The function defined by $f(x)=\sin (\pi x)$ has zeroes at every integer.

Question

The function defined by $f(x)=\sin (\pi x)$ has zeroes at every integer. Show that when $-1<a<0$ and $2<b<3$, the bisection method converges to $0$, if $a+b<2$.

How to approach this? Any ideas?

Answer 1

We have $f (a) < 0, f(b) >0 $.

Also, $-1<a<0$ and $2<b<3$, hence $1< a+b<3$.

If $a+b<2$, then $1<a+b<2$. That is $\frac12 < \frac{a+b}2<1$. Hence $f\left( \frac{a+b}2\right)>0.$

Hence after a single step, the interval to consider is $\left(a, \frac{a+b}2\right)$. Note the function is continuous and the interval end point has opposite signs.

You just have to check that there are no other integers besides $0$ in $\left(a, \frac{a+b}2\right)$.