How to get a very small value rather than 0?

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Lorenzo
Lorenzo 2012-11-19
Dear all; I'm trying to work out a script which should give a ver small value as a result. The problem is that I keep getting 0 instead…
Any help would be really appreciated.
Here's the script:
rho=7860;
B=-1.00000000011701;
BL=23.562;
Bj=46.9362549800797;
k=1;
int1(k)=(exp(-2*BL(k))*(2*exp(2*BL(k))*sin(2*BL(k))+(-4*exp(3*BL(k))-4*exp(BL(k)))*sin(BL(k))+(4*exp(BL(k))-4*exp(3*BL(k)))*cos(BL(k))+exp(4*BL(k))+8*BL(k)*exp(2*BL(k))-1))/(8*Bj(k));
int2(k)=-(exp(-2*BL(k))*(2*exp(2*BL(k))*sin(2*BL(k))+(4*exp(3*BL(k))+4*exp(BL(k)))*sin(BL(k))+(4*exp(BL(k))-4*exp(3*BL(k)))*cos(BL(k))-exp(4*BL(k))+1))/(8*Bj(k));
int3(k)=-(exp(-BL(k))*((exp(2*BL(k))-1)*sin(BL(k))-exp(BL(k))*((exp(BL(k))+exp(-BL(k)))/2)^2+exp(BL(k))*cos(BL(k))^2))/(2*Bj(k));
m_cap1=rho.*(int1+B.^2.*int2+2.*B.*int3);

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回答（2 个）

Matt J
Matt J 2012-11-19
编辑：Matt J 2012-11-19
You haven't said what variable in your code is the problematic one.
Regardless, computers are inevitably discrete things. If you get too close to 0, then the number will have to get rounded to zero in order to be represented in floating point.
  2 个评论
Matt Fig
Matt Fig 2012-11-21
Lorenzo Lore comments:
Sorry, the variable is m_cap1
Matt J
Matt J 2012-11-21
编辑：Matt J 2012-11-21
If Matt Fig's suggestion doesn't work for you (because you don't have the Symbolic Toolbox), then there is this and
http://www.mathworks.com/matlabcentral/fileexchange/22725-variable-precision-integer-arithmetic
and others like it on the FEX.
However, it would be wise to re-evaluate your approach. It looks very dubious to be working with numbers as huge as
>> exp(4*BL)
ans =
8.5386e+40
and still requiring precision<1.

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Matt Fig
Matt Fig 2012-11-19
Use the symbolic toolbox if you need more precision than double.

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