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Dec
28
comment How to visualize 3D fit
+1. It is useful at this stage to plot the residuals against the fit: that will make your comments more precise. BTW, fit["FitResiduals"] is a built-in way to obtain the residuals.
Dec
28
comment How to visualize 3D fit
A standard--and very effective--way to show the fit is to plot the residuals (equal to actual - estimated) against $x$ and $y$. For a good fit they should cluster evenly and randomly around the $xy$ plane. Because there's no question about where that plane lies, you don't even need the third dimension. For instance, many people map out the residuals in 2D using scaled and/or colored point symbols to represent their sizes and signs.
Dec
26
revised Define Log so that negative reals evaluate on lower edge of branch
added 154 characters in body
Dec
26
comment Using Transpose with a list as the second argument
+1 I am thinking the illustrations might work even better with some color coding. E.g., colors[1, 1] = Darker[Red]; colors[1, 2] = Darker[Green]; colors[1, 3] = Darker[Blue]; colors[2, 1] = Lighter[Red]; colors[2, 2] = Lighter[Green]; colors[2, 3] = Lighter[Blue]; f[x_] := Replace[MatrixForm[x], Subscript[a, i_, j_, k___] -> Style[Subscript[a, i, j, k], colors[i, j]], -1]; f[A], etc.
Dec
26
comment How do I expand a sum?
You are requesting two changes: first to Expand the products in the summation and second to Distribute the action of Sum over the addition. Consulting the help pages for Expand and Distribute will answer your question.
Dec
26
answered Define Log so that negative reals evaluate on lower edge of branch
Dec
24
revised How do I solve this equation?
Removed a potentially deceptive term from the title (this is not a functional equation).
Dec
21
comment Can Depth be used as an equivalent for MatrixQ?
@jVincent I do not know why you are addressing that last comment to me: I have not written anything here about lack of knowledge, intuition, FullForm, f[x_], or anything else you mention, nor have I ever suggested that the status of this question should be resolved based on our guesses about the O.P.'s state of mind. I have explained why I think this question, based on its merits, needs improvement, without resorting to any speculation about the knowledge or background of the person who asked it. (And--unlike 7 other people--I have not downvoted it.)
Dec
21
reviewed Leave Closed Can Depth be used as an equivalent for MatrixQ?
Dec
21
comment Creating an animation illustrating the time-evolution of a pre-computed orbit
+1 The tube is a great idea for giving a 3D visual cue.
Dec
21
comment Can Depth be used as an equivalent for MatrixQ?
I appreciate your comment, @jVincent. I would vote to reopen if the O.P. would at least explain what they think each expression is supposed to be doing. Without that, I maintain there is no question here except as each reader individually chooses to imagine it. Your personal construction of the question, as reflected in your answer, is reasonable, but whether it is a unique interpretation or the intended interpretation is difficult to determine.
Dec
21
comment An recurrence equation that can be solved in version 7 but not in version 8
It is interesting that the same problem, with $5$ changed to $4$, is solved in less than $0.1$ seconds with MMA 8, and with $5$ changed to $7$, the solution also takes less than $0.1$ seconds (although it's trivially expressed in terms of DifferenceRoot, which is just another way of giving up). With $5$ changed to $6$, a useful solution is obtained in $6$ seconds.
Dec
21
comment Can Depth be used as an equivalent for MatrixQ?
I confess to being mystified by this question. First, what is a "phrase," exactly? Second, if we take these to be Mathematica expressions, it is obvious they do completely different things, leaving me wondering what you might mean by "interchangeable." I am voting to close this as a non-question.
Dec
21
awarded  Nice Answer
Dec
20
comment How to calculate this sum?
There is a story, well known to mathematicians, about a very young Karl Gauss being set the problem of summing $1+2+\cdots+100$. He obtained the solution almost immediately by observing that by taking the sum from both ends, it "folds" into $(1+100)+(2+99)+\cdots+(49+52)+(50+51)$ = $101+101+\cdots+101$ = $50\times 101$. Summing $f(i/n)$ is accomplished with the same idea.
Dec
20
comment How to find lattice points on a line segment?
Nice idea. Incidentally, I noticed, in editing the question, that it stipulates the solutions should have positive coordinates.
Dec
20
revised How to find lattice points on a line segment?
deleted 1 characters in body; edited title
Dec
20
answered How to calculate this sum?
Dec
20
reviewed Close Function with different set of arguments
Dec
19
reviewed Close How to pass arguments between functions