7 Use the idea in the proof of the IVT to estimate root 2 by

7. Use the idea in the proof of the IVT to estimate root 2 by coming up with the first 6 terms of the sequences (an) and (bn) for f: [0,2] - > R defined by f(x) = x^2

Solution

as per intermediate value theorm:

    f(x) is continuous on [a,b],

   and M is any number between f(a) and f(b).

Then

there is at least one number c in [a, b] such that f(c) = M.

Consequently, if

(i) f(x) is continuous on [a, b],

and

(ii) f(a) and f(b) are of opposite signs.

If ck is not the desired root, test if f(ck)f(ak) < 0. If so, set bk+1 = ck and ak+1 = ak.

Otherwise, set ak+1 = ck, bk+1 = bk.

Then there is a root x = c of f(x) = 0 or in [a, b].

Now for f(x)=x².

              f : [0,2] -> R

    Now we must try to find that interval first. This can be done using IVT.

Choose a0 = 0 and b0 = 2.

We now show that both hypotheses of IVT are satisfied for f(x) in [0, 2].

(i) f(x) = x² is continuous on [0,2].

(ii) f(0)f(2) < 0.

Thus, by IVT, there is a root of f(x) = 0 in [0,2].

Here input Data:

          f(x) = x³

             a₀ = 0, b₀ = 2

Solution.

Iteration 1. (k = 0):

c₀ =(a₀ + b₀)/2

      =0+2/2=1

Since f(c0)f(a0) = f(1)f(0) < 0,

Iteration 2. (k = 1):

          c1 =1 + 2 /2 =1.5

Since f(c1)f(a1) > 0, set  a2 = c1, b2 = b1.

Iteration 3. (k = 2):

c2 =1.5+1 / 2

      =1.25

Since f(c2)f(a2) < 0, set b3 = c2, a3 = a2.

Iteration 4. (k = 3):

And c4 = (a3+b3) / 2

             =(1.5+1.25)/2 =1.375

Since f(c4)f(a4) < 0, set b5 = c4, a5 = a4.

Iteration5. (k = 4):

And c5 = (a4+b4) / 2

             =(1.5+1.375)/2 =1.4375

It is clear that the root x = 1.437