Given `f(x)=root3 (1+3x)` at a=0 and use it to estimate the value of the `root3( 1.03)`?

There are a couple of approaches to get a really good approximation

Method 1 - Binomial Series (Power Series) Expansion:

The first is to use the Binomial series expansion, which is essentially the Taylor Series about `a=0` (ie the Maclaurin series), which is I assume what questioner intended.

So using the Binomial Series we have;

`f(x) = root(3)(1+3x)`
`" " = (1+3x)^(1/3)`
`" " = 1 +(1/3)(3x) + (1/3)(-2/3)(3x)^2/(2!) +`
`" "(1/3)(-2/3)(-5/3)(3x)^3/(3!) + ...`
`" " = 1 + x -x^2+5/3x^3 + ...`

And if we put `x=0.01` then we have:

`f(0.01) = root(3)(1+0.03)`
`" " = root(3)(1.03)`
`" " = 1 + 0.01 -0.01^2+5/3(0.01)^3 + ...`
`" " = 1 + 0.01 -0.0001+5/3(0.000001) + ...`
`" " ~~ 1 + 0.01 -0.0001+0.000001667` (ignoring higher terms)
`" " ~~ 1.009901667`

And so it would be reasonable to conclude that `root(3)(1.03)=1.009902` to 6dp.

And in fact using a calculator we find `root(3)(1.03)=1.00990163...`

Method 2 - Newton-Rhapson

Another approach is to use Newton Rhapson to solve an equation of which `root(3)(1.03)` is a solution. So we require

`\ \ \ \ \ \ \x = root(3)(1.03)`
`:. x^3 = 1.03`
`:. x^3 - 1.03 = 0`

Let `f(x) = x^3 - 1.03` Then our aim is to solve `f(x)=0` in the interval `0 le x le 2`

First let us look at the graphs:
graph{x^3 - 1.03 [-5, 5, -15, 15]}

We can see there is one solution in the interval `0 le x le 2`.

We can find the solution numerically, using Newton-Rhapson method

`\ \ \ \ \ \ \f(x) = x^3 - 1.03`
`:. f'(x) = 3x^2`

The Newton-Rhapson method uses the following iterative sequence

`{ (x_0,=1), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)) ) :}`

Then using excel working to 8dp we can tabulate the iterations as follows:

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 `x=1.009902` to 6dp