V12(:,:,h)​=(((tau2(:​,:,h)-tau2​R(:,:,h)).​^2+((tau3(​:,:,h)- tau3R(:,:,​h)).^2).^0​.5)./7)));

How to fix its brackets.

Answer 1

[Walter Roberson]

You have a number of "zero-width space" characters in your code, such as immediately before the "=" character. You have to fix that.

S = 'V12(:,:,h)​=(((tau2(:​,:,h)-tau2​R(:,:,h)).​^2+((tau3(​:,:,h)- tau3R(:,:,​h)).^2).^0​.5)./7)));' ;
find(S == 8203)

ans = 1×7

11 22 33 44 55 74 85

V12(:,:,h)​=(((tau2(:​,:,h)-tau2​R(:,:,h)).​^2+((tau3(​:,:,h)- tau3R(:,:,​h)).^2).^0​.5)./7)));
   1     0 123    4     3      4     32    34    5     4       5     43   2     1   0??

You have two more ) than you have (

[Namrata Chandel]

Thanks alot for your response sir.0.5 is the whole power of first two terms.

[Walter Roberson]

V12(:,:,h) = sqrt((tau2(:,:,h) - tau2R(:,:,h)).^2 + (tau3(:,:,h) - tau3R(:,:,h)).^2) ./ 7;

[Namrata Chandel]

Please see the attchment sir,how to modify above code.I have to multiply with ni also.

[Walter Roberson]

V = sqrt(sum( n .* (tau - tauR).^2 ) ./ sum(n))

This assumes that tau and n are both vectors with the same orientation. If they are 2D arrays then you might need to be more specific about which dimension you sum() over.

[Namrata Chandel]

Thanks alot sir.