Increase precision of digits

I am dealing with numbers around 10^-18. Matlab type double only handles up to 10^-15. Is there a way to increase the number of digits? Many thanks.

Answer 1

[Tobi12]

Thanks for your replies. Sorry, maybe I was not precise enough in my question.

I am dealing with numbers like -0.897214128288675 * (1.0e+03).

For my calculation I need more accuracy than above -> 18 digits after the comma would be excellent.

Also, I am working with arrays...

[Iain]

Depending on what you're doing, 64 bit integers can represent almost 19 digits.

Why do you need more than 16 digits of accuracy?

[Tobi12]

I am numerically trying to solve an integral equation. For this purpose I had to substitute a log-function.

After my iteration in the "log-domain", I need to re-substitute with exp-function in order to find the results.

It turns out that the iteration in the "log-domain" is quite good (input/output identical up to 10^-14), however, in the "exp-domain" is not good enough.

...

I just tried the uint64() function for an arrays. Unfortunately the exp() function does not work then anymore...

[Titus Edelhofer]

Integers will not help you, at least not without effort. I would recommend to try the symbolic toolbox with VPA.

Or rethink your algorithm ;-).

Titus

[Tobi12]

I am now using the sym. tool box. That makes my algorithm increadibly slow.

Isn't there another solution?!?!

[Titus Edelhofer]

Hmm, not really. That's why numerical analysts have been working for decades to think of numerically stable algorithms. As said before, not often you have problems where adding digits is the brute force attempt to save a bad/unstable algorithm from failing (not saying that this is indeed the case for you!).

Sorry,

Titus

Answer 2

[Titus Edelhofer]

Hi Tobi,

this is not really true:

x = 2e-127;
y = 3e-126;
x*y
ans =
  6.0000e-253

I think you mix up the size of numbers and precision? Could it be that your problem is more like

a = 1+1e-18
a =
   1

Titus

[adi]

hey,

i have the exact problem you described.

how can i fix it?

thanks

[Rik]

Do you really need such a large range of decimal precision?

Possible solutions if you indeed need this are linked in several comments in this thread.

[adi]

i have 1+7e-6

and it gives me 1 , im dealing with electric fields and i need to have high resolution.

thanks

[Rik]

No, it only displays as a 1. Try the code below to see that 7 digits is not an issue at all.

val=1+7e-6;
clc
fprintf('%.8f\n',val)

[adi]

thank you!

Answer 3

[Guillaume]

Due to the way double values are stored the accuracy of a double value depends on its magnitude. At 1e-18, that accuracy is around 2e-34:

>> eps(1e-18)
ans =
    1.92592994438724e-34

Of course, if the magnitude increases, the accuracy decreases.

>>eps(10)
ans = 
    1.77635683940025e-15

There is no floating-point data type in matlab with more accuracy than double.

If you want fixed accuracy regardless of the magnitude of the number, you have to use a fixed-point data type (double is floating point). I believe there is a matlab toolbox for that but I don't know anything about it.

Another option is to use java's BigDecimal:

>>n = java.math.BigDecimal(java.math.BigInteger('1'), 24)
n = 
   1E-24

>>m = n.add(java.math.BigDecimal(10)) 
m = 
   10.000000000000000000000001

[Titus Edelhofer]

Or use variable precision arithmetic from symbolic toolbox, if you need more (relative) accuracy. http://www.mathworks.com/help/releases/R2014b/symbolic/vpa.html

But I would suggest to let us know some more details, most often 16 digits should be enough ...

Titus

[John D'Errico]

Or use my HPF toolbox , which does allow your choice of higher precision in a floating point form. It is on the file exchange.

So in 100 decimal digits of precision...

exp(hpf('1.23',100))
ans =
3.421229536289673573790152351452246322159257631512470074140348979345525540446010876820674847912376597

log(ans)
ans =
1.23