The Chop feature in Mathematica replaces approximate real numbers in expression that are close to zero by the exact integer. I am searching for an equivalent function in Maple. Kindly note that I am NOT in search of trunc, round, frac etc. as they round off other floats to integers. I am only interested in "chopping" or rounding the near to 0 numbers to 0.
Any help is highly appreciated!
Check out Maple's command fnormal.
Related
I couldn't find anything in the API. Converting the number to a math/big.Int and back is not an option because the fractional component is significant to my calculation.
I'll end up repeatedly multiplying if there's no API, but that's a dissatisfying solution (math/big.Int.Exp is just O(log(n))) which might not be practical when I run into this problem again.
Thanks!
You may use MantExp() to take the exponent of a big.Float for a particular base/mantissa. Note that the formula for calculating the exponent for a given mantissa is:
x == mant × 2**exp
The sensor that I'm using will return 16-bit word and convert it to an actual value I need to use an expression,
The expression is ((175.72*16b_word)/65536)-46.85.
Can I divide by right shifting 16 positions?
I have searched for a couple of hours now and I still have no clue how to do with decimal representation! Does anyone have an example of how to solve it out?
Yes, shifting a binary number by 16 positions to the right is the same as a division by 65536 (with poor rounding however if your drop the shifted digits).
I'm looking for easiest way to divide two floating point numbers using VHDL. I need the code to be synthesizable (I'll be implementing it on Spartan 3 FPGA).
First operand will always be a fixed number (e.g. 600), and second one will be integer, let's say between 0 and 99999. Fixed number is dividend, and the integer one is divisor. So I'll have to calculate something like this: 600/124.
Or any other number instead of 124, of course that is in range between 0 and 99999. Second number (the one that is changing) will always be integer !! (there won't be something like 123.45).
After division, I need to convert the result into integer (round it up or just ignore numbers after decimal point, which ever is faster).
Any ideas ? Thanks !
There are many ways to do this, with the easiest being a ROM. You don't need floating point anywhere since doing an integer divide and compensating for a non-zero remainder can give you the same results. I'd suggest calculating the first 600 results in MATLAB or a spreadsheet so you can see that handling values up to 99999 isn't necessary.
Also, some common nomenclature for range and precision is QI.F where I is the number of integer bits and F is the number of fractional bits. Thus 0..99999 would be Q17.0 and your output would be Q10.0.
There's an FP divide function in this VHDL file from this site.
I tried a few minutes ago simple math operation
<%=((3+2+1)/100).round(8)%>
The result is 0.06, but the result of the ruby code above is 0.0. I would expect the result should be 0.060000.
Why not?
Thanks
(3+2+1)/100
is 0 because the division is integer. Try
(3+2+1)/100.0
You see, if both arguments of / are integer, the result of the division is an integer (the whole part). If at least one of the arguments is floating-point, then the result is also floating-point.
The dreadful integer arithmetic attacks again!
When you calculate ((3+2+1)/100), since all the operands are integers, Ruby uses integer arithmetic rather than floating point arithmetic.
If you do 7/100 it will also return 0, as it's rounded down to the nearest integer, which is 0.
Operations involving only integer data are done in integer (and then 6/100 is 0). Converting that 0 to float later (by round) does not bring you back the already discarded fractional part.
Change either of the values to float (e.g. 3.0) and you are done.
What is the recommended way to convert a floating point type to an integer type, truncating everything after the decimal point? CLng rounds, apparently, and the documentation for the = operator doesn't mention the subject.
UseFix or Int depending on the treatment you wish for negative numbers.
Microsoft article Q196652 discusses rounding in incredible detail. Here is an excerpt
The VB Fix() function is an example
of truncation. For example, Fix(3.5)
is 3, and Fix(-3.5) is -3.
The Int() function rounds down to
the highest integer less than the
value. Both Int() and Fix() act
the same way with positive numbers -
truncating - but give different
results for negative numbers:
Int(-3.5) gives -4.
Full disclosure: I referred to this nice answer by elo80ka
see
this
Undocumented behavior of the CInt() function
The CInt() function rounds to the nearest integer value. In other words, CInt(2.4) returns 2, and CInt(2.6) returns 3.
This function exhibits an under-documented behavior when the fractional part is equal to 0.5. In this case, this function rounds down if the integer portion of the argument is even, but it rounds up if the integer portion is an odd number. For example, CInt(2.5) returns 2, but CInt(3.5) returns 4.
This behavior shouldn't be considered as a bug, because it helps not to introduce errors when doing statistical calculations. UPDATE: Matthew Wills let us know that this behavior is indeed documented in VB6's help file: When the fractional part is exactly 0.5, CInt and CLng always round it to the nearest even number. For example, 0.5 rounds to 0, and 1.5 rounds to 2. CInt and CLng differ from the Fix and Int functions, which truncate, rather than round, the fractional part of a number. Also, Fix and Int always return a value of the same type as is passed in.
For positive numbers you would use
truncated = Int(value)
If used on negative numbers it would go down, i.e. -7.2 would become -8.