# how to decrease the tolerence or accuracy of my results

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Youseff on 14 Feb 2014
Answered: Mike Hosea on 14 Feb 2014
i have a hard integral problem , and i would like to decrease its tolerence , i read about abstol and reltol but ive been unable to use them. the value of the 4371875181445801/9007199254740992 , which is rediculous what i need is to probably 3 decimal places. my code is below only the last coulple of lines are the actual double integration.
TT= T/4; for t=0:T: TT %time is equal to zero since the power is measured , and later on this value might be changed making a 3d pressure dist over time. ask for timeperiod imaginary= sqrt(-1); for qqq= 1 : N u_n= Amplitude*(exp(imaginary*((w*t)+phi_i(qqq)))); %complex excitation of each element
UU= [u_n u_n]; U= [U;UU];
U(1, :)= []; %this will cut out the first row which is random numb and one of the colomns U(:, 2)= []; %%Finding h Width_x= (xmaxfield-xminfield)/(no_grid_x); Width_y= 0; Width_z= (zmaxfield-zminfield)/(no_grid_z); Area= Width_x*Width_z; y_maxtwo= sqrt(((diameter_element/2)^2)-((x-x_element(qqq))^2)) -2;
for ppp= 1: no_grid_x for pppp= 1: no_grid_z %q for numerical value %summation sum(f, i = a..b) tries to find a closed form representation of the sum . syms zz rmn dr_dy dr_dx x y zz= @(x,y) r_a-(r_a cos((asin(((sqrt((x.^2)+(y.^2)))-b)/r_a)))); INT = ((i*density*c1*k)/(2*pi)); %constant in hmn before the integration rmn= @(x,y) sqrt((((x-x_field(ppp)).^2)+((y-y_field).^2))+((zz(x,y) -z_field(pppp)).^2)); dr_dy= @(x,y) ((y)/(((x)^2)+((y)^2))); %dr/dy dr_dx= @(x,y) (x)/(((x)^2)+((y)^2)); %dr/dx %p_mn1= integral2((exp((-imaginary*k*rmn)-phi_i(qqq)/rmn))(sqrt(((dr_dy)^2)+((dr_dx)^2)+1)),x_min,x_max) %p= dblquad(@(x_t,y_t)((exp((-imfunaginary*k*rmn)-phi_i(qqq)/rmn))*(sqrt(((dr_dy)^2)+((dr_dx)^2)+1))), x_min(qqq), x_max(qqq),y_min(qqq), sqrt((((diameter_element/2)^2)-(x_t-x_element)^2))-2 )
%p_mn2 = @(x,y)exp((-imaginary*k*sqrt(((x-x_field(ppp)).^2)+((y-y_field).^2)+((r_a- (r_a cos((asin(((sqrt((x.^2)+(y.^2)))-b)/r_a))))-z_field(pppp)).^2)))-phi_i(qqq)/sqrt(((x-x_field(ppp)).^2)+((y-y_field).^2)+((r_a- (r_a *cos((asin(((sqrt((x.^2)+(y.^2)))-b)/r_a))))-z_field(pppp)).^2)))(sqrt(((y/(((x)^2)+((y)^2)))^2)+((x/(((x)^2)+((y)^2)))^2)+1)); %p_mn1 = integral2(p_mn2, x_min(qqq), x_max(qqq),y_min(qqq), y_maxtwo); %integration q = integral2(fun,xmin,xmax,ymin,ymax) fun(x,y) , first_int= int((exp(((-imaginary*k*rmn(x,y))-phi_i(qqq))/rmn(x,y)))*(sqrt(((dr_dy(x,y))^2)+((dr_dx(x,y))^2)+1)),x, x_min(qqq),x_max(qqq)); second_int= int(first_int, y, y_min(qqq), y_max(qqq)); p_mn = INT*second_int; PMN_N= [p_mn p_mn]; PMN= [PMN ;PMN_N];
end end end end
Jan on 14 Feb 2014

Mike Hosea on 14 Feb 2014
Youseff, I did not try to read your code, but I see integral2() in there, and you mention not knowing how to use reltol and abstol. Check out Loren's blog on the subject here:
Also, there's a follow-up now that might be of some interest as well: