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A numerical calculation problem leading to Inf or NaN in matlab

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Hancheng Zhu
Hancheng Zhu el 20 de Jul. de 2024 a las 15:02
Comentada: David Goodmanson el 21 de Jul. de 2024 a las 20:46
I want to calculate the exact value of , where and λ is a very large positive number. Obviously, we have the bound ,and therefore .
However, in reality, for example, if , due to the large λ, we have and the matlab will treat it as 0 and .
On the other hand, if , due to the large λ, we have a very large and matlab will treat the sum as Inf and . So how to avoid the above two cases and get the exact value of F in matlab?
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David Goodmanson
David Goodmanson el 21 de Jul. de 2024 a las 20:46
Hi HZ,
(1/lam) log( (x1^lam)*(1 + (x2/x1)^lam + (xn/x1)^lam) )
= log(x1) + (1/lam)*log(1 + (x2/x1)^lam + (xn/x1)^lam))

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Respuestas (2)

Torsten
Torsten el 20 de Jul. de 2024 a las 15:35
Movida: Torsten el 20 de Jul. de 2024 a las 15:35
log2(norm(x,lambda))
does not work ?
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Walter Roberson
Walter Roberson el 21 de Jul. de 2024 a las 4:34
Use the Symbolic Toolbox
Torsten
Torsten el 21 de Jul. de 2024 a las 14:59
Editada: Torsten el 21 de Jul. de 2024 a las 15:01
Maybe rewriting the expression as
1 / (1 + (x2/x1)^lambda + ... + (xn/x1)^lambda)*u1 +
(x2/x1)^lambda / (1 + (x2/x1)^lambda + ... + (xn/x1)^lambda)*u2 +
(x3/x1)^lambda / (1 + (x2/x1)^lambda + ... + (xn/x1)^lambda)*u3 +
...
(xn/x1)^lambda / (1 + (x2/x1)^lambda + ... + (xn/x1)^lambda)*un
can help.
If not, please give an example for x, u and lambda where the computation fails.

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Walter Roberson
Walter Roberson el 20 de Jul. de 2024 a las 21:18
If you need the exact value, calculate using the Symbolic Toolbox.
However, it is questionable what meaning to assign to the exact value of log2 of an expression. It is highly likely that log2 will be an transcendental number -- something that you cannot calculate the exact decimal representation for.

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