# Using symbolic math to retain accuracy. Arrays

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daniel adams on 1 Dec 2021
Hi folks. I am trying to avoid numerical underflow using symbolic math.
Basically at some point in my algorithm I have a matrix Which I call C. The values of C are very lagre. Now I would like to perform some operations on C by using some inbuilt matlab functions, but without loosing numerical accuracy.
The operations in words, and then in code are as follows : 1. Take the exponential of each element of ( pointwise )
a=exp(-C)
2. Sum the rows of C, so we are left with a array of the sum of each row of
b=sum(a,2)
3. Take the pointwise logarithm of each element in the array .
ans=log(b)
Of course if the elements of C are too large then matlab reads the elements of as 0. I think there may be a way around this ( since in the end we take logarithms ) by using symbolic math, but I dont know how (im very new to matlab). Anyone have any ideas?
I would like to mention that in practice my matrix C is of dimension , and these operations are performed at every iteration of a for loop, hence Im worried that using symbolic math will increases the run time of my code too much.
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Bjorn Gustavsson on 1 Dec 2021
@John D'Errico, ops, should've caught that one - factoring out the max of each row is better. But maybe rows with all zeros in C is not that bad - if the factoring out of the max will only lead to a term of the type of log(exp(0)) then this should be OK? (The original formulation seemed a bit vague as I speed-read it...)
daniel adams on 1 Dec 2021
@John D'Errico I dont think that strategy works here because we take exponential pointwise before summing. ( If im worng could you show me how the code would work ).

Bjorn Gustavsson on 1 Dec 2021
Factor out the average of each row of C. That should give you one term of the row-averages of C and then the sum of the logs of the exponential of the deviations relative to the average - these are hopefully small enough that your worries about nummerical accuracy are quenched...
HTH
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Bjorn Gustavsson on 2 Dec 2021
TRY!
exp([1 2 3]) = exp(3).*exp([-2 -1 0]);
Sum and take the log of both expressions and compare the results:
S1 = log(sum(exp([1 2 3])));
S2 = log(exp(3)) + log(sum(exp([-2 -1 0])));
isequal(S1,S2) % Yeah this is a numerically dodgy comparison
TRY, if you "think", dont, check and work things out instead.
If this is not the expression you want your problem description is either too vague or inaccurate.
HTH
Bjorn Gustavsson on 7 Dec 2021