# chemetronics model 601 manual

You can enter two three-dimensional coordinate systems of a line and a circle, as well as choose two of them as the input, and obtain a list of coordinates representing the line and circle. Furthermore, it is possible to enter coordinates of a point of intersection of the line and the circle.
Simple Application:
Enter the line:
Line: y=3x+2
X1: 1
Y1: -1
X2: -1
Y2: -1
And the circle:
Circle: (x-4)2 + (y-8)2 = 4
X1: -4
Y1: 4
X2: 4
Y2: -2
Point of intersection:
Point: (-4 -8)
X: -6
Y: 3

The code source:
import org.geomajas.geometry.Line;
import org.geomajas.geometry.Point;
import org.geomajas.geometry.Rectangle;
import org.geomajas.gml.GML;
import org.geomajas.gml.LineString;
import org.geomajas.gml.Point;
import org.geomajas.gml.Rectangle;
import org.geomajas.gml.Vector;

public class SFSG {
public static void main(String[] args) {
try {
Line l1 = GML.createLineString(“line”, new Vector[] { new Vector(1, -1), new Vector(-1, -1) });
Line l2 = GML.createLineString(“line”, new Vector[] { new Vector(3, 2), new Vector(-1, -1) });
Point p = GML.createPoint(“point”, new Vector(-4, -8));
Rectangle r1 = GML.createRectangle(“rectangle”, new Vector[] { new Vector(4, 8), new Vector(-2, -8) });
Rect 0cd6e936a3

Source:

Q:

Is the Fourier Transform of a Convolution The Convolution of the Fourier Transform?

The Fourier Transform of a convolution of functions $f_1,f_2$ is $F(f_1,f_2)(\omega)=F_1(\omega)F_2(\omega)$
Does the Fourier Transform of the convolution of the Fourier Transforms $\hat{f_1}(\omega),\hat{f_2}(\omega)$ follow the above relation?

A:

No. There is no single general relation between the Fourier transforms of two functions, and the Fourier transform of the convolution of two functions. However, the answer will depend on what it means for $\hat{f_1}$ and $\hat{f_2}$ to be the Fourier transform of the functions $f_1$ and $f_2$, respectively. If they are, as they are in the paper, the Fourier transforms of the functions $f_1$ and $f_2$ then the answer is “no” because
$$\hat{f_1}(\omega)\hat{f_2}(\omega) eq\hat{f_1*f_2}(\omega)$$
unless $f_1$ and $f_2$ are both invertible.

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