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

Question

Aurora Alvarado · Accepted Answer

#x=1.009902# to 6dp

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

Emilia Aguilar · Answer

undefined To estimate the value of $ \sqrt[3]{1.03} $ using the given function $ f(x) = \sqrt[3]{1+3x} $ at $ a=0 $, we can use linear approximation (the tangent line at $ a=0 $). The linear approximation is given by:

$$ f(x) \approx f(a) + f'(a)(x-a) $$

where $ f'(x) $ denotes the derivative of $ f(x) $ with respect to $ x $.

First, find the derivative of $ f(x) $ and evaluate it at $ a=0 $. Then, plug in $ a=0 $ and $ x=0.03 $ into the linear approximation formula to estimate the value of $ \sqrt[3]{1.03} $.