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

Question

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

(a) $0$, if $a+b<2$
(b) $2$, if $a+b>2$
(c) $1$, if $a+b=2$

$a$ denotes lower limit & $b$ denotes upper limit of the interval $[a,b]$

How to consider the intervals fulfilling the conditions?

I have thought by the following way:

(a) $1<a+b<2$ . but if i put $f(1)=\sin\pi=0$ and $f(2)=\sin 2\pi=0$

it is not providing the different signs.

I haven't idea to solve the questions.

Answer 1

To help the OP, let me suggest a corrected version of the question:

Consider the function $f$ defined by $f(x)=\sin(\pi x)$, and the bisection method based on $f$ and starting from some points $a$ and $b$ such that $-1<a<0$ and $2<b<3$. Let $c$ denote the root of $f$ this algorithm converges to. Show that $c=0$ if $a+b<2$, $c=2$ if $a+b>2$, and $c=1$ if $a+b=2$.

The first thing I would do to solve this question is to draw a sketch of the graph of $f$ on the interval $(-1,2)$. And it happens that already the first midpoint of the bisection method determines $c$... Hint: this first midpoint is either in the interval $(\frac12,1)$ or equal to $1$ or in the interval $(1,\frac32)$. What happens next, in each case?

Answer 2

one can notice that $a_0$ lies in the blue region while $b_0$ hence we can establish that : $ 1\leq\frac{a_0 + b_0}{2}\leq\frac{3}{2}$ ; it is clear that after the first iteration if $a_0 + b_0 = 2$ then we get a point $c$ with $f(c) = 0$ and iteration will stop and converge to $1$.

if we say that $a+b < 2$ then we will have the point $c$ for the next iteration to the left of $1$ and the iteration will clearly keep on choosing points $c_n < 1$ and so the iteration converges to $0$.

a similar argument can be checked out for the case when $a+b > 2$ in which iteration converges to $2$ as we always stay on the right of point $(1, 0)$