20190215 00:45:50 0600  answered a question  Conditional Plot, plot f(x+iy) if g(x+iy) satisfies some kind of condition No elaborate solution (some knowledge of

20190213 11:04:55 0600  answered a question  exponential equation solve problem Well...
Maybe looking for other solutions ? (Hint, hint...) EDIT : Further hint : the logarithm is a "multivalued function" in the complex field (i. e. not a function strictly speaking). Any time you take a log, you introduce further, possibly spurious, solutions... EDIT 2: Full solution, since no one seemed to see the problem : Original problem: if the members of this equations are equal, so do their logs. So we might try to solve : But the converse is not true !. More specifically : Therefore, we have to consider the solutions of : for any integer values of i. e. $$\left[x = \frac{2 i \pi z_{1} + 2 i \pi z_{2} + \log\left(5\right) + \log\left(3\right) + 2 \log\left(2\right)}{2 {\left(\log\left(3\right)  \log\left(2\right)\right)}}\right]$$ which is unique for any difference $z=z_1z_2$. Checking these solutions is not as direct as one could wish. But one can check that the ratio of the two members is one : One can note that the nonreal roots of this equation are somehow missed
by Sage (and Maxima). This is also true for i. e. $$c_1\in \mathbb{Z}\land x=\frac{2 i \pi c_1+\log (5)+\log (3)+2 \log (2)}{2 (\log (2)\log (3))}$$ HTH, 
20190209 04:09:53 0600  answered a question  Is there an easy way to get the matrix of coefficients from a product of a matrix and a vector? I beg to differ with the commenters. Staying at highschool level, your system boils down to a system of three linear equations with nine unknowns. It has therefore a sextuple infinity of solutions. And sage knows that : Or, more clearly, the only solution is: $${a_{1,1}} = \frac{r_{24} {x_{2}} + r_{21} {x_{3}}  {x_{1}}}{{x_{1}}}$$', '$${a_{1,2}} = r_{24}$$', '$${a_{1,3}} = r_{21}$$', '$${a_{2,1}} = \frac{r_{23} {x_{2}} + r_{22} {x_{3}}  {x_{2}}}{{x_{1}}}$$', '$${a_{2,2}} = r_{23}$$', '$${a_{2,3}} = r_{22}$$', '$${a_{3,1}} = \frac{r_{20} {x_{2}} + r_{19} {x_{3}}  {x_{3}}}{{x_{1}}}$$', '$${a_{3,2}} = r_{20}$$', '$${a_{3,3}} = r_{19}$$' In other words, your solution is a vector space of dimension six. This means that you could pick six of the elements of your and sole for the three last. For example: '$${a_{1,1}} = \frac{{a_{1,2}} {x_{2}} + {a_{1,3}} {x_{3}}  {x_{1}}}{{x_{1}}}$$', '$${a_{2,2}} = \frac{{a_{2,1}} {x_{1}} + {a_{2,3}} {x_{3}}  {x_{2}}}{{x_{2}}}$$', '$${a_{3,3}} = \frac{{a_{3,1}} {x_{1}} + {a_{3,2}} {x_{2}}  {x_{3}}}{{x_{3}}}$$' Sage has other, higher level, methods for solving this kind of linear algebra problems, whose discovery is left to the reader as an exercise ;). 
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20190205 16:15:48 0600  answered a question  How can I avoit the scientific notation on the yaxis? Well... Do yoy really need 14digits wide yaxis markers ? 
20190205 04:00:34 0600  answered a question  Problem with Padé approximation Well... By the way, the Padé approximation is slightly less precise than the Taylor development of the equivalent degree: And Your Sage version (which is antique, by Sage standards...) gives you bullshit. 
20190203 07:24:53 0600  answered a question  Piecewise in SageTeX Somebody called But the implementation isn't trivial. I also note that the 
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20190131 09:11:41 0600  answered a question  why gives a wrong result? Problem similar to this previous question, which generated the bug report Trac#26563. BTW, you should try to create questions a bit less err... elliptic (pun not intended but kept...). Your question is more a riddle than a question... 
20190127 12:01:07 0600  answered a question  Bug in series expansion? What about : 
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20190121 10:52:15 0600  answered a question  SageMath export to latex Document Foreword :
That said, LaTeX care and feeding may be considered superfluous at the beginning of a curriculum. A few possibility come to mind :
Another possibility is to use Its largest drawback (and, possibly, advantage) is that it is (currently) Emacsbound (although ports to other editors have been started, I'm not sure that Orgmode can be effectively compiled by anything else than emacs) : using it entails the Matterhornlike learning curve (and benefits) of Emacs... But the real points to Babel are :
So this may be an interesting tool for the instructor, but probably less so for the students. Note that, in any case, your students will have to use a markup language to be able to do about anything ; learning to leave the damned mouse alone (i. e. renounce WYSIWYG) is probably the hardest step of the learning process. Th exact markup language is probably not too important : markdown is less verbose than LaTeX, but has less possibilities (and, as far as I know, is not Turingcomplete...). 
20190119 13:31:36 0600  commented question  Problem with boring message I'm running H. sapiens sapiens L., without any upgrades such as divination coprocessor or telepathic interface. Without any details about what you attempting, it's quite hard for me to guess if LaTeX is or not needed for it. What in hell are you trying to do ? 
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20190118 01:10:01 0600  answered a question  Recovering numbers from continued fraction What's wrong with this : 
20190117 15:18:49 0600  answered a question  Multiplying Roots of a Polynomial Let's see... The roots are : Therefore, the factorized polynom is : And the roots' products is: Whereas moving to a polynomial ring, more specialized, is often useful, it is not necessary here... 
20190117 14:21:32 0600  answered a question  How can I solve the following (linear) differential equation? [ Not really an answer, but refinements needing more than the 500 characters of a comment ] calc314's answer is an explicit solution : The function If symbolic solutions are needed, you can also extract the characteristic polynomial and solve it symbolically : But I think that the "formal root" form s a better expression in a lot of situations. 
20190116 02:16:32 0600  commented question  Wrong solution/output for differential equation

20190115 01:42:01 0600  commented question  Wrong solution/output for differential equation

20190113 05:25:08 0600  commented answer  Combine plots with builtin Maxima, trajectory in Sage available? I see what you mean : your initial condition set admits both 1 and 1 as roots for Film at 11... BTW : sympy complains of Workaround :

20190112 07:06:52 0600  answered a question  Combine plots with builtin Maxima, trajectory in Sage available? [ Not really an answer, but the damn server software refuses this as a comment... ] I don't understand your beef with Therefore, S1 is a set of solutions of E1 (though possibly not the set of solutions of S1, but this is another question). BTW, both

20190108 01:18:02 0600  commented question  Sage Notebook IOPub data rate exceeded error? You should check that the documentation your research has led you to is applicable to Sage's jupyter and jupyter modules' version : Sage is not up to date on ths respect (and won't until the port to Python 3 is complete : recent jupyter versions need Python 3). 
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20181204 14:42:46 0600  commented question  Sage program for 40!+k project Homework ? 
20181203 09:49:22 0600  answered a question  Sage doesn't simplify a fraction if it's multiplied by 2 Probable 
20181203 09:21:10 0600  answered a question  How to get the collection of all functions from X to Y in SageMath Disclaimer : I know zilch about combinatirics. This problem might have elegant (and already programmed) solutions among the combinatorics tools of Sage. However, the problem seems simple enough to attempt a "naïve" solution. A function from $X$ to $Y$ is the set of tuples $[ (x_i,y_j),...,i=1,\dots,n,j\in[0..n]]$, meaning that $x_i$ has $y_j$ as image, the notation $(x_i,y_0)$ denoting the case where $x_i$ has no image. Let $S_{n,m}$ the set of all functions from $X$ to $Y$ when X has $n$ elements and Y has $m$ elements. 1): the set $S_{1,m}$ is simply $[(x_1,y_j),j=0,...,m]$. 2): knowing $S_{n,m}$, one can simply build $S_{n+1,m}$ by adding (i. e. concatenating) to each of the functions in $S_{n,m}$ a member of $[(x_{n+1},y_j),j=0,...,m]$. In other words, S_{n+1,m}$ is the cartesian product of $S_{n,m}$ and $[(x_{n+1},y_j),j=0,...,m]$. 3): 1) and 2) are sufficient to build $S_{n,m}$ by recurrence on n. It results immediately that the cardinal is $(m+1)^n$, including the degenerate case where no element of $X$ has an image in $Y$. The implementation, whose details depend of the planned use of the functions, is, as usual, left to the reader as an exercise ;). 
20181202 16:29:01 0600  answered a question  Variable Not Found while Plotting Finite Sum

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20181129 11:54:54 0600  answered a question  log base 2 in sagemath What's wrong with : Note that this isn't an exact answer (the exact answer is log(1000)/log(2)). But it's a reasonable numerical approximation of a quantity whose "exact" numerical expression would be infinite... Further discussion deserves reading part III of this excellent book as a prerequisite... 
20181124 13:26:11 0600  commented answer  finding solutions with exponential series expansion Neither Fricas nor Giac can solve this symbolically : Mathematica gives an answer subsuming Sympy's and asserting the existence of an infinity of (complex) solutions : (Reformatted from Mathematica's help) : From Sage's This function satisfies the equation Same difference... 