also consider that the general solution of ##F_{n+2} = F_{n+1} + F_n## is $$\begin{split} F_n &= A \left( \frac{1 + \sqrt{5}}{2}\right)^n + B\left( \frac{1 - \sqrt{5}}{2}\right)^n \\ &= A \varphi^n + B (-\varphi^{-1})^n \\ &= \varphi^n \left(A + B (-\varphi^{-2})^{n}\right)...

A slight correction is indeed required. From ##a_n \geq 1## we have ##a_n \varphi \geq \varphi ## so that $$ |a_{n+1} - \varphi| \leq \frac{1}{\varphi} |a_n - \varphi|$$ which together with ##\varphi > 1## is sufficient for ##a_n \to \varphi##.

There is a much easier way to arrive at these results. From ##a_{n+1} = 1 + a_n^{-1}## and the definition of ##\varphi## we can obtain $$ a_{n+1} - \varphi = (-1)\frac{a_n - \varphi}{a_n\varphi}.$$ Assuming ##a_n \geq 1## we have ##a_n \varphi \geq \varphi > 1## so that $$ |a_{n+1} - \varphi| <...

So that we understand the proofs, and can satisfy ourselves that they are indeed correct proofs. We don't take other people's word for it in mathematics.

It is a fact of basic group theory that conjugation preserves the order of an element (because conjugation is an isomorphism from a group to itself). The order of a cycle is the same as its length: the identity (generally written as a single 1-cycle) has order 1. a transposition (1 2) has order...

A generalization is to fix ##g: [0,1] \to \mathbb{R}## and let $$f_n : [0,1] \to \mathbb{R} : x \mapsto \begin{cases} ng(nx) & x < \frac 1n \\ 0 & x \geq \frac 1n \end{cases}.$$ Then $$\int_0^1 f_n(x)\,dx = \int_0^1 g(x)\,dx.$$ The pointwise limit is again zero for ##x > 0##, but at zero we have...

I was going to append this to @chwala's thread here, but thought it deserved a new thread. For ##n \geq 1##, define $$f_n : [0,1] \to \mathbb{R} : x \mapsto \begin{cases} 1 & x = 0, \\ n(1 - nx) & x \in (0, \tfrac 1n], \\ 0 & x \in (\tfrac 1n, 1]. \end{cases}$$ (Note that ##f_n## is...

Consider also ##e^{-n\sin \theta}## for ##\theta \in [0, \pi]##.