Qfunction with unreal argument

hi,

i want to calculate the qfunction value qfunc(x) where x is not a real value instead it is a function of x.

can this be calculated in MATLAB?

for eg

sym x

f=(40*log10(x/7))/4

qfunc(f);

i am getting an error saying "The argument of the Q function must be a real array."

Regards,

Karki

Answer 1

[Walter Roberson]

According to the documentation,

Q(x) = 1/2 * erfc(x / sqrt(2))

So for symbolic use I would say

Qfunc = @(x) erfc(x / sqrt(sym(2))) / 2;

then

Qfunc(f)

[Suwas Karki]

now it is generating error as -->

Undefined function or method 'erfc' for input arguments of type 'sym'.

[Walter Roberson]

erfc is documented for symbolic. Which MATLAB version are you using?

[Suwas Karki]

MATLAB 7.12.0(R2011a)

[Walter Roberson]

erfc() was introduced in R2011b. You can use the integral form of it instead.

[Suwas Karki]

 I am done with previous problem but now I got new problem. After series of calculation and simplification I came to following inline function,

f00(L) = sqrt(2.0).*sqrt(pi).*(erf(sqrt(2.0).*sqrt(3.0).*sqrt(1.31e2).*(sqrt((log(L.*(7.837318882787328e15./1.964805411538173e17)).*(-1.31e2./6.075e3))./log(1.0e1)+1.31e2./2.0e2)-1.31e2./1.0e2).*(2.5e1./1.31e2))-1.0).*(6.403354352296725e18./3.602879701896397e18)+7.91e2./1.0e2

here f00 is nothing but an inline function which is a function of L and I have to solve the expression to get the value of L using iteration method like Newton Raphson Method. But when solving it is generating error as

??? Error using ==> inlineeval at 15 Error in inline expression ==> sqrt(2.0).*sqrt(pi).*(erf(sqrt(2.0).*sqrt(3.0).*sqrt(1.31e2).*(sqrt((log(L.*(7.837318882787328e15./1.964805411538173e17)).*(-1.31e2./6.075e3))./log(1.0e1)+1.31e2./2.0e2)-1.31e2./1.0e2).*(2.5e1./1.31e2))-1.0).*(6.403354352296725e18./3.602879701896397e18)+7.91e2./1.0e2

Input must be real and full.

Error in ==> inline.subsref at 27 INLINE_OUT_ = inlineeval(INLINE_INPUTS_, INLINE_OBJ_.inputExpr, INLINE_OBJ_.expr);

I think the parameter inside erf function became complex that's why i am getting this error. If this is the problem then tell me how can i calculate erf of complex number if not then give me any suggestion how can i tackle this problem ? Thanks in advance

[Walter Roberson]

L would have to reach 10^31 for the argument to erf() to become complex.

The solution to L is real valued and positive.

Have you let your test L values become 0? That would involve log(0) ...

[Suwas Karki]

As I told you before I have to calculate L by using Newton Raphson Iteration method. I will write part of code here, and A is the expression which is the function of L. If you want I can provide my full code which is not longer (only 62 lines).

%solving A using Newton Raphson Method

A= (6403354352296725483*2^(1/2)*pi^(1/2)*(erf((25*2^(1/2)*3^(1/2)*131^(1/2)*((131/200 - (131*log((7837318882787328*L)/196480541153817325))/(6075*log(10)))^(1/2) - 131/100))/131) - 1))/3602879701896396800 + 791/100

f00=inline(A)

f01=inline(diff(A))

L=2 %initial guess

for i=1:inf

    y=L
    disp('i = ')
    disp(i)
    L=y-[f00(L)./f01(L)]
    if L==y
        break
    end
end
L=disp(L)

After running I am getting above errors.

[Walter Roberson]

f01 and f00 are both negative near L=2, so you are going to project towards 0. There is a very narrow zone just slightly above 0 where the ratio goes negative, but by the time you get near there you are likely to have projected below 0. And f01 is complex for L negative, leading you to complex L. At that point it seems unlikely you would escape from complex values, as the division of complex numbers seldom will give exact real multiples.

It would be better for your purposes if you could constrain your L to be non-negative.

[Suwas Karki]

How can I constrain L to a non negative value, because I have to find the value for L and I am using iteration method for solving it. Is there any other method or way so that I can solve A for the value of L ? Actually I am in the middle of my thesis and I am stuck here. If you can help me here it would be great help for me.

[Walter Roberson]

Earlier you indicated that you have to use Newton-Raphson; is that not the case? The symbolic toolbox might be able to solve it (note: it might take a bit of time.) The answer is in the range of L = 1E-30

If you do use an iterative method, you could modify it slightly to add

L = max(L, realmin);

after calculating L.

[Suwas Karki]

Actually in my paper which I am taking as base paper for my thesis, the writer has mentioned that we can use various fast and efficient numerical root finding methods such as the Newton Raphson Method, Fixed Point Iteration Method to solve it, so I am thinking about Newton Raphson method but if with Newton Raphson method I am getting error then I think I should change the method. The goal is to find the value for L and I think I can use any method.

And what is this symbolic toolbox is all about how can I use it here ? The expression for A has came from above series of calculation and simplification code now can I use this expression with this symbolic toolbox ?

[Walter Roberson]

The symbolic toolbox is the tool that allows "sym" and "diff".

Try this:

feval(symengine, 'numeric::solve', A, L)

[Suwas Karki]

I used both of your suggestions. After using your first suggestion i.e

L = max(L, realmin)

after calculating L,

I got the answer L= 3.1783e-030

But after using your second suggestion

feval(symengine, 'numeric::solve', A, L)

I got the answer = [ empty sym ]

I think the answer with the first suggestion is right. So I should go with your first suggestion.

Thank you so much !!