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In the course of doing mathematics, I make extensive use of computer-based calculations. There's one CAS that I use mostly, even though I occasionally come across out-and-out wrong answers.

After googling around a bit, I am unable to find a list of such bugs. Having such a list would help us remain skeptical and help our students become skeptical. So here's the question:

What are some mathematical bugs in computer algebra systems?

Please include a specific version of the software that has the bug. Please note that I'm not asking for bad design decisions, and I'm not asking for a discussion of the relative merits of different CAS's.

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5  
Judging by the answers below, maybe a better question would be not "What are some bugs?" but "Which websites have the most useful/comprehensive lists of bugs?". –  David Speyer Jan 12 '10 at 15:36
11  
This is possibly some sort of record: Richard Parker told me that he once typed "isprime(2)" as his first ever query to a certain computer algebra system, and got the reply "2 is not prime". He also claimed, probably correctly, that he could find a bug in any computer algebra system within 5 minutes. –  Richard Borcherds Aug 9 '10 at 14:27
3  
Richard's story is really surprising, because all systems I know will look up small primes in a table, so someone left off 2 from that table. It's possible, I guess, but really a silly goof. –  Thierry Zell Apr 27 '11 at 16:51

33 Answers 33

In the paper The Misfortunes of a Trio of Mathematicians Using Computer Algebra Systems. Can We Trust in Them?, the authors report a bug in Mathematica (which is still present in the version 10) that happens when computing determinants of matrices with large integers as entries.

The strangest thing of this bug is that for some matrices, the determinant function can give different values. The Mathematica notebook which reproduces the bug is available here.

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1  
That's a great article! –  Kevin O'Bryant Oct 16 at 16:03

$2^{4^{4^4}} < 4^{4^{4^4}}$

WA: False

Update: it seems to be fixed now

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2  
I think they have fixed it know. I tried and it says "True". –  Sasha Patotski Dec 17 '13 at 2:23

According to Wolfram Alpha

$$ (\log{(5+i)}+\log{(5-i)})^4= 10\,000$$

When one clicks on "10 000" WA spells it as integer.

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David Bailey and Jonathan Borwein said in a talk yesterday that the most recent editions of both Maple and Mathematica give the nonsensical result $$\int_0^1\int_0^1|e^{2\pi ix}+e^{2\pi iy}|\,dx\,dy=0$$ I later verified this for Maple 17, entering int(int(abs(exp(2*Pi*I*x)+exp(2*Pi*I*y)),x=0..1),y=0..1).

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From the sage-support mailing list.

Sage 5.10 claims $$\forall a,b \in \mathbb{R}, \; \sqrt{(a+b)^2}=\sqrt{a^2}+\sqrt{b^2} $$

though it contradicts it numerically for $a=1,b= -1$.

Session:

sage: var('a,b');assume(a,'real');assume(b,'real');ex=sqrt( (a+b)^2 ) - (sqrt(a^2)+sqrt(b^2));ex
(a, b)
sqrt((a + b)^2) - sqrt(a^2) - sqrt(b^2)
sage: ex.full_simplify()
0
sage: ex.subs(a=1,b=-1)
-2
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I don't know any interesting bugs in symbolic algebra packages but I know a true, enlightening and entertaining story about something that looked like a bug but wasn't.$\def\sinc{\operatorname{sinc}}$

Define $\sinc x = (\sin x)/x$.

Someone found the following result in an algebra package: $\int_0^\infty dx \sinc x = \pi/2$

They then found the following results:

$\int_0^\infty dx \sinc x \; \sinc (x/3)= \pi/2$

$\int_0^\infty dx \sinc x \; \sinc (x/3) \; \sinc (x/5)= \pi/2$

and so on up to

$\int_0^\infty dx \sinc x \; \sinc (x/3) \; \sinc (x/5) \; \cdots \; \sinc (x/13)= \pi/2$

So of course when they got:

$\int_0^\infty dx \sinc x \; \sinc (x/3) \sinc (x/5) \; \cdots \; \sinc (x/15)$$= \frac{467807924713440738696537864469}{935615849440640907310521750000}\pi$

they knew they had to report the bug. The poor vendor struggled for a long time trying to fix it but eventually came to the stunning realisation that this result is correct.

These are now known as Borwein Integrals.

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6  
Which means that nobody knows Fourier analysis nowdays. Very sad and discouraging story... –  fedja Jan 29 '10 at 18:47
26  
@fedja I disagree. It merely shows the reasonable state of affairs that someone expert in one field, having to borrow theorems from another field, may get a surprise. –  Dan Piponi Feb 16 '10 at 19:29
129  
The actual person at that "poor vendor" was me. I must have spent 3 days on this problem before I figured out that Jon had tricked me. And, indeed, I am an expert in computer algebra, but do not know much Fourier analysis. But Jon's proof for why this is 'correct' is quite geometrical. –  Jacques Carette Feb 17 '10 at 4:03
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@Jacques It's wonderful to hear from the 'poor vendor'. Thanks for replying. –  Dan Piponi Feb 17 '10 at 4:19
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@Harald, the integral in the link has the term $x^{-(n+1)/2}$, while the integral in the answer does not. Perhaps this is the reason for the differences in values? –  Joel Reyes Noche Apr 27 '11 at 2:31

The PARI/GP Thue equations solver gives wrong results when they are conditional on GRH.

Affected are at least versions 2.5.1 (latest) and 2.4.3.

? p=x^3 - 18*x^2 + 81*x + 1;a=3^3
%1 = 27
? t=thue(thueinit(p,0),a);[#t,t] \\ conditional on GRH
%2 = [3, [[0, 3], [3, 0], [19, 2]]]
? t=thue(thueinit(p,1),a);[#t,t] \\ uncoditional
%3 = [4, [[0, 3], [3, 0], [27, 3], [19, 2]]]

Found on the pari-dev mailing list http://permalink.gmane.org/gmane.comp.mathematics.pari.devel/3629.

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As was noted for Sage, for any open source CAS you can just look up the issue tracker. For example, here's the list if all the issues in SymPy tracker with the WrongResult label: http://code.google.com/p/sympy/issues/list?q=label:WrongResult. Most of them are pretty rare. You're much more likely to hit a bug that just gives an error when it shouldn't, or that gives an unexpected, but not technically wrong (mathematically), result.

My advice is to double check your answer in some other way. The chances of the same bug manifesting itself in two different ways are almost zero. For example, you can check a result numerically, which will use a completely different algorithm from the symbolic version. Many CASs even have built in functions that do this for you.

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We found some interesting bugs in Mathematica's integration software on this thread.

To wit, set

integral[m_,n_] = Integrate[Log[2+Cos[2Pi x]+Cos[2Pi y]] Cos[2Pi m x] Cos[2Pi n y],
                      {x, 0, 1}, {y, 0, 1}];

Then integral[1,1] should be $1/2-2/\pi$, but Mathematica 8.0.1 returns $1/2+2/\pi$. Values for other $m$ and $n$ are also wrong (see the question linked above), as can be quickly verified by replacing the "Integrate" command with "NIntegrate".

Curiously, if one changes the limits of integration to {x,-1/2,1/2} and {y,-1/2,1/2}, then the correct answers appear.

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1  
I don't know how Mathematica handles this integral, but I wonder if it's somehow related to the problems mentioned in Kurt's answer in dealing with branch cuts symbolically...in this case the problem is at the point $(1/2, 1/2)$ and moving the limits around moves this to a corner instead of the middle of the interval. –  Kevin P. Costello Mar 7 '12 at 20:56

Here are some results where different CAS give conflicting results:

  1. $\int_{y}^{\infty} \frac{e^{-x}}{x}{d x}$ for $y \in \mathbb{R}$ and $y>0$. Wolfram Alpha gives $$\log{y}+\Gamma(0,y)$$ and sage 4.7.1 gives $$ -{\rm Ei}\left(-y\right) $$

  2. For all integers $n$, Coq proves $$n \mod 0 \equiv 0$$ and Isabelle proves $$n \mod 0 \equiv n$$ (The proofs are just stated in theorems, I can give the exact theorems if needed). Interesting, both proofs doesn't seem to lead to inconsistency though AFAICT they depict the usual mod.

[Added] I am a fan of sage, but this bug annoyed me.

sage 4.7.2 incorrectly reports the girth of a 7 vertex graph:

H=Graph([(0, 1), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 6), (2, 5), (3, 4), (5, 6)]) 
H.girth()
4
H.is_triangle_free()
False

sage 4.3 and 4.6.2 return correct value.

sage session in the notebook and a plot of the graph

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1  
The mod 0 example is interesting. One could argue that $mod 0$ is "not defined" and so one has a choice. E.g. some CA systems will raise an error if you use $mod 0$. Yet the definition I know is: $n \equiv m mod k$ iff $n + k\mathbb{Z} = m + k\mathbb{Z}$ (and $a mod k$ is the least integer $r\geq 0$ such that $a=kx+r$ for some $x$). This appears in many places, e.g. in group theory $C_k$ is the cyclic group of order $k$, but often $C_0$ then denotes the infinite cyclic group. Which makes sense if you set $C_k:=\mathbb{Z}/k\mathbb{Z}$. Isabelle agrees. Anybody know a why Coq does what it does? –  Max Horn Oct 25 '11 at 11:35

This might get fixed in the future, but at the time of this writing, Wolfram Alpha gets apparently sometimes confused by logarithms of complex numbers:

Wolfram Alpha -- $\log(1+ \frac{1}{2}i) - \log(1 - \frac{1}{2} i)$

For reference, should the problem get fixed: it claims that $2i = 2i\cot^{-1}(2) \approx 0.9272$.

Curiously, the numerical approximation is correct, but the symbolic form seems to be wrong.

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Wolfram alpha is saying that the series of $\sum_k\sin(2 k \arctan(k^2))$ does not converge:

http://www.wolframalpha.com/input/?i=sum+sin%282+k+atan%28k%5E2%29%29

instead it converges! Seems that mathematica is only dealing with limits of functions not with limit of sequences.

Another simpler example is $\sum_k \sin(2k \pi + 1/k^2)$:

http://www.wolframalpha.com/input/?i=sum+sin%282k+pi+%2B+1%2Fk%5E2%29

E.

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2  
The first example is very nice, even without considering the original question; a fun exercise in calculus. The second example is really shocking! –  Zen Harper May 16 '11 at 0:36

Wolfram Mathematica 7 routinely confuses sums with integrals.

Example 1:

DSolve[(-Log[Log[a]] f'[x] + f''[x])/(Log[a] f'[x]) == D[Sum[f[x], x], x], f[x], x]

g[x_] := f[x] /. s
g[x]

Checking the result by inserting it into the equation shows the result is incorrect:

(-Log[Log[a]] g'[x] + g''[x])/(Log[a] g'[x]) - D[Sum[g[x], x], x]

Example 2:

s=NDSolve[{0.9159460564995328*Derivative[1][f][x] == f[x]*Product[f[x], x], f[0] == 1}, f, {x, -1.9, 15}]

Plot[Evaluate[f[x] /. s], {x, -0.4, 1.5}, AspectRatio -> Automatic, AxesOrigin -> {0, 0}]

In Mathematica 8.0 this has been fixed (i.e. it will report inability to solve the equations.

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This error affects all versions of Mathematica from 6 to 8. The result of a function depends on what letter is chosen for argument when calling it. In the simplest case it can be illustrated as follows:

in:

$A[\text{x_}]\text{:=}\sum _{k=0}^{x-1} x $

$A[k]$

$A[z]$

out:

$1/2 (-1 + k) k$

$z^2$

The correct answer is evidently, the later. This behavior affects not only sums but also integrals, so one have to check so that the letter user for the argument not to coincide with the index variable used for definition. In the case of recursion this becomes very difficult. The following example shows that moving a factor not dependent on the index variable out of the sum sign changes the result:

in:

A1[0,x_]:=1
A2[0,x_]:=1

A1[n_,x_]:=Sum[A1[-1 - j + n, x]*Sum[A1[j, k], {k, 0, -1 + x}], {j, 0, -1 + n}]
A2[n_,x_]:=Sum[Sum[A2[j, k]*A2[-1 - j + n, x], {k, 0, -1 + x}], {j, 0, -1 + n}]

A1[1,x]/.x->2
A1[2,x]/.x->2
A1[3,x]/.x->2

A2[1,x]/.x->2
A2[2,x]/.x->2
A2[3,x]/.x->2

A2[1,2]
A2[2,2]
A2[3,2]

out:

2
5
13

2
5
12

2
5
13
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12  
OK, so you should not use global variables as auxiliary variables in your definitions. Now why do you call this an error of Mathematica? –  Wilberd van der Kallen May 3 '11 at 9:23
2  
More specifically, use A1[n_, x_] := Module[{k, j}, Sum[A1[-1 - j + n, x]*Sum[A1[j, k], {k, 0, -1 + x}], {j, 0, -1 + n}]]; A2[n_, x_] := Module[{k, j}, Sum[Sum[A2[j, k]*A2[-1 - j + n, x], {k, 0, -1 + x}], {j, 0, -1 + n}]] –  Wilberd van der Kallen May 6 '11 at 8:09

Here is an example in Wolfram Alpha. A student had been given the assignment of finding the limit as $n$ tends to infinity of $\frac{1}{1+\frac{(-1)^n}{log(n)}}$. He had correctly arrived at the answer 1. Now he used WA to check if he was correct. WA returned 0 (the command lim n-> inf 1/(1-(-1)^n/log(n)) ). On examining the steps, it turned out that WA had manipulated a bit and used L'Hopital on the expression $\frac{log(n)}{(-1)^n+log(n)}$.

Note that if one instead asks for the limit of $\frac{1}{1-\frac{(-1)^n}{log(n)}}$ WA correctly returns 1, using the same method one usually would.

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5  
They must be reading this; it seems to work now. –  Kevin O'Bryant Apr 27 '11 at 13:07

Not a bug but a difficulty for users:

I do often not really understand how assignements work for CAS:

Given a variable $a$ with value, say, $\pi$, set $b:=a$ and set now $a$ to, say, $e$. What is the value of $b$?

As I understand the answer depends sometimes on the context (working with symbolic variables, vectors, floating numbers etc.) and the exact behaviour is sometimes difficult to guess for me.

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2  
Normally this should not be a matter of guessing. Somewhere the documentation should state whether the evaluation is call-by-value ($b=\pi$) or call-by-reference ($b=e$). –  darij grinberg Apr 27 '11 at 9:02
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Maple and Mathematica are both call-by-value. What Roland is probably referring to is that Maple has some variables which have an entirely 'new' calling convention,last-name-evaluation: a cross between call-by-value and call-by-name. An LNE variable (like a table) will 'evaluate' all the way to a value and then BACKTRACK one level and return the last name encountered! The reason for this is purely for display purposes, as the name is preferred over a large value. This decision was made in 1982 (or so), when it made some sense, but now Maple is stuck with this. MMA has similar oddities too –  Jacques Carette Apr 27 '11 at 12:46
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@Roland, the problem with leaving the choice to the user would be that the same program would give different results for different users. –  Joel Reyes Noche Apr 28 '11 at 1:04

This story heard from Enrico Bombieri. I do not know if it qualifies, since it is not a CAS bug, and in addition it is second-hand. However it might be quite effective in casting doubt in the mind of your students, if that's your purpose :)

E.B. was doing some Riemann zeta zero crunching on his PC some years ago, the software he wrote seemed ok, and the next step was to run it on a mainframe to get some serious data. He was given the privilege to try it on the first Cray supercomputer. Most of the time results were nice, but every now and then he got really weird results. He and his coworkers spent some awful weeks trying to catch the bug. In the end, they cornered the problem: when the Cray divided 1 by 12 the result was a negative number...

EDIT: I double checked, it was not a Cray supercomputer but a computer based on an early iteration of the Pentium chip (I guess an IBM one), and the Pentium bug was also encountered by others of course. Sorry for the inaccuracy.

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22  
Clearly, it calculated $1/12$ by summing the series $-1 - 2 - 3 - \cdots$. :) –  Tanner Swett Mar 8 '12 at 2:32

During some experimentation on Mathematica, attempting to symbolically evaluate an alternating series for the Stieltjes constants (formula 16 in the link) returns "Indeterminate", apparently due to the software attempting to evaluate the derivative of Hurwitz zeta where it shouldn't.

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Just found this in Mathematica 7.0 for Mac OS X x86 (64-bit) (November 11, 2008):

x=Exp[Pi Sqrt[163] ];
N[x-Round[x] ]
N[x-Floor[x] ]
N[x-Ceiling[x] ]
N[x - Round[x], 2]
N[x - Floor[x], 2]
N[x - Ceiling[x], 2]

The functions Round, Floor, and Ceiling are the obvious functions, while "N" converts the infinite-precision expression to a floating point number (the last three lines are aimed at 2-digit precision, while the first three should be 16-digit).

The first three calculations turn up as "-480." The last three give more correct values of -$7.5*10^{-13}, 1.0, -7.5*10^{-13}$.

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1  
Indeed N[x - Round[x], MachinePrecision] returns -480.\\ But N[x - Round[x], \$MachinePrecision] returns -7.499274028018143*10^-13\\ The decision to make N[x - Round[x], MachinePrecision] mean N[x - Round[x]] is indeed a bad one. –  Wilberd van der Kallen Jun 10 '10 at 15:37
4  
Very peculiar. So N[x] does its work with certain precision, while N[x,k] does its work aiming for a certain precision. I've been using Mathematica a long time, and never realized this. –  Kevin O'Bryant Jun 11 '10 at 17:23

In 1999, when I first bought an HP49G, whose major selling point was a CAS, I thought I'd try summing the harmonic series $\sum_{n=1}^\infty \frac{1}{n}$. I was a bit surprised to see the result 1151.8697216.

It turned out that it knew how to numerically compute the discrete antiderivative $\Psi(m) := \sum_{n=1}^m \frac{1}{n} \approx \ln m + \gamma$, and in the particular mode that it happened to be in, it would replace $\infty$ with the largest floating-point number it could represent, which was just under $10^{500}$. Indeed, $\Psi(10^{500}) \approx 500\ln 10 + \gamma \approx 1151.8697216$.

The story has a happy ending: after changing some flags, it returned $+\infty$.

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8  
For the record, the HP49G CAS source code is actually available from the author! See www-fourier.ujf-grenoble.fr/~parisse/english.html for the download link and whatnot. Unfortunately (?), it is written in RPL, which I find a little ugly. –  Quadrescence Jan 10 '11 at 0:05

Mathematica 7.0.1 says that Sum[1/(k*Length[Divisors[k]]), {k, 1, n}] is the harmonic number $H_n$, which is clearly wrong. The correct answer is at An elementary number theoretic infinite series

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Sometimes a CAS cannot get the right branch of inverse trig functions when calculating integrals symbolically. See for instance: https://pantherfile.uwm.edu/sorbello/www/classes/mathematica_badintegral.pdf

Apparently this is an unsolved problem in computer algebra.

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9  
Correct: this is an unsolved problem. There is no theory of 'analytic' integration in closed-form, the only theory that exists is purely algebraic. And these algebraic theories all ignore branch cuts. –  Jacques Carette Mar 3 '10 at 21:36
1  
It does not seem to have been mentioned in the paper linked above that one reason behind the mishandling of these trigonometric integrals is in mishandling the Weierstrass substitution. I have seen the following paper: doi.acm.org/10.1145/174603.174409 but I have no idea what the state of the art is nowadays for such things. –  J. M. Aug 2 '10 at 2:07
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@J.M.: as far as I know, David's paper that you refer to above is the 'most recent' on this topic. –  Jacques Carette Apr 27 '11 at 12:49

Over the summer I came across an elementary bug in Magma when working with congruence subgroups of SL_2(Z). The isEquivalent function, which is supposed to tell whether two points are identified by a congruence subgroup, would miss a lot of identifications. For example:

G := CongruenceSubgroup(2); % \Gamma(2)

H := UpperHalfPlaneWithCusps();

(G! [-11,4,8,-3]) in G; % Cast this matrix into \Gamma(2)

true % It's really in \Gamma(2)!

(G! [-11,4,8,-3]) * (H! 3/8); % Have the matrix act on the point 3/8

oo % Magma correctly computes that it gets sent to infinity

IsEquivalent(G, H! 3/8, H! Infinity()); % Are 3/8 and infinity equivalent under the action of \Gamma(2), and specifically, can you given me a matrix representing an element of \Gamma(2) sending the former to the latter?

false [1 0] [0 1] % Doh!

It's a pretty simple computation, and it was pretty clear what loop it was leaving out. We may have been running an old version of Magma, but anyway we reported the error to them, and they fixed it quickly, but I've never trusted computer algebra systems since!

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7  
My favorite Magma bug was when NthPrime(4) was 6. Supposedly, with NthPrime they implemented a system involving checkpoints and $x/log(x)$ estimations for $x\ge 10$, and then hard-coded the first few primes as: 2, 3, 5, 6. Oops... –  Junkie May 25 '10 at 6:20

Here's one I came across just now, in Maple 12. The code

with(combinat):
F := fibonacci:
phi := (1+sqrt(5))/2:
G := k -> F(k+1)/phi^k;
limit(G(n), n=infinity);

returns 0. But from the usual explicit formula for the Fibonacci numbers, which gives $F(n) \sim \phi^n/\sqrt{5}$, the output should be $\phi/\sqrt{5}$, or $(5+\sqrt{5})/10$. Replacing the built-in Fibonacci function with one that gives the explicit formula, and running the code

F := n -> 1/sqrt(5)*(((1+sqrt(5))/2)^n-((1-sqrt(5))/2)^n);
phi := (1+sqrt(5))/2:
G := k -> F(k+1)/phi^k;
limit(G(n), n=infinity);

gives the correct answer. I've encountered things like this fairly frequently when using the built-in routine for Fibonacci numbers; presumably this routine doesn't "know" the asymptotics.

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5  
Actually, for any function unknown function P (where fibonacci is 'unknown' in this context since it returns unevaluated for symbolic arguments), <code>limit(P(n)*exp(-n),n=infinity)</code> returns 0. Polynomially descent to 0 is not enough, that will return unevaluated. I've reported the bug. The problem is that there is an implicit assumption in the implementation that unknown functions do not "grow too fast" (or go to 0 to fast or ...), which is clearly wrong. The theory for computing limits does not deal with unknown functions at all. –  Jacques Carette Mar 3 '10 at 21:28

A quite serious error in Mathematica 7 in my opinion is that it thinks $ \sqrt{x^2} =x$, not $|x|$, leading for example to 2 solutions to the following differential equation: $$ y'(x) = 2 y(x) (x \sqrt{y(x)} - 1) \quad y(0) =1$$ Mathematica happily gives the following solutions: $$ y(x) \rightarrow \frac{1}{(1-2 e^x +x)^2}, \quad y(x) \rightarrow \frac{1}{(1+x)^2} $$ Of course, it is a theorem that there is a unique solution to a differential equation of this type, but that doesn't mean my students hand in the wrong answer in droves...

Mathematica code: FullSimplify[DSolve[{y'[x] == 2 y[x] (x Sqrt[y[x]] - 1), y[0] == 1}, y[x], x]]

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14  
Maple gets that one right. If the issues really is sqrt(x^2)=x, then Maple gets it right because I am the one who removed that assumption for the library in fall 1994 (Mike Monagan removed the buggy transformation from the kernel earlier that summer). –  Jacques Carette Feb 26 '10 at 16:36
3  
Mathematica also gets it right: Simplify[Sqrt[z^2], Element[z, Reals]] returns Abs[z], while Simplify[Sqrt[z^2]] returns Sqrt[z^2] unevaluated. The bug must be in the code of Dsolve, that somehow does not use this fact. –  Julián Aguirre Feb 27 '10 at 9:10
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See groups.google.com/group/comp.soft-sys.math.mathematica/msg/… for a workaround. This is indeed a bug. –  Mariano Suárez-Alvarez Mar 1 '10 at 14:00

Because the most popular systems are all commercial, they tend to guard their bug database rather closely -- making them public would seriously cut their sales. For example, for the open source project Sage (which is quite young), you can get a list of all the known bugs from this page. 1582 known issues on Feb.16th 2010 (which includes feature requests, problems with documentation, etc).

That is an order of magnitude less than the commercial systems. And it's not because it is better, it is because it is younger and smaller. It might be better, but until SAGE does a lot of analysis (about 40% of CAS bugs are there) and a fancy user interface (another 40%), it is too hard to compare.

I once ran a graduate course whose core topic was studying the fundamental disconnect between the algebraic nature of CAS and the analytic nature of the what it is mostly used for. There are issues of logic -- CASes work more or less in an intensional logic, while most of analysis is stated in a purely extensional fashion. There is no well-defined 'denotational semantics' for expressions-as-functions, which strongly contributes to the deeper bugs in CASes.

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3  
I'm not claiming Sage is bug-free, but a lot of those bugs aren't mathematical errors -- there are plenty of compilation issues, documentation problems, etc., not to mention nearly 700 tickets classified as "enhancement" rather than "defect" -- so claiming 1582 known bugs is a little misleading. –  Steven Sivek Feb 17 '10 at 14:38
7  
That is why SAGE and other OPEN SOURCE initiatives works great here: because implementation detail are publicly available ( as well as binaries, which allows testing for free). No secret methods, no unproven improvements without peer review... –  kakaz Feb 25 '10 at 15:20
10  
But 'peer review' has essentially established (over 15 years ago) that the fundamental design of using untyped expressions when doing symbolic calculus (i.e. computing closed-forms) is unsound. SAGE does not fix that - so the fact that it is open source and publicly available changes that how? SAGE is, by explicit design, just as bad as the closed-source systems at analysis. While I am a definite fan of open source, I do not see how, in this case, that is actually relevant. –  Jacques Carette Feb 26 '10 at 18:07

A friend of mine told me about his experience with Maple (version 5 or 6, I think) when dealing with matrices over $\mathbb{Q}(\sqrt{2},\sqrt{3})$. When he computed the rank and the determinant for one particular $3\times3$-matrix, he was told that the rank was 3, and the determinant was equal to zero. The answer to this paradox is, that by default, for determinants the symbolic computation methods were used for radicals, and for ranks, the floating point representations of matrix elements!

This can be thought of as either a bug or his naiveness (for not checking out how to represent elements of number fields so that floating point representations never appear), but in any case is a serious argument for treating the computer algebra software with care...

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8  
Wait a minute, how the hell can a software written by someone with at least half a brain even try to compute rank using float computations? –  darij grinberg Feb 26 '10 at 15:51
5  
I wish I knew of this example when I was teaching linear algebra! I like to emphasize that someone must write all that magic software that makes the world go round, and it may well be them. Students nod sagely. Some of them later become programmers... –  Victor Protsak May 25 '10 at 5:41
5  
Just this week I graded a paper where a student started out with a 3x3 matrix with 2 repeated rows, very resourcefully did some row operations including division by 3, used rounding from a calculator, and managed to find an inverse for the original matrix... I had to go back to find out what the mistake was! –  Pietro KC Jun 10 '10 at 9:33

If you are performing numerical computations, then a more likely source of error is in roundoff or over/underflow. In these cases, I wouldn't say that the CAS is necessarily in the wrong, just that you need to know the properties of the underlying algorithm and either recast your input or reimplement it in a more numerically robust way. In such cases, decent introductions to numerical analysis should give you a feel for the types of issues you need to worry about.

Of course, on the matter of symbolics, then there are no excuses for errors.

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In Mathematica 7, the command

Table[DirichletCharacter[4, 2, n], {n, 0, 8}]

should return a list of values of the character with modulus 4 and index 2, evaluated at 0, 1, 2, ..., 8. Instead, it returns the decidedly non-multiplicative:

{0,1,0,-1,0,-1,0,-1,0}

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