finding root using bysection method (error)

Question

buxZED 2011-2-23

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1900-finding-root-using-bysection-method-error

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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Paulo Silva 2011-2-23

What's stu_id ?

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Answer 1

Walter Roberson 2011-2-23

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1900-finding-root-using-bysection-method-error#answer_2833

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.

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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 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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