when i run the following code its showing an error that dimensions must agree

1 次查看(过去 30 天)
w=(2.3.^-u.*log(1.+(0.5615./u)-0.4385*(((1.0421.*u))+(1/(1.+u.^1.5))+(1.0801./(1.+(2.35.*u.^(-1.0919))))))^-2./(0.5616+(0.4385+2.35.*2.3^(-2.283.*u))))
i want to calculate 'w' for a set of 'u' values.i'm confused while arranging '.' when i run this i get "error usin + and matrix dimensions must agree".

采纳的回答

madhan ravi
madhan ravi 2018-12-11
编辑:madhan ravi 2018-12-11
If you still have problem then use bsxfun() with mtimes and rdivide appropriately if your usig version prior to 2016b
w=(2.3.^-u.*log(1.+(0.5615./u)-0.4385.*(((1.0421.*u))+(1/(1.+u.^1.5))+(1.0801./(1.+(2.35.*u.^(-1.0919))))))^-2./(0.5616+(0.4385+2.35.*2.3^(-2.283.*u))))
^---dot missed
  5 个评论
vaishnavi potharaju
vaishnavi potharaju 2018-12-12
编辑:madhan ravi 2018-12-12
sir i tried to run my code with bsxfnc but as it is a very complex equation i couldnt do it.so i split the equation into small parts as follows and now im able to run it.
w=((2.3).^-u.*log(1.+(0.5615./u)-0.4385.*(((1.0421.*u))+(1./(1.+u.^1.5))+(1.0801./(1.+(2.35.*u^(-1.0919)))))).^-2./(0.5616+(0.4385*2.3.^(-2.2803.*u))))
w1=1.+(0.5615./u)
w2=1.0801./(1.+(2.35.*u.^(-1.0919)))
w3=1./(1.+u.^1.5)
w4=1.0421.*u
w5=2.3.^-u
w6=0.4385.*2.3.^(-2.2803.*u)
w7=0.4385.*(w4+w3+w2).^-2
w8=log(1+w1-w7)
w9=w8.*w5
w10=0.5616+w6
wu=w9./w10

请先登录,再进行评论。

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Numeric Types 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by