# Storing small numbers and logarithms.

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daniel adams 2021 年 11 月 30 日
コメント済み: daniel adams 2021 年 12 月 1 日
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?
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Walter Roberson 2021 年 11 月 30 日
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

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### 回答 (2 件)

Walter Roberson 2021 年 11 月 30 日
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 =
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daniel adams 2021 年 12 月 1 日
Yes cheers for the try though :)

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Steven Lord 2021 年 11 月 30 日
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 =
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daniel adams 2021 年 12 月 1 日
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 .

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