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In Matlab, I am trying to solve an equation that has single integral. For certain parameters the solution goes very close one even beyond the 64 bit double precision of the matlab.

I am not able use "vpa" kind of symbolic math as I need to use this high precision value in the numerical integration's later.

Is there any way to do numerical computations with this high precision in matlab?

sar1729
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  • It's better to ask your question in scicomp.stackexchange.com – MohammadSh Jul 06 '17 at 09:47
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    There appears to exist a multiple precision toolbox. If you like using toolboxes. – mathreadler Jul 06 '17 at 09:47
  • Of course you can also compile your own mex-files with the latest SSE instructions supported which usually include higher precision instructions. But that is only if you enjoy or would benefit from the low level coding it would include. – mathreadler Jul 06 '17 at 10:01
  • Edit the question to include the exact equation which you are trying to solve as well as the algorithm you are using. It is entirely possible that your problem is quite different from what you think it is. – Carl Christian Jul 06 '17 at 12:05

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One mathematical approach would be to build a matrix representation for the multiplication algorithm.

As we know from elementary school when multiplying numbers of different digits, we can do this as a convolution. For example:

$$123*234 = 3*234 + 20*234 + 100*234$$ Each digit in first number multiplied by the other number. We can determine exactly the position of the resulting products.

Now let's try and treat the range for a floating point number the same way as a "digit".

With matrices we represent convolution with Toeplitz matrices:

$${\bf M_1} = \left[\begin{array}{ccccc}3&2&1&0&0\\0&3&2&1&0\\0&0&3&2&1\\0&0&0&3&2\\0&0&0&0&3\end{array}\right], {\bf M_2} = \left[\begin{array}{ccccc}4&3&2&0&0\\0&4&3&2&0\\0&0&4&3&2\\0&0&0&4&3\\0&0&0&0&4\end{array}\right]$$

Now $$ {\bf M_1M_2} = \left[\begin{array}{ccccc}12&17&16&7&2\\0&12&17&16&7\\0&0&12&17&16\\0&0&0&12&17\\0&0&0&0&12 \end{array}\right]$$

We see we get values larger than 9, and that's when a carry digit comes into play. We can implement carry as an exercise and check with the real result $123\times 234 = 28782$.

Now all we need to do is translate all of this into double precision arithmetics.

EDIT: Implementing this in octave for decimal digits, and trying out the following calculations: $$3^{32} \cdot 3 \textrm{ ( no error ) }\\3^{33} \cdot 3 \textrm{ ( error last digit )}$$ This is reasonable because: $$\log_2(3^{33}) = 52.3\\\log_2(3^{34}) = 53.9$$ And the mantissa of a double precision number is just about 54 bits and the number 3 has a rather nasty representation in the binary base.

Here the numbers are:

$$3^{33} : 5 5 5 9 0 6 0 5 6 6 5 5 5 5 2 3\\ 3^{34} : 1 6 6 7 7 1 8 1 6 9 9 6 6 6 5 6 (9/\color{red}{8})$$

So the first time a digit differs it is by 1 from 9 to 8. And our approach gets the correct digit.

mathreadler
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