how to write a bisection loop?

y = (-0.50598*10^-10)*x.^3 + (0.38292*10^-7)*x.^2 + (0.74363*10^-4)*x + 0.88318*10^-2

and I am trying to use bisection to find three point for this function, here is what I did.

For bisection one: xl = -1000 and xu = -500, xm = (-1000+(-500))/2 = -750

If xl*xu < 0
Xu=xm
Xm=(xl+xu)/2
error = ((xm-xu)/xm)*100
end
If xl*xu >0
xl = xm
Xm=(xl+xu)/2
error = ((xm-xu)/xm)*100
end
 
If error < 5
end

this is what I plan to do in word I dont know how to make a funtion to start it.

[wenchong chen]

I tried to put them in matlab but nothing happened

[Jan]

"find three point for this function" - which 3 points?

What do you expect to happen for your code?

[wenchong chen]

I am tring do bisection between -1000 to -500, -500 to 0 and 1500 – 2000.

[wenchong chen]

I want a for loop, for error >5, If xl*xu < 0

Xu=xm

Xm=(xl+xu)/2

error = ((xm-xu)/xm)*100

end

If xl*xu >0

xl = xm

Xm=(xl+xu)/2

error = ((xm-xu)/xm)*100

end

[wenchong chen]

I think I need to create a while loop but how and how to use it ?

[wenchong chen]

while error >5
if xl*xu<0
xu = xm
xm=(xl+xu)/2
error = ((xm-xu)/xm)*100
end
 
if xl*xu >0
xl = xm
Xm=(xl+xu)/2
error = ((xm-xu)/xm)*100
end
 
end

here is my new code it is runing but not working as I think

[Jan]

Please use the tools for formatting code. This improves the readability.

It is still not clear, what you want to achieve. "bisection between -1000 to -500, -500 to 0 and 1500 – 2000" does not explain this, because you can divide anything into anything. I guess, you want to find a minimum or zero value?

Your code contains an "Xm" adn "xm" with different uppercase X.

If you check for "if xl*xu<0" you need an "else" to handle the opposite case. Changing the value of xu and checking again "if xl*xu>0" is not suffcient.

The case that "xl*xu==0" is not considered.

Most likely I assume you want to create a function, which is evaluated at the points xl and xu. Comparing the x values is not meaningful. You can find many bisection codes in Matlab, if you ask an internet search engine.

Answer 1

[Matt J]

Because y is a polynomial, it is much more expedient to use roots(),

y = [(-0.50598*10^-10) , (0.38292*10^-7) , (0.74363*10^-4), 0.88318*10^-2];

r=roots(y)

r = 3×1

1.6884 -0.8029 -0.1288

fplot(@(x)polyval(y,x)); hold on
plot(r,0*r,'*'); hold off
xlim([min(r), max(r)])