How can I use integral to solve a equation with two function handle?

Hello, I want to solve the equation below solution = fzero(@(Cpk_hat) fun2,0.01) but I get the feedback :

'Undefined operator '==' for input arguments of type 'function_handle'.'

Error in fzero (line 314)
    if fx == 0

Does anyone can tell me how to solve the problem?

thanks!

===========================================below is my code ======================================

n = 150; 

w = 1.33; 

p = 0.95;

s = 0.00969;

delta = 0.103; 

alpha = ( n-1 )/2; beta = ( ((n-1)*s^2)./2 )^-1;

b1_of_y = @(y) 3*sqrt(n)*( Cpk_hat*sqrt( 2./((n-1)*y)) - w );

b2_of_y = @(y) 3*sqrt(n)*( (Cpk_hat + 2*delta./3)*sqrt( 2./((n-1)*y)) - w );

fun = @(y,Cpk_hat) ( (1./( gamma(alpha).*(y.^(alpha+1)) ) ).*exp(-1./y).*( normcdf(b1_of_y,0,1))+normcdf(b2_of_y,0,1) - 1);

fun2 = @(Cpk_hat) integral(@(y) fun,0,inf) - p ; 

solution = fzero(@(Cpk_hat) fun2,0.01) ;

Answer 1

[Walter Roberson]

solution = fzero(fun2, 0.01) ;

or

solution = fzero(@(Cpk_hat) fun2(Cpk_hat), 0.01) ;

[Louis Liu]

@Walter Roberson hi, both the answer above will get the feedback:

Do you have any other advices?

_Error using fzero (line 306) FZERO cannot continue because user-supplied function_handle ==> @(Cpk_hat)fun2(Cpk_hat) failed with the error below.

 Input function must return 'double' or 'single' values. Found 'function_handle'._

[Walter Roberson]

n = 150; 

w = 1.33; 
p = 0.95;
s = 0.00969;
delta = 0.103; 

alpha = ( n-1 )/2; beta = ( ((n-1)*s^2)./2 )^-1;
b1_of_y = @(y, Cpk_hat) 3*sqrt(n)*( Cpk_hat*sqrt( 2./((n-1)*y)) - w );
b2_of_y = @(y, Cpk_hat) 3*sqrt(n)*( (Cpk_hat + 2*delta./3)*sqrt( 2./((n-1)*y)) - w );
fun = @(y,Cpk_hat) ( (1./( gamma(alpha).*(y.^(alpha+1)) ) ).*exp(-1./y).*( normcdf(b1_of_y(y, Cpk_hat), 0, 1)) + normcdf(b2_of_y(y, Cpk_hat), 0, 1) - 1);
fun2 = @(Cpk_hat) integral(@(y) fun(y, Cpk_hat), 0, inf) - p ; 

solution = fzero(fun2, 0.01) ;

However, you are not going to get a solution.

If you use the symbolic toolbox to put in y and Cpk_hat values, then you can get to a single formula,

int(-(1/2)*erfc((15/2)*2^(1/2)*6^(1/2)*((1/149)*2^(1/2)*149^(1/2)*(1/Y)^(1/2)*(CPK_HAT+103/1500)-133/100))-(1/38388487458436191731876003581520911053353221075554252123409483883291285488049656907094870486907296398966784)*exp(-1/Y)*((1/2)*erfc((15/2)*2^(1/2)*6^(1/2)*((1/149)*CPK_HAT*2^(1/2)*149^(1/2)*(1/Y)^(1/2)-133/100))-1)/Y^(151/2), Y = 0 .. infinity)-19/20

If you take the function being integrated,

-(1/2)*erfc((15/2)*2^(1/2)*6^(1/2)*((1/149)*2^(1/2)*149^(1/2)*(1/Y)^(1/2)*(CPK_HAT+103/1500)-133/100))-(1/38388487458436191731876003581520911053353221075554252123409483883291285488049656907094870486907296398966784)*exp(-1/Y)*((1/2)*erfc((15/2)*2^(1/2)*6^(1/2)*((1/149)*CPK_HAT*2^(1/2)*149^(1/2)*(1/Y)^(1/2)-133/100))-1)/Y^(151/2)

then limit() as Y approaches infinity is -1. There do appear to be some positive values, but they are within a rather narrow band near Y = 0, like Y < 1/500 or so, and none of them are very large. A minor positive contribution from that area at most. The integral is therefore adding an infinite number of non-zero negative values that get close to -1 very quickly, and the result is therefore going to be -infinity, for any Cpk_hat. There is no zero crossing based upon Cpk_hat

[Louis Liu]

@Walter Robertson

hi, so I couldn't use this method to get the value of Cpk_hat, right?

[Walter Roberson]

Let me put it this way: with that formula, you are attempting to do the equivalent of

solution = fzero(@(Cpk_hat) -infinity, 0.01) ;

The value of Cpk_hat does not matter: whatever numeric value you give to Cpk_hat, the result of the integral is going to be -infinity . It will never equal 0.

[Walter Roberson]

fzero can only solve functions that cross zero. Your function does not cross zero.