Why do I get the message NaN?

Question

Max 2016-1-30

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/265870-why-do-i-get-the-message-nan

回答： Walter Roberson 2016-2-3

采纳的回答： Walter Roberson

在 MATLAB Online 中打开

Hello,

I would like to calculate the expression below. But I always get: NaN. I can´t find my mistake.

Can somebody help me?

I4=5.718534314;
I5=-0.001546715453;    
Cov__B_C_direct =1125899906842624*exp(I4^(4831891999187463/1125899906842624))*exp(8187485477687969/1715238139330560)*((230090095199403*...
    exp(-I4^(4831891999187463/1125899906842624))*exp(-8187485477687969/1715238139330560)*I4^...
    (3705992092344839/1125899906842624))/1233348233694652400 - (2558136220007576158378628793351*...
    exp(-I4^(4831891999187463/1125899906842624))*exp(-8187485477687969/1715238139330560)*I5*I4^(2580092185502215/1125899906842624))/10681743549394799937933824491520 + ...
    (3335311470258831134378888053767*exp(-I4^(4831891999187463/1125899906842624))*exp(-8187485477687969/1715238139330560)*...
    I5* I4^(3705992092344839/562949953421312))/10681743549394799937933824491520)

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Star Strider 2016-1-30

在 MATLAB Online 中打开

When I converted it into a symbolic by putting:

syms I4 I5

between the assignments to those and your equation, I got:

Error using mupadengine/feval (line 163)
Exponent overflow.

I suspect you’re dividing Inf/Inf, which will evaluate to NaN.

This is an observation, not an Answer, so I’m not posting it as one.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2016-2-3

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/265870-why-do-i-get-the-message-nan#answer_208328

在 MATLAB Online 中打开

When I test in a different software package, I find no problem with the representation, even if I convert the floating point numbers to rationals. I do not find any portion of the expression to be particularly large.

An algebraic equivalent to the expression is

(16191151046901145333338734592/77084264605915775) * I4^(3705992092344839/1125899906842624) - (2558136220007576158378628793351/9487294105343480) * I5 * I4^(2580092185502215/1125899906842624) + (3335311470258831134378888053767/9487294105343480) * I5 * I4^(3705992092344839/562949953421312)

Caution: remember that all of those fractions are going to be evaluated to floating point numbers, losing precision in the process. If you do not want that, then you need to use the Symbolic Toolkit, with code such as

Cov__B_C_direct = sym('1125899906842624') * exp(I4^(sym('4831891999187463') / sym('1125899906842624'))) * exp(sym('8187485477687969') / sym('1715238139330560')) * ((sym('230090095199403') * exp(-I4^(sym('4831891999187463') / sym('1125899906842624'))) * exp(-sym('8187485477687969') / sym('1715238139330560')) * I4^(sym('3705992092344839') / sym('1125899906842624'))) / sym('1233348233694652400') - (sym('2558136220007576158378628793351') * exp(-I4^(sym('4831891999187463') / sym('1125899906842624'))) * exp(-sym('8187485477687969') / sym('1715238139330560')) * I5 * I4^(sym('2580092185502215') / sym('1125899906842624'))) / sym('10681743549394799937933824491520') + (sym('3335311470258831134378888053767') * exp(-I4^(sym('4831891999187463') / sym('1125899906842624'))) * exp(-sym('8187485477687969') / sym('1715238139330560')) * I5 * I4^(sym('3705992092344839') / sym('562949953421312'))) / sym('10681743549394799937933824491520'))

Or in the reduced form,

Cov__B_C_direct = (sym('16191151046901145333338734592') / sym('77084264605915775')) * I4^(sym('3705992092344839') / sym('1125899906842624')) - (sym('2558136220007576158378628793351') / sym('9487294105343480')) * I5 * I4^(sym('2580092185502215') / sym('1125899906842624')) + (sym('3335311470258831134378888053767') / sym('9487294105343480')) * I5 * I4^(sym('3705992092344839') / sym('562949953421312'))

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Answer 2

Abhisek Roy 2016-2-3

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/265870-why-do-i-get-the-message-nan#answer_208319

Hi Max,

It looks like some part of the expression reaches the maximum representable value for a double precision floating point number (which you can find using the function 'realmax'), it overflows, so it is represented as 'inf'. At this point, the numerical result no longer correctly represents the value of the function.

As a workaround for this type of issue try to re-express the function in a different form which does not contain quantities that overflow. In this case, simplify it if possible.