Is this a symbolic math bug?

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Igor
Igor 2011-2-9
1. Let's a=log(2), I want create with only 5 digits:
>>digits(5);b=sym(a,'d')
b = 0.69315
But what is another way? This way isn't good while - by command digits I effect global changes -- and I am forced to return: digits(32)
2. Perhaps one wants to include periodic ratio Suppose a working example: >> b=sym(0.69696969696969,'r') true result 69/99=23/33, but many signs are needed -- on contrary: >> b=sym(0.6969,'r') b = 6969/10000
How to easy include, for example, 1.23(45) as ratio?
3. >> syms x positive;
>> x=solve('y=-1')
x = -1
But why was quiet reaction? How to clarify the attribute "positive" (in order to machine feel)?
4. And critical and scandal bug:
>> sym(0.6915 - 0.69,'d')
ans = 0.0015000000000000568434188608080149
at this: >>0.6915000-0.69
ans = 0.001500000000000
and also you may see:
>> sym(0.6915 - 0.69,'r')
ans = 211106232533/140737488355328
>> 211106232533/140737488355328
ans = 0.001500000000000
instead of sholar simple 15/10000
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Doug Hull
Doug Hull 2011-2-9
It is a best practice to post ONE question per question. It keeps things orderly.
James Tursa
James Tursa 2011-2-9
This is the decimal representation of the exact value you are using:
>> num2strexact(0.6915 - 0.69)
ans =
1.50000000000005684341886080801486968994140625e-3
The fact that MATLAB only displays 15 significant decimal digits of this does not mean that the underlying number is exactly the displayed result.

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Igor
Igor 2011-2-10
Thanks. To continue:
Q1 Solution based on your idea (oh.. I am stupid :)- ): >>a=log(2);a=sym(vpa(a,5))
Q2: to clarify -- notice for developer I want such reaction: >>a=sym('1.23(45)','r')
a= 679/550
MATLAB MUST tranform periodic kind of rational numbers in math: 1.234545454545..=1.23(45)=1.23+0.01*45/99=...=679/550
A valid script on this topic from another forum user (but I don't like it)
----
function numrat = forMatigor(numch)
% numch -> char ?????: numch = '1.23(45)'
I = regexp(numch,'[.(]');
J = length(numch)-1;
numrat = rats(eval([numch(1:I(2)-1) repmat( numch( I(2)+1:J),1,1+ceil((20-diff(I)+1)/(J-I(2))))]));
----
Also MATLAB would realize reverse: for example -- '1/3' -> '0.(3)'
Q3: yes, here is different (>>syms x positive):
solve('x=-1') VS solve('y=-1')
if both commands run without assigning to x -- it seems results are equavalent:
ans
-1
I don't understand difference... YES -- I HAVE CLARIFYED solve() uses directly noted variable or create it on flying, and assigning x=.. gives us new var. at same name
Q4: The conclusion become obvious. But no good construction. In general:
Let's x,y real anyway created, needed (x-y) as symbolic ratio. I wish be guaranteed:
sym(0.5-0.8) --> -3/10 %normal
sym(0.6915 - 0.69) --> 211106232533/140737488355328 %wrong
sym(rats(0.5)-rats(0.8)) --> [ 0, 0, 0, 0, 0, 0, -3, 0, -3, 0, 0, 0, 0, 0] %what is it?
sym(0.69315)-sym(0.69) --> 63/20000 % best way
>>a=log(2);a=sym(vpa(a,5));a-sym(0.69)
0.0031471805598357605049386620521545 % again wrong
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Andrew Newell
Andrew Newell 2011-2-10
Igor, can you please make a new question for each of these issues? This sprawling post is hard to deal with.
James Tursa
James Tursa 2011-2-10
In the case of (0.5-0.8) the sym function is trying to "guess" what you meant by the input. Again, the input is not exactly 0.3, it is:
>> num2strexact(0.5-0.8)
ans =
-0.3000000000000000444089209850062616169452667236328125
So in this case it guessed that you really wanted an exact 0.3 result even though the input was *not* exactly 0.3. But in the other case:
>> num2strexact(0.6915 - 0.69)
ans =
1.50000000000005684341886080801486968994140625e-3
the sym function just wasn't clairvoyant enough to guess what you wanted for a result. Don't blame syms for being "wrong" just because it didn't guess what you wanted as well as you had hoped. Reformulate your code properly instead.

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更多回答（1 个）

Andrew Newell
Andrew Newell 2011-2-9
I think what you want in question 1 is
b = vpa(log(sym('2')),5);
For question 3, x should be the variable in SOLVE:
syms x positive
x = solve('x=-1',x)
For question 4, try
vpa(sym('0.6915')-sym('0.69'),5)
ans =
0.0015
To get the right answer for the fraction, use:
vpa(sym('211106232533/140737488355328'))
You were calculating it in double precision, which isn't accurate enough.
EDIT: I have incorporated a couple of suggestions from @Walter and another from @James.
  2 个评论
Walter Roberson
Walter Roberson 2011-2-9
Andrew, you are letting things be evaluated by Matlab in double precision before having them pass in to the symbolic engine.
b = vpa(log(sym('2')),5);
vpa(sym('211106232533/140737488355328'))
Andrew Newell
Andrew Newell 2011-2-9
Good point, @Walter. I prefer doing my symbolic calculations directly in Maple - fewer pitfalls.

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