If I evaluate the following input in Mathematica 12:
SetPrecision[DecimalForm[123.432654/54.1122356, 130], 130]
The result is:
2.2810488724291406725797060062177479267120361328125000000000000000000000000000000000000000000000000000000000000000000000000000000000
When I run the same calculation in other calculators, the results are equal until the 15th digit of the Mathematica result: 2,281048872429140. However, as of the 16th digit, the other calculators show an equal result whereas Mathematica is showing a different result:
Windows Calculator:
2,281048872429140591633586101550
https://keisan.casio.com/calculator:
2.281048872429140591633586101550[.....]
https://www.mathsisfun.com/calculator-precision.html:
2.281048872429140591633586101550[.....]
Mathematica:
2.281048872429140672579706006217[.....].
Why is (only) Mathematica ending up with a different result?
Can Mathematica somehow end up with the same result as the other calculators (supposing that these unanimous results are the correct ones)?
Mathematica's model of approximate decimal numbers is different from almost everyone else's model of approximate decimal numbers.
Because of the number of digits you supplied for each of 123.432654 and 54.1122356 these are assumed to be and treated as MachinePrecision numbers. That means they have the usual "about 16 digits of precision as supplied by the CPU floating point hardware in your computer, but it is a little more complicated than that.
Because of precedence rules Mathematica first evaluated each of those numbers and converted them to the internal floating point form, with the limited accuracy and all the problems that brings and all the speed of being able to perform calculations in hardware instead of software.
Then it did the division using the internal floating point hardware which resulted in another MachinePrecision number with only about 16 digits of precision.
Then with DecimalForm you asked Mathematica to extrapolate that result with only about 16 good digits into a 130 digit display.
All, or almost all with some very subtle things in dark corners, of the *Form functions are intended to and only used to produce something that can be displayed and not used for further calculations. For example, new users routinely do m=MatrixForm[mymatrix] to see a pretty formatting of the matrix and then proceed to try to do calculations with m, which fails.
Then you asked Mathematica to perform the SetPrecision function on that display to try to turn that into a 130 bit precision number. I can't even guess what that really did internally.
It seems those other calculators assume that the precision of the entered numbers is infinite. WL does not. You can specify what precision the entered numbers have e.g.
123.432654`30/54.1122356`30
2.28104887242914059163358610155
Related
My application requires a fractional quantity multiplied by a monetary value.
For example, $65.50 × 0.55 hours = $36.025 (rounded to $36.03).
I know that floats should not be used to represent money, so I'm storing all of my monetary values as cents. $65.50 in the above equation is stored as 6550 (integer).
For the fractional coefficient, my issue is that 0.55 does not have a 32-bit float representation. In the use case above, 0.55 hours == 33 minutes, so 0.55 is an example of a specific value that my application will need to account for exactly. The floating point representation of 0.550000012 is insufficient, because the user will not understand where the additional 0.000000012 came from. I cannot simply call a rounding function on 0.550000012 because it will round to the whole number.
Multiplication solution
To solve this, my first idea was to store all quantities as integers and multiply × 1000. So 0.55 entered by the user would become 550 (integer) when stored. All calculations would happen without floats, and then simply divide by 1000 (integer division, not float) when presenting the result to the user.
I realize that this would permanently limit me to 3 decimal places of
precision. If I decide that 3 is adequate for the lifetime of my
application, does this approach make sense?
Are there potential rounding issues if I were to use integer division?
Is there a name for this process? EDIT: As indicated by #SergGr, this is fixed-point arithmetic.
Is there a better approach?
EDIT:
I should have clarified, this is not time-specific. It is for generic quantities like 1.256 pounds of flour, 1 sofa, or 0.25 hours (think invoices).
What I'm trying to replicate here is a more exact version of Postgres's extra_float_digits = 0 functionality, where if the user enters 0.55 (float32), the database stores 0.550000012 but when queried for the result returns 0.55 which appears to be exactly what the user typed.
I am willing to limit this application's precision to 3 decimal places (it's business, not scientific), so that's what made me consider the × 1000 approach.
I'm using the Go programming language, but I'm interested in generic cross-language solutions.
Another solution to store the result is using the rational form of the value. You can explain the number by two integer value which the number is equal p/q, such that both p and q are integers. Hence, you can have more precision for your numbers and do some math with the rational numbers in the format of two integers.
Note: This is an attempt to merge different comments into one coherent answer as was requested by Matt.
TL;DR
Yes, this approach makes sense but most probably is not the best choice
Yes, there are rounding issues but there inevitably will be some no matter what representation you use
What you suggest using is called Decimal fixed point numbers
I'd argue yes, there is a better approach and it is to use some standard or popular decimal floating point numbers library for your language (Go is not my native language so I can't recommend one)
In PostgreSQL it is better to use Numeric (something like Numeric(15,3) for example) rather than a combination of float4/float8 and extra_float_digits. Actually this is what the first item in the PostgreSQL doc on Floating-Point Types suggests:
If you require exact storage and calculations (such as for monetary amounts), use the numeric type instead.
Some more details on how non-integer numbers can be stored
First of all there is a fundamental fact that there are infinitely many numbers in the range [0;1] so you obviously can't store every number there in any finite data structure. It means you have to make some compromises: no matter what way you choose, there will be some numbers you can't store exactly so you'll have to round.
Another important point is that people are used to 10-based system and in that system only results of division by numbers in a form of 2^a*5^b can be represented using a finite number of digits. For every other rational number even if you somehow store it in the exact form, you will have to do some truncation and rounding at the formatting for human usage stage.
Potentially there are infinitely many ways to store numbers. In practice only a few are widely used:
floating point numbers with two major branches of binary (this is what most today's hardware natively implements and what is support by most of the languages as float or double) and decimal. This is the format that store mantissa and exponent (can be negative), so the number is mantissa * base^exponent (I omit sign and just say it is logically a part of the mantissa although in practice it is usually stored separately). Binary vs. decimal is specified by the base. For example 0.5 will be stored in binary as a pair (1,-1) i.e. 1*2^-1 and in decimal as a pair (5,-1) i.e. 5*10^-1. Theoretically you can use any other base as well but in practice only 2 and 10 make sense as the bases.
fixed point numbers with the same division in binary and decimal. The idea is the same as in floating point numbers but some fixed exponent is used for all the numbers. What you suggests is actually a decimal fixed point number with the exponent fixed at -3. I've seen a usage of binary fixed-point numbers on some embedded hardware where there is no built-in support of floating point numbers, because binary fixed-point numbers can be implemented with reasonable efficiency using integer arithmetic. As for decimal fixed-point numbers, in practice they are not much easier to implement that decimal floating-point numbers but provide much less flexibility.
rational numbers format i.e. the value is stored as a pair of (p, q) which represents p/q (and usually q>0 so sign stored in p and either p=0, q=1 for 0 or gcd(p,q) = 1 for every other number). Usually this requires some big integer arithmetic to be useful in the first place (here is a Go example of math.big.Rat). Actually this might be an useful format for some problems and people often forget about this possibility, probably because it is often not a part of a standard library. Another obvious drawback is that as I said people are not used to think in rational numbers (can you easily compare which is greater 123/456 or 213/789?) so you'll have to convert the final results to some other form. Another drawback is that if you have a long chain of computations, internal numbers (p and q) might easily become very big values so computations will be slow. Still it may be useful to store intermediate results of calculations.
In practical terms there is also a division into arbitrary length and fixed length representations. For example:
IEEE 754 float or double are fixed length floating-point binary representations,
Go math.big.Float is an arbitrary length floating-point binary representations
.Net decimal is a fixed length floating-point decimal representations
Java BigDecimal is an arbitrary length floating-point decimal representations
In practical terms I'd says that the best solution for your problem is some big enough fixed length floating point decimal representations (like .Net decimal). An arbitrary length implementation would also work. If you have to make an implementation from scratch, than your idea of a fixed length fixed point decimal representation might be OK because it is the easiest thing to implement yourself (a bit easier than the previous alternatives) but it may become a burden at some point.
As mentioned in the comments, it would be best to use some builtin Decimal module in your language to handle exact arithmetic. However, since you haven't specified a language, we cannot be certain that your language may even have such a module. If it does not, here is how to go about doing so.
Consider using Binary Coded Decimal to store your values. The way it works is by restricting the values that can be stored per byte to 0 through 9 (inclusive), "wasting" the rest. You can encode a decimal representation of a number byte by byte that way. For example, 613 would become
6 -> 0000 0110
1 -> 0000 0001
3 -> 0000 0011
613 -> 0000 0110 0000 0001 0000 0011
Where each grouping of 4 digits above is a "nibble" of a byte. In practice, a packed variant is used, where two decimal digits are packed into a byte (one per nibble) to be less "wasteful". You can then implement a few methods to do your basic addition, subtract, multiplication, etc. Just iterate over an array of bytes, and perform your classic grade school addition / multiplication algorithms (keep in mind for the packed variant that you may need to pad a zero to get an even number of nibbles). You just need to keep a variable to store where the decimal point is, and remember to carry where necessary to preserve the encoding.
double r = 11.631;
double theta = 21.4;
In the debugger, these are shown as 11.631000000000000 and 21.399999618530273.
How can I avoid this?
These accuracy problems are due to the internal representation of floating point numbers and there's not much you can do to avoid it.
By the way, printing these values at run-time often still leads to the correct results, at least using modern C++ compilers. For most operations, this isn't much of an issue.
I liked Joel's explanation, which deals with a similar binary floating point precision issue in Excel 2007:
See how there's a lot of 0110 0110 0110 there at the end? That's because 0.1 has no exact representation in binary... it's a repeating binary number. It's sort of like how 1/3 has no representation in decimal. 1/3 is 0.33333333 and you have to keep writing 3's forever. If you lose patience, you get something inexact.
So you can imagine how, in decimal, if you tried to do 3*1/3, and you didn't have time to write 3's forever, the result you would get would be 0.99999999, not 1, and people would get angry with you for being wrong.
If you have a value like:
double theta = 21.4;
And you want to do:
if (theta == 21.4)
{
}
You have to be a bit clever, you will need to check if the value of theta is really close to 21.4, but not necessarily that value.
if (fabs(theta - 21.4) <= 1e-6)
{
}
This is partly platform-specific - and we don't know what platform you're using.
It's also partly a case of knowing what you actually want to see. The debugger is showing you - to some extent, anyway - the precise value stored in your variable. In my article on binary floating point numbers in .NET, there's a C# class which lets you see the absolutely exact number stored in a double. The online version isn't working at the moment - I'll try to put one up on another site.
Given that the debugger sees the "actual" value, it's got to make a judgement call about what to display - it could show you the value rounded to a few decimal places, or a more precise value. Some debuggers do a better job than others at reading developers' minds, but it's a fundamental problem with binary floating point numbers.
Use the fixed-point decimal type if you want stability at the limits of precision. There are overheads, and you must explicitly cast if you wish to convert to floating point. If you do convert to floating point you will reintroduce the instabilities that seem to bother you.
Alternately you can get over it and learn to work with the limited precision of floating point arithmetic. For example you can use rounding to make values converge, or you can use epsilon comparisons to describe a tolerance. "Epsilon" is a constant you set up that defines a tolerance. For example, you may choose to regard two values as being equal if they are within 0.0001 of each other.
It occurs to me that you could use operator overloading to make epsilon comparisons transparent. That would be very cool.
For mantissa-exponent representations EPSILON must be computed to remain within the representable precision. For a number N, Epsilon = N / 10E+14
System.Double.Epsilon is the smallest representable positive value for the Double type. It is too small for our purpose. Read Microsoft's advice on equality testing
I've come across this before (on my blog) - I think the surprise tends to be that the 'irrational' numbers are different.
By 'irrational' here I'm just referring to the fact that they can't be accurately represented in this format. Real irrational numbers (like π - pi) can't be accurately represented at all.
Most people are familiar with 1/3 not working in decimal: 0.3333333333333...
The odd thing is that 1.1 doesn't work in floats. People expect decimal values to work in floating point numbers because of how they think of them:
1.1 is 11 x 10^-1
When actually they're in base-2
1.1 is 154811237190861 x 2^-47
You can't avoid it, you just have to get used to the fact that some floats are 'irrational', in the same way that 1/3 is.
One way you can avoid this is to use a library that uses an alternative method of representing decimal numbers, such as BCD
If you are using Java and you need accuracy, use the BigDecimal class for floating point calculations. It is slower but safer.
Seems to me that 21.399999618530273 is the single precision (float) representation of 21.4. Looks like the debugger is casting down from double to float somewhere.
You cant avoid this as you're using floating point numbers with fixed quantity of bytes. There's simply no isomorphism possible between real numbers and its limited notation.
But most of the time you can simply ignore it. 21.4==21.4 would still be true because it is still the same numbers with the same error. But 21.4f==21.4 may not be true because the error for float and double are different.
If you need fixed precision, perhaps you should try fixed point numbers. Or even integers. I for example often use int(1000*x) for passing to debug pager.
Dangers of computer arithmetic
If it bothers you, you can customize the way some values are displayed during debug. Use it with care :-)
Enhancing Debugging with the Debugger Display Attributes
Refer to General Decimal Arithmetic
Also take note when comparing floats, see this answer for more information.
According to the javadoc
"If at least one of the operands to a numerical operator is of type double, then the
operation is carried out using 64-bit floating-point arithmetic, and the result of the
numerical operator is a value of type double. If the other operand is not a double, it is
first widened (§5.1.5) to type double by numeric promotion (§5.6)."
Here is the Source
Mathematica 8.0.1
Any one could explain what would be the logic behind this result
In[24]:= Round[10.75, .1]
Out[24]= 10.8
In[29]:= Round[2.75, .1]
Out[29]= 2.8000000000000003
I have expected the second result above to be 2.8?
EDIT 1:
I was trying to do the above for formatting purposes only to make the number fit in the space. I ended up doing the following to get the result I want:
In[41]:= NumberForm[2.75,2]
Out[41] 2.8
I wish Mathematica has printf() like formatting function. I find formatting numbers in Mathematica for exact field width and form a little awkward compared to using printf() formatting rules.
EDIT 2:
I tried $MaxExtraPrecision=1000 on some number I was trying for format/round, but it did not work, that is why I posted this question. Here it is
In[42]:= $MaxExtraPrecision=1000;
Round[2035.7520395261859,.1]
Out[43]= 2035.8000000000002
In[46]:= $MaxExtraPrecision=50;
Round[2.75,.1]
Out[47]= 2.8000000000000003
EDIT 3:
I found this way, to format a number to one decimal point only. Use Numberform, but first need to find what n-digit precision to use by counting the number of digits to the left of the decimal point, then adding 1.
In[56]:= x=2035.7520395261859;
NumberForm[x,IntegerLength[Round#x]+1]
Out[57]//NumberForm= 2035.8
EDIT 4:
The above (Edit 3) did not work for numbers such as
a=2.67301785 10^7
After some trials, I found Accounting Form to do what I want. AccountingForm gets rid of the 10^n form which NumberForm did not:
In[76]:= x=2035.7520395261859;
AccountingForm[x,IntegerLength[Round#x]+1]
Out[77]//AccountingForm= 2035.8
In[78]:= x=2.67301785 10^7;
AccountingForm[x,IntegerLength[Round#x]+1]
Out[79]//AccountingForm= 26730178.5
For formatting numerical values, the best language I found was Fortran, followed COBOL and also by those languages that use or support printf() standard formatting. With Mathematica, one can do such formatting I am sure, but it sure seems too complicated to me. I never understood why Mathematics does not have Printf[].
Not all decimal (base 10) numbers with a finite number of digits are representable in binary (base 2) with a finite number of digits. E.g. 0.1 is not representable in binary, just like 1/3 ~= 0.33333... is not representable in decimal. Mathematica (and other software) will only use a limited number of decimal digits when showing the number to hide this effect. However, occasionally it might happen that enough decimal digits are shown that the mismatch becomes visible.
http://en.wikipedia.org/wiki/Floating_point#Representable_numbers.2C_conversion_and_rounding
EDIT
This command will show you what happens when you find the closes binary representation of 0.1 using 20 binary digits, then convert it back to decimal:
RealDigits[FromDigits[RealDigits[1/10, 2, 20], 2], 10]
The number is stored in base 2, rather than base 10 (decimal). It's impossible to represent 2.8 in base 2, so it uses the closest value: 2.8000000000000003
Number/AccountingForm can take a list in the second argument, the second item of which is how many digits after the decimal place to show:
In[61]:= x=2035.7520395261859;
In[62]:= AccountingForm[x,{Infinity,3}]
Out[62]//AccountingForm= 2035.752
Perhaps this is useful.
In Mma, for example, I want to calculate
1.0492843824838929890231*0.2323432432432432^3
But it does not show the full precision. I tried N or various other functions but none seemed to work. How to achieve this? Many thanks.
When you specify numbers using decimal point, it takes them to have MachinePrecision, roughly 16 digits, hence the results typically have less than 16 meaningful digits. You can do infinite precision by using rational/algebraic numbers. If you want finite precision that's better than default, specify your numbers like this
123.23`100
This makes Mathematica interpret the number as having 100 digits of precision. So you can do
ans=1.0492843824838929890231`100*0.2323432432432432`100^3
Check precision of the final answer using Precision
Precision[ans]
Check tutorial/ArbitraryPrecisionNumbers for more details
You may do:
r[x_]:=Rationalize[x,0];
n = r#1.0492843824838929890231 (r#0.2323432432432432)^3
Out:
228598965838025665886943284771018147212124/17369643723462006556253010609136949809542531
And now, for example
N[n,100]
0.01316083216659453615093767083090600540780118249299143245357391544869\
928014026433963352910151464006549
Sometimes you just want to see more of the machine precision result. These are a few methods.
(1) Put the cursor at the end of the output line, and press Enter (not on the numeric keypad) to copy the output to a new input line, showing all digits.
(2) Use InputForm as in InputForm[1.0/7]
(3) Change the setting of PrintPrecision using the Options Inspector.
Why this code 7.30 - 7.20 in ruby returns 0.0999999999999996, not 0.10?
But if i'll write 7.30 - 7.16, for example, everything will be ok, i'll get 0.14.
What the problem, and how can i solve it?
What Every Computer Scientist Should Know About Floating-Point Arithmetic
The problem is that some numbers we can easily write in decimal don't have an exact representation in the particular floating point format implemented by current hardware. A casual way of stating this is that all the integers do, but not all of the fractions, because we normally store the fraction with a 2**e exponent. So, you have 3 choices:
Round off appropriately. The unrounded result is always really really close, so a rounded result is invariably "perfect". This is what Javascript does and lots of people don't even realize that JS does everything in floating point.
Use fixed point arithmetic. Ruby actually makes this really easy; it's one of the only languages that seamlessly shifts to Class Bignum from Fixnum as numbers get bigger.
Use a class that is designed to solve this problem, like BigDecimal
To look at the problem in more detail, we can try to represent your "7.3" in binary. The 7 part is easy, 111, but how do we do .3? 111.1 is 7.5, too big, 111.01 is 7.25, getting closer. Turns out, 111.010011 is the "next closest smaller number", 7.296875, and when we try to fill in the missing .003125 eventually we find out that it's just 111.010011001100110011... forever, not representable in our chosen encoding in a finite bit string.
The problem is that floating point is inaccurate. You can solve it by using Rational, BigDecimal or just plain integers (for example if you want to store money you can store the number of cents as an int instead of the number of dollars as a float).
BigDecimal can accurately store any number that has a finite number of digits in base 10 and rounds numbers that don't (so three thirds aren't one whole).
Rational can accurately store any rational number and can't store irrational numbers at all.
That is a common error from how float point numbers are represented in memory.
Use BigDecimal if you need exact results.
result=BigDecimal.new("7.3")-BigDecimal("7.2")
puts "%2.2f" % result
It is interesting to note that a number that has few decimals in one base may typically have a very large number of decimals in another. For instance, it takes an infinite number of decimals to express 1/3 (=0.3333...) in the base 10, but only one decimal in the base 3. Similarly, it takes many decimals to express the number 1/10 (=0.1) in the base 2.
Since you are doing floating point math then the number returned is what your computer uses for precision.
If you want a closer answer, to a set precision, just multiple the float by that (such as by 100), convert it to an int, do the math, then divide.
There are other solutions, but I find this to be the simplest since rounding always seems a bit iffy to me.
This has been asked before here, you may want to look for some of the answers given before, such as this one:
Dealing with accuracy problems in floating-point numbers