The error message "DOUBLE cannot convert the input expression into a double array"

 syms x i z;
 offset=3;
 n=10;
 lambda=1/(10.^((20 - (128.1+37.6*log10(150.*10^-3)))./10));
 beta=10^(offset/10);
 fx=(lambda^n)*(x^(n-1))*exp(-lambda*x)/factorial(n-1);
 P_cond=1-symsum(1/factorial(i)*exp(-lambda*x/beta)*(lambda*x/beta)^i,i,0,n-1);
 P_HO=symsum(factorial(8)/(factorial(i)*factorial(8-i))*(P_cond^i)*(1-P_cond)^(8-i),i,5,8);
 P_HOt=double(int(fx*P_HO,x,0,inf))

What i like to calculate is last term of above code.

But there is some strange error.

(Error:DOUBLE cannot convert the input expression into a double array If the input expression contains a symbolic variable, use VPA. 오류 발생: sym/double (line 588) Xstr = mupadmex('symobj::double', S.s,0);)

When i set 'lambda = 1', above code doesn`t show me error.

But when purely run above code, error occurs.

I don`t know what is wrong.

Thank you!

Answer 1

[Walter Roberson]

When that happens, the meaning is that it was not able to find a numeric integral for the expression.

When I plot the expression, I find that it is all fairly small (or extremely small) values, except between about 1.3E-7 and 9.0E-7, with the peak at about 3.5559578143593540162E-7 where it reaches about 85885.334826. It is essentially not worth integrating outside that range. The overall integral marginally exceeds 0.0106219

You could use matlabFunction() to convert the symbolic expression to a function handle, and then use integral() on that. However, if you integrate much further out than 0 to 0.01 then it will not notice the hump near 1E-7 and will give you a totally inaccurate answer.

[h kim]

Finally i find that above integral`s success depends on scale of lambda.

And above integral result is independent to lambda value.