Storing small numbers and logarithms.

Hi I have the following problem, in my code I need to calculate for very large. The largeness causes matlab to store and to be zero. Does anyone know a way around this? Is there a way of taking logarithms first?

[Walter Roberson]

What precision do you need for output? If a and b differ by more than about 36.04 then to numeric precision, the answer is going to be the same as -min(a,b) . If they differ by N then approximately the first (3/2)*abs(N) bits are conserved

Answer 1

[Walter Roberson]

Depending on your range of values and your accuracy needs, you could try a double taylor approximation.

syms a b positive
E = log(exp(-a) + exp(-b))

E = 

Ta = taylor(E, a, 'ExpansionPoint', 10^5, 'Order', 10)

Ta = 

Tb = taylor(Ta, b, 'ExpansionPoint', 10^5, 'Order', 10)

Tb = 

Ts = simplify(expand(Tb))

Ts = 

Tsa = collect(Ts, [a])

Tsa = 

[Walter Roberson]

Well, this turns out to be quite inaccurate in my tests !

[daniel adams]

Yes cheers for the try though :)

Answer 2

[Steven Lord]

How large is "very large"? Can you perform the calculation symbolically (by converting a and b into sym values before trying to pass them into exp -- if you try to perform the exp calculations in double precision then convert to sym you've already underflowed) and convert the result back to double afterwards?

a = 12345;
as = sym(a);
L = log(sym(exp(-a))) % exp(-a) underflowed in double before being converted to sym

L = 

Ls = log(exp(-as)) % No underflow

Ls = 

[daniel adams]

Would this method still work if a and b were very large dimensional vectors and I was applying the exp and log functions pointwise? Will using sym take my code longer to run?

[Steven Lord]

1) Yes, the symbolic exp and log functions operate element-wise just like the numeric exp and log functions.

a = 1:10;
as = sym(a);
e = exp(a);
es = exp(as);
format longg
double(e-es) % Small differences

ans = 1×10

1.0e+00 * 2.99524520677138e-16 1.79711394978391e-16 1.82756255255129e-16 -2.87415780158441e-15 -3.48635149004642e-15 -1.23596280244504e-14 -9.86975264043409e-14 2.71032958168736e-14 2.15308776210672e-13 1.37801347005174e-12

2) Yes, there is some additional overhead for the symbolic calculations. How large is "very large" in "very large dimensional vectors"?

[daniel adams]

Hi @Steven Lord thank you greatly for your help so far this is a problem that has really been bothering me! For more context can you see my new question ( which is the full version of what I was asking on this post ) https://uk.mathworks.com/matlabcentral/answers/1600419-using-symbolic-math-to-retain-accuracy-arrays By very large the matrices will be of size .