Matlab function with inf in end points

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george veropoulos 2022-9-4

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https://ww2.mathworks.cn/matlabcentral/answers/1794980-matlab-function-with-inf-in-end-points

评论： george veropoulos 2022-9-7

采纳的回答： Walter Roberson

hi

I have a function define in an range [ a, b ] the function is ιnf at end point F(a)=inf F(b)=inf

there a method to eliminate this singular point?

thank

George

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Matt J 2022-9-4

编辑：Matt J 2022-9-4

在 MATLAB Online 中打开

You can rewrite F to give whatever alternative value you wish at a and b, e.g,

tand(89)
ans = 57.2900
tand(90)
ans = Inf
F(89)
ans = 57.2900
F(90)
ans = 0
function f=F(x)
 f=tand(x);
 if isinf(f), f=0; end
end

but whether that will solve your ultimate problem, we've no way of knowing, without more context.

george veropoulos 2022-9-4

移动：Matt J 2022-9-4

Thank tou I. Move the and point a liitle. I add a small number an i substrat a small number. I want cokpare with a. Monte catlo simulation

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Answer 1

Walter Roberson 2022-9-4

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https://ww2.mathworks.cn/matlabcentral/answers/1794980-matlab-function-with-inf-in-end-points#answer_1041910

Possibly

In some cases, inf shows up due to loss of precision. For example, if you had A = 1e100, B = 1e10 then A-B is going to evaluate in double precision as being the same as A, in a context where you end up with an infinity as a consequence. These kind of situations can be reduced by using the Symbolic Toolbox.

In some cases, you are working with expressions that mathematically evaluate to something but when evaluated naïvely in floating point numbers, evaluate to inf or nan. Such situations can be reduced by using the Symbolic Toolbox and taking limit() of evaluating an expression at a particular value, instead of just evaluating the expression at the value directly.

In some cases, when you are operating in floating point, temporary expressions can overflow to infinity as far as floating point is concerned, whereas they might mathematically be well defined. For example, log(exp(x)+1) when x > roughly 709.7827. Such situations can be reduced by using the Symbolic Toolbox.

In some cases, you can get around problems by rewriting the order of operations, such as 0.1 + 0.2 - 0.3 being different than 0.1 -0.3 + 0.2 .

In some cases, you can get around problems by doing a log transformation, such as log(y) = log(x1) + log(x2) + log(x3) - log(x4) instead of y = x1 * x2 * x3 / x4