Question

x= 2sin x, solve in false-position method
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Answer 1

We want to solve:

   2sinx=2−x

Let:

   f(x)=2sinx+x−2

It is important to understand that we cannot use Newton's Method to determine the existence of a root, or establish the number of roots of f(x)=0 in any particular interval. A graphical approach is recommended initially to establish this:

graph{2sinx + x -2 [-10, 10, -10, 10]}

From which we can be confident that there is a single root, α where 0<α<2. Our aim is to solve f(x)=0. To find the solution numerically, using Newton-Rhapson method we use the following iterative sequence

   {x1=x0xn+1=xn−f(xn)f'(xn)n>1

Therefore we need the derivative:

         f(x)=2sinx+x−2

   ∴f'(x)=2cosx+1

So our iterative formula is:

   {x1=x0xn+1=xn−2sinx+x−22cosx+1x>1

Then using excel working to 8dp with x0=1, we can tabulate the iterations as follows:

Steve M using Microsoft ExcelSteve M using Microsoft Excel

We get convergence to 6 dp with 6 iterations.

We could equally use a modern scientific graphing calculator as most new calculators have an "Ans" button that allows the last calculated result to be used as the input of an iterated expression.

And we conclude that the solution is (to 6dp):

   x=0.704577