6,900 reputation
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visits member for 2 years, 9 months
seen 8 hours ago

I received my Ph.D in mathematics from Ohio State in 1994 under the direction of Gerald Edgar and have been a professor of mathematics at The University of North Carolina - Asheville since 1997. In recent years, I've also worked as a part-time consultant to Wolfram Research focusing on development of mathematical content for WolframAlpha.


19h
comment Why such iteration leads to fractal?
@Mr.Wizard Not planning to - I think it should be closed.
19h
comment Is there a way to rewrite integrals in Mathematica using u substitution?
Perhaps you should ask over on mathematica.stackexchange.com? If you do, be sure to a specific example you'd like to see worked and some code indicating some effort on your part.
19h
comment Why such iteration leads to fractal?
It might be worth mentioning that the code is not particularly good. I have a package that generates the image, with all 100000 points and colors, in well under half a second.
2d
comment What is linearity?
There are some answers here.
Sep
13
answered The projection of a vector value function onto the xz-plane.
Sep
10
comment Will the Newton's method be convergent to the root of the following function: $f(x)=\frac{-x}{x^2-1}$?
Well, if you're going to use WolframAlpha, then you can just ask it to perform Newton's method for you! That link starts at $x_0=1/2$. Starting at $x_0=2$, WolframAlpha says there are no valid results.
Sep
10
answered Will the Newton's method be convergent to the root of the following function: $f(x)=\frac{-x}{x^2-1}$?
Sep
5
awarded  Cleanup
Sep
5
revised Solve $\cos{z}+\sin{z}=2$
rolled back to a previous revision
Sep
4
comment Solve $\cos{z}+\sin{z}=2$
@Did If there is a problem with this approach, I'd love to hear it. The only "problem" that you've alleged is that the solution set has period $\pi$ but, then, I never claimed that it does.
Sep
4
comment Solve $\cos{z}+\sin{z}=2$
@Did That's not true at all. Checking the solution against the original equation is quite a genuine suggestion. The original, $\sin(z)+\cos(z)$ has period $2\pi$, while the squared, $(\sin(z)+\cos(z))^2$, has period $\pi$ - thus the spurious solutions. Honestly, this is so elementary and encountered so frequently in pre-calculus, I assumed you saw it.
Sep
4
comment Solve $\cos{z}+\sin{z}=2$
@Did And they worked - awesome! Now, $\sin(z)+\cos(z)$ has period $2\pi$ - done. :)
Sep
4
comment Solve $\cos{z}+\sin{z}=2$
@Did Have you tried plugging my solutions (the only two actually written down) into the equation?
Sep
4
comment Solve $\cos{z}+\sin{z}=2$
@Did Have you tried plugging the solution into the equation? You'll find that both solutions work just fine, if you do. Now I never claimed or wanted to present a full solution - the answer started as a hint. In particular, I never claimed that the solution set had period $\pi$, merely that the period is related to the period of the tangent. In squaring both sides, it's likely that some spurious solutions are introduced.
Sep
4
comment Solve $\cos{z}+\sin{z}=2$
@ahorn You have a mistake in your 3rd line. The $-3$ should be a $+3$.
Sep
4
comment Solve $\cos{z}+\sin{z}=2$
@Did I guess we just apply the arc tangent.
Sep
4
comment Solve $\cos{z}+\sin{z}=2$
@ahorn I think you made a mistake in your solution of the quadratic.
Sep
4
revised Solve $\cos{z}+\sin{z}=2$
added 261 characters in body
Sep
4
answered Solve $\cos{z}+\sin{z}=2$
Sep
4
answered The order of convergence and the asymptotic error constant of the sequence $p_n=g(p_{n-1})$