Recent content by mutton

I think this will work: Pick 3 points on the line I am z = 2 that go in a certain direction. Say you pick them such that the half plane is on their right. For their images under the Mobius transformation, pick 3 points on the circle |w - 2| = 3 such that the interior of the circle is also on...

Since the question asks for evaluation by areas, graph the integrand from -3 to 0 and look at the resulting shape. Don't worry about summation.

Do your notes say that Mobius transformations are determined by 3 images and preserve orientation?

a1 -a2 -a3 -b1 b2 b3 c1 -c2 -c3 Find "patterns" in the negative signs so you can multiply certain rows or columns by -1 to get rid of them.

The midpoint between u and v is (u + v)/2. Convince yourself by drawing a parallelogram. This fact can be used to find conditions on c, d such that the linear combinations cu + dv fill in the line segment between u and v. A better way to think about this: The vector from u to v is v - u...

I think you actually want to find planes containing at least 2 of the 3 lines, and all my comments above are based on this. Otherwise, your answers don't make sense. No. If the 3 lines do not lie on one plane, then 3 planes are determined. For example, consider these lines. Define line A by...

This is actually only a subcase, so let's split into Cases V and VI. Case V, as you drew, has 3 intersecting lines all on one plane. If you wrote the slopes of the 3 lines in vector form, one of the vectors would be a linear combination of the other two. Case VI: What if the 3 intersecting...

Given ab = bc for all a, b, can you prove (ab)c = a(bc) for all a, b, c? Manipulate these equations for a while. To think of examples, try defining specific binary operations on small sets like {0, 1}. For an answer that's sort of intuitive, search Wikipedia for "commutative non-associative...

Cases II and III are essentially the same, and I think they both determine 1 plane only by two lines, even though the third line intersects that plane because the third line doesn't determine a plane with either of the other 2. It will help to know exactly what is meant by "determine a plane". I...

There is no other solution to f(x) = 1 because f is nondecreasing everywhere and increasing at 0 (use f' to show this).

Looks right. Doesn't matter how many base cases you have as long as you cover all natural numbers in the end.

5. Show that the points 1, -\frac{1}{2} + i \frac{\sqrt{3}}{2} and -\frac{1}{2} - i \frac{\sqrt{3}}{2} are the vertices of an equilateral triangle. 6. Show that the points 3 + i, 6 and 4 + 4i are the vertices of a right triangle. 7. Describe the set of points z in the complex plane that...

1. The slope of the line tangent to a differentiable function at a point is the derivative at that point. 2. a) This is a composite function of x, so use the chain rule. b) It's hard to isolate for y, so use implicit differentiation. 3. The slope of the line and the curve at any point can be...

d) Notice that (2, -4) is not on the curve. We want the equations of all lines passing through (2, -4) that are tangent to the curve. There could be 0, 1, or 2 of these lines because y is a quadratic. Let (x, y) be a point on one such line that's also on the curve. Then you know what y is in...

Infinity/infinity is an indeterminate form. Consider for example \lim_{x \to \infty} \frac{x}{x^2}. Both top and bottom approach infinity, but the limit is not 1.