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6

In this answer I first give a solution to the great circle distance triangulation problem of the OP using spherical trigonometry. I then present a numerical method to solve the geodesic distance triangulation problem on the WGS84 reference ellipsoid (the ellipsoid used in the Global Positioning System). This numerical method requires an initial approximate ...


4

As the points are fromsampled data, we may ignore border cases such as that three of the four points are collinear. For a point $(x,y)$ you can compute the expression $$ f(x,y)=(x-x_1)(y_2-y_1)-(y-y_1)(x_2-x_1)$$ and this changes sign precisely when $(x,y)$ moves across the line given by $(x_1,y_1)$ and $(x_2,y_2)$. Hence if $f(x_3,y_3)$ and $f(x_4,y_4)$ ...


3

One possible way is to rewrite the equality as $$ 3\sin^3(x)-2\sin(x)=-2\cos^3(x)+\cos(x). $$ WARNING: The following step may introduce extra solutions which will need to be checked at the end. Now, square both sides. $$ 9\sin^6(x)-12\sin^4(x)+4\sin^2(x)=4\cos^6(x)-4\cos^4(x)+\cos^2(x). $$ Using the trig identity $\cos^2(x)=1-\sin^2(x)$, you can rewrite ...


2

HINT: $$\sin x(3\sin^2x-2)=\cos x(1-2\cos^2x)$$ Dividing both sides by $\cos^3x,$ $$\tan x(3\tan^2x-2\sec^2x)=\sec^2x-2$$ $\tan x=u$ $$\implies u\{3u^2-2(1+u^2)\}=1+u^2-2\iff u^3-u^2-2u+1=0$$


2

If you want to fit the function $$y=a+b\:X^c$$ with a set of data $(X_1,y_1),(X_2,y_2),...,(X_k,y_k),...,(X_n,y_n)$, this is possible on various ways. The usual methods are iterative and require guessed initial values $a_0,b_0,c_0$ to start the process. A non-usual method is described in the paper : ...


2

Here is what I do to use numerical methods to find the root of a real function $f(x)$. Note that numerical methods practically always assume that the function is continuous near the root. First I use a graph or other methods to find points $a$ and $b$ where $f(a)<0<f(b)$ and $f$ is continuous between $a$ and $b$. Then I am guaranteed that a root ...


1

Let $G$ be a rectangular domain $[0, a] \times [0, b]$ with equally spaced grid $N+1 \times M+1$. Also let $h_x = \frac{a}{N}, h_y = \frac{b}{M}$. Also let the known graient be $\mathbf{f}_{i,j}, \; i = 0,\dots,N,\;j = 0,\dots,M$. We hope that $\mathbf{f} = (f_x,f_y) $ really is a conservative field. Approach 1. Direct gradient approximation For every ...


1

If you want to determine $p$ instead of assuming a value for $p$ and fitting $a$ and $b$, you have moved from linear curve fitting to non-linear curve fitting. For linear curve fitting it is not required that the curve be a straight line, but that the model be linear in the parameters. Fitting data to $y=ax^2+bx+c$ is linear because $y$ depends linearly on ...


1

No, because if you know $5^{1/2}$ to use in $f(x)$, then clearly you know $\dfrac{1}{5^{1/2}}$. The usual Newton's method for finding the reciprocal of the square root of $a$ uses $f(x)=\dfrac{1}{x^2}-a$, which has the advantage that the Newton iteration does not need a division: $$ x_{n+1} = \frac{x_{n}(3-ax_n^2)}{2} $$


1

We see that two points $P_1$, $P_2$ are an the same (opposite) sides of the line $QR$ if an only if the $\it{oriented}$ areas $\text{Area}(P_1 QR) $ and $\text{Area}(P_2 QR)$ have the same (opposite) sign. The expression that appear are just determinants. This is easily generalized in $n$-space, when inquiring whether two points $P_1$, $P_2$ are on the ...


1

Assuming you're using method of lines. Let the original initial-boundary problem be $$ u_t = u_{xx}\\ u(0, x) = f(x)\\ u(t, 0) = a(t)\\ u(t, 1) = b(t). $$ Introduce a set of points $x_j = jh,\; j = 0,1,\dots, N,\;Nh = 1$. Let $u_j(t) = u(t, x_j)$. Note that $u_0(t) = a(t),\; u_N(t) = b(t)$ are already known. Unknown are $u_j(t),\; j = 1, 2, \dots, N-1$. ...



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