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location Champaign, IL
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visits member for 3 years
seen 8 hours ago

13h
answered Too high differential order in boundary conditions?
13h
comment Too high differential order in boundary conditions?
BVP as opposed to IVP is not relevant to what is going wrong. NDSolve wants initial/boundary conditions to have order strictly less than that of the differential equation. You could work around this by taking another derivative and also giving another condition say by evaluating the original DE at a point. (That verbiage is not easy to understand. In actual code: NDSolve[{x'''[t] - x'[t] == 0, x''[0] - x[0] == 0, x''[0] == 0, x[1] == 1}, x, {t, 0, 1}])
18h
comment Rings modulo n, units
Q: What does it mean for n to be a unit in integers modulo 60? A: n must not be a zero divisor. Q: Which numbers (as in, how to characterize them) in this ring are zero divisors? A: The ones not relatively prime to 60. Q: How do we count how many numbers in the ring are not relatively prime to 60? A: there's a well known function for that...
21h
comment Is the divergence between Round and setting an explicit precision intentional or a bug?
"The question is: Why does <even>.5 round to <even> with Round, while it rounds to <even+1>. when setting precision." To respond, I'll note that I only get evens for that latter scenario. In[1]:= Round[{2.5`1, 3.5`1, 4.5`1, 5.5`1, 12.5`1, 13.5`1}] Out[1]= {2, 4, 4, 6, 12, 14}. When one instead uses evenorodd.501`1 then the guard bits (which lie beyond the given precision boundary) influence the decision and rounding is upward (with evenorodd.499\1` it woud of course be sownward).
1d
comment Proving large inequality in 6 variables
If the integers can only lie in a finite set e.g. 0<=a+b+c<=10 and all nonnegative. could try brute force with all possible substitutions. Then you have a trivariate that might be amenable to either Reduce or FindInstance[opposite inequality].
1d
comment Is the divergence between Round and setting an explicit precision intentional or a bug?
Almost certainly Round is using guard digits, so the ones ending .501`1 will round up whereas the ones with .5`1 will round to even.
1d
comment ordered tree data structure
Define fInv as you proceed through the list X. If f[x]==k for some integer k then define fInv[k]=x (this is where you have created a DownValue) . Every time you encounter a new y in the list, first check if fInv[f[y]] is define (if not, then Head[fInv[f[y]]===fInv] will be satisfied because it will not evaluate). On the other hand, if fInv[f[y]] evaluates to some x then you have found a pair {x,y} for which f evaluates to the same integer.
2d
comment Unexpected pattern matching behaviour: PatternSequence vs. Optional
@Mr.Wizard Working on it (and the issue from the original post).
2d
answered Replacement of terms/Pattern matching involving products of derivatives of a function
2d
comment ordered tree data structure
Have a look at DownValues. I suspect that will serve you better. Basically you'd be storing values to give an "inverse" of f (quotes because it is multivalued), and any time one was encountered for some x', you'd have your corresponding x stored.
2d
comment Why does Mathematica add Hold to the result of my user defined symbol?
When I see a result like that, I make sure to pay attention to the warning messages that preceded it. I see no indication of those above. I don't suppose you thought they were optional? (I sometimes see people who treat stop signs that way, but that's a different matter.)
2d
comment Using Jacobi Method
Your matrix is far from diagonally dominant. I doubt the Jacobi method will give a good result.
Jan
22
comment How to find Lagrange multipliers corresponding to constraints?
From the FindMinimum result you can determine which are the binding constraints, and then write down the corresponding Lagrangian equation(s) and use NSolve or FindRoot to solve for the multipliers.
Jan
22
comment Find the smallest eigenvalue (not absolute value ) for a generalized eigenvalue problem
I guess I had not realized the general case would behave differently. I sent a link to the question to someone more knowledgeable, so maybe a better answer will emerge.
Jan
22
comment Find the smallest eigenvalue (not absolute value ) for a generalized eigenvalue problem
One approach is to get an a priori estimate of largest in magnitude, and shift along positive real axis with a multiple of identity matrix so that the most negative eigenvalue becomes also the largest. This will only work if there are no complex ones that become larger still.
Jan
22
comment $L = \lim_{z \rightarrow 1} \frac{1-z}{1-z^*}$ Evaluates incorrectly
I don't think Limit will handle Conjugate very well. Probably better to use explicitly real variables e.g. x+/-I*y. Then specify the path because, as you and others have noted, it matters. To approach from above fix x at 1 and have y->0. To approach along the unit circle, from the top, reparametrize as Cos[t]+I*Sin[t] and have t->0. Etc.
Jan
22
comment The 1st approximation of multi-viables which contains the differential
Your svrs2 might not be what you expect, and this could cause the result to have more terms than expected.
Jan
22
comment Unexpected pattern matching behaviour: PatternSequence vs. Optional
Interesting. Yes, probably a bug. I'll try to have a look when I get a chance.
Jan
22
comment How do I write nested for-loops?
Heh. Somehow strangely reminds me of myself on MSE.
Jan
22
comment Weird behavior when solving a simple physics problem
I cannot replicate the indicated results. I guess some input is missing. By eyeball I do not see 'V' anywhere, for example.