How to change the accuracy of calculations? For example: >> x = log(2) x =0.6931 >> vpa(x) ans = 0.69314718055994528622676398299518041312694549560546875 I see only 53 symbols after dot. But I need to more. Help, please.
MATLAB: Change length of mantissa
mantissaMATLAB
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vpa(pi,100)evalutes pi at the Matlab level, to a double precision number, and passes that to vpa.
vpa pi 100is equivalent, because of command / function duality, to
vpa('pi','100')which would evaluate pi at the symbolic level according to the given number of digits.
Likewise, when you evaluate
vpa(log(2),5)you are evaluating log(2) at the Matlab level, to a double precision number. You don't want to do that if you care about precision.
vpa(log(sym(2)),5)I do not see at the moment a command to convert a decimal symbolic number to a fraction. Possibly
sym(vpa(log(2),5))but I do not have the toolbox to test this with.
(I know the command in Maple, but it does not appear to exist in MuPad.)
First, when you do this:
vpa(0.123456789123456789123456789)
MATLAB stores the input to VPA as a DOUBLE. MATLAB does all computations using doubles, stores all numbers by default as doubles. The symbolic toolbox is different. But MATLAB itself works on doubles.
So as far as MATLAB is concerned,
0.123456789123456789123456789
is just a number, a double precision number. And doubles are stored in a binary form, with 52 bits of precision in the mantissa. So, while you gave MATLAB a number with lots of decimal digits, only about the first 16 of them will be stored correctly. The rest have now been lost, thrown into the bit bucket.
sprintf('%0.55f',0.123456789123456789123456789)
ans =
'0.1234567891234567837965840908509562723338603973388671875'
VPA is just a function. MATLAB does not know what you will do with any number. It stores it temporarily in a double, then passes the results to the function being called, in this case, VPA.
Of course, VPA takes the number, and sees a DOUBLE. All it sees is what it was passed. It sees that binary representation, and converts those binary bits into a number. So, this is actually what VPA sees:
sprintf('%0.55f',0.123456789123456789123456789)
ans =
'0.1234567891234567837965840908509562723338603973388671875'
So, if I ask for at least 55 digits for VPA, I'll get the complete number stored.
vpa(0.123456789123456789123456789,55) ans = 0.1234567891234567837965840908509562723338603973388671875
By the way, you get those extra digits because binary fractions essentially gain an extra digit each time you divide by 2. So the first few powers of 1/2 are:
1 ./ 2.^(0:10)'
ans =
1
0.5
0.25
0.125
0.0625
0.03125
0.015625
0.0078125
0.00390625
0.001953125
0.0009765625
Remember, the mantissa of a double has 52 binary bits in it. While some binary fractions can be stored exactly, like 1/2 and 1/4, most decimals are not stored exactly as decimals.
Now, VPA is SOMETIMES a bit smarter than you might think. It can recognize some numbers even when they have been converted to approximate doubles.
x = 1/3
x =
0.333333333333333
sprintf('%0.55f',x)
ans =
'0.3333333333333333148296162562473909929394721984863281250'
vpa(x,55) ans = 0.3333333333333333333333333333333333333333333333333333333
But, for the most part, VPA is pretty dumb. It takes what it has been passed. And MATLAB is not smarter than that, because it works in terms of doubles by default.
Ok, so you CANNOT pass in a number as a double in general to VPA, and expect it to get things right. But you can pass in the number as a string.
digits 40
vpa('0.123456789123456789123456789')
ans =
0.123456789123456789123456789
Once MATLAB understands that you entered that number into symbolic form, it has no problem. It stores it in high precision form.
digits 40
x = vpa('2.38987696573583658648235686');
y = vpa('5.6387649837698237687649879932');
x + y ans = 8.0286419495056603552473448532
Once x and y are symbolic numbers, they will stay as such.
exp(x)*y^2 ans = 346.9591673851377328054605380636824962144
So you are now working in 40 digits of precision. Computations will be a lot slower of course, but you cannot expect everything.
I would note that you could have done exactly the same thing using my HPF toolbox, found on the file exchange, but the symbolic toolbox is entirely sufficient here.
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