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From: Kyle on 29 Mar 2010 23:25
I cannot figure out why I keep getting this error:

??? Error using ==> fzero
FZERO cannot continue because user supplied function_handle ==> @(Velocity)(0.5*C_D1*p*A_o*(Velocity^2)*((1000^2)/(3600^2)))+F_grav+F_rolling+Force_Car
failed with the error below.

Error using ==> plus
Matrix dimensions must agree.

here is part of my code:

fun=@(t)8*x*(t^2)+Growth+Length+Width
x = fzero(fun,1)

Where growth, length, and width are the result of other equations in my code. I have this code in a for loop to find the value at which this function is zero at 30 different values for a particular parameter that each equation is dependent upon. What am I doing wrong?
From: Matt Fig on 30 Mar 2010 00:03
Seems to work here.

% Data
Growth = 7;
Length = 5;
Width = 3;
x = -1;

% Find new x
fun=@(t)8*x*(t^2)+Growth+Length+Width
x = fzero(fun,1)
x =
1.36930639376292
From: Kyle on 30 Mar 2010 00:13
Well, growth, width, and length are arbitrary equations that are matrices in my actual code. I guess I'm confused on why exactly I'm getting the error and what it means.
From: Kyle on 30 Mar 2010 00:46
I think I found where I my error is. Growth is a 20X1 matrix while length and width are both 12X1 matrices. Any suggestions on how to tackle this?
From: Matt Fig on 30 Mar 2010 00:53
Well, like was mentioned in another of your threads, fun must take scalar inputs and return a scalar.


% Data, based on latest information
Growth = rand(2);
Length = rand(2);
Width = rand(2);
x = -1;
fun=@(t)8*x*(t^2)+Growth+Length+Width


% Run and check if fun takes a scalar (it does) and returns a scalar (it doesn't).
>> fun(4)
ans =
-126.104734437492 -126.555874461904
-126.615771738023 -126.243465573257

So it takes a scalar, but does not return a scalar. You may need to break the function down even further.
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