Intermediate value theorem.
Let be continuous. Suppose for
,
. If c is a real number between
and
, then there is an
between
and
such that
.
Proof.
Define by
. Then, g is a continuous function such that
and
are of opposite signs. The assertion of the theorem amounts to saying that there is a point
between
and
such that
. Without loss of generality, we may take
and
(otherwise replace g by -g). If
, we write
and
; otherwise, write
and
so that we have
,
. Now if
, then
.
If , write
and
, otherwise write
and
, so that we have
and
. We could continue this process and find sequences
,
with
and
and
,
.
Since is a monotonically non-decreasing sequence bounded above, it must converge. Suppose it converges to
. Similarly,
is monotonically non-increasing, bounded below and therefore converges to, say,
. We further note that
as
implying
. Let us call this
. By the continuity of g, we have
, and since
for all n, we must have
and at the same time since
for all n, we must also have
. This implies
. QED.
Corollary.
If f is a continuous function in an interval I and for some
, then there is a point c between a and b for which
. (Exercise).
The above result is often used to locate the roots of equations of the form .
For example, consider the equation: .
Note that whereas
. This shows that the above equation has a root between 0 and 1. Now try with 0.5.
. So there must be a root of the equation between 1 and 0.5. Try
.
, which means that the root is between 0.5 and 0.75. So, we may try
.
. So the root is between 0.75 and 0.625. Now, if we take the approximate root to be 0.6875, then we are away from the exact root at most a distance of 0.0625. If we continue this process further, we shall get better and better approximations to the root of the equation.
Exercise.
Find the cube root of 10 using the above method correct to 4 places of decimal.
More later,
Nalin Pithwa