finding root using bysection method (error)

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buxZED
buxZED 2011-2-23
x = 0:.01:1;%used to generate the x values
y=(2.8*x.^3)-(3.5*x.^2)+(1.5*x)-(0.15+(0.1*stu_id));%Provided function
plot(x,y)
xl=0;%lower limit set as 0
xr=1;%upper limit set as 1
xc=(xl+xr)/2;
while abs(y(xc)) > 0.00001
if (y(xc) * y(xr)) < 0
xl = xc
else
xr = xc
end
xc = (xl + xr)/2;
end
fprintf('the root is %g\n' , xc)
tihis is my attempt to find the root between 0 and 1 but the answer comes to be = 2 pleese help me identyfi the error

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Walter Roberson
Walter Roberson 2011-2-23
Define your y as:
y = @(x) (2.8*x.^3)-(3.5*x.^2)+(1.5*x)-(0.15+(0.1*stu_id));
and make your plot
plot(x, y(x));
I'm surprised that the program didn't bomb out on you complaining that indices must be logical or positive integers.
  2 个评论
buxZED
buxZED 2011-2-23
thanks for the answer
can you claryfy the the difference and why we write the function in such format
y=f(x)
y=@x f(x)
Walter Roberson
Walter Roberson 2011-2-23
y = f(x) evaluates f(x) with the current values of x and creates a matrix (or vector) of results, which it stores in y. That matrix (or vector) of results is indexed by integer indices -- y(1) for the first, y(2) for the second, and so on.
y = @(x) f(x)
makes y an anonymous function. You can pass y an array (or vector) of values, and the function f will be evaluated on those values.
For example, with the code you had,
y(pi) would have tried to index the already-calculated vector to find the pi'th element, which would be an error. But with the alternate version, pi would be substituted at run time as x in the expression for f(x), and that particular value would be returned.

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