Understanding the Value 135 and Advanced Taylor Series Approximation Techniques

The problem at hand is to find the value of 135, which is 5 when 135 5 and 555 125. This might seem like a simple problem, but it can be approached using a more sophisticated mathematical technique called the Taylor series approximation. By utilizing this powerful tool, we can approximate the value of the cube root of 130 and other similar problems with higher precision. Let's dive into the details.

Using Taylor Series Expansion for Value Calculation

The Taylor series expansion is a fundamental concept in calculus that allows us to approximate functions around a point. Given a function ( f(x) sqrt[3]{x} ), we can use the Taylor series expansion to find the value of ( f(130) ) using the following formula:

[ f(x) sum_{n0}^{infty} frac{f^{(n)}(a)(x-a)^n}{n!} ]

Step-by-Step Calculation Using Taylor Series

We will start by setting ( a 125 ) and calculating the necessary derivatives of ( f(x) ) at ( a 125 ).

1. First Derivative

The first derivative of ( f(x) sqrt[3]{x} ) is:

[ f'(x) frac{x^{-frac{2}{3}}}{3} ]

Evaluating this at ( x 125 ):

[ f'(125) frac{1}{75} ]

2. Second Derivative

The second derivative of ( f(x) sqrt[3]{x} ) is:

[ f''(x) frac{-2x^{-frac{5}{3}}}{9} ]

Evaluating this at ( x 125 ):

[ f''(125) frac{-2}{28125} ]

3. Focusing on Three Terms

We will now use the first three terms of the Taylor series approximation to find ( f(130) ):

[ f(x) f(a) f'(a)(x-a) frac{f''(a)(x-a)^2}{2!} ]

Plugging in ( x 130 ), ( a 125 ), ( f(125) 5 ), ( f'(125) frac{1}{75} ), and ( f''(125) frac{-2}{28125} ):

[ f(130) 5 frac{1}{75}(130-125) frac{frac{-2}{28125}(130-125)^2}{2} ]

Simplifying this expression:

[ f(130) 5 frac{1}{75}(5) frac{frac{-2}{28125}(25)}{2} ]

This further simplifies to:

[ f(130) 5 frac{5}{75} frac{-25}{2 times 28125} ]

Given that ( frac{5}{75} frac{1}{15} ) and ( frac{25}{56250} frac{1}{2250} ):

[ f(130) 5 frac{1}{15} - frac{1}{2250} ]

To combine these terms, we find a common denominator:

[ f(130) 5 frac{150}{2250} - frac{1}{2250} 5 frac{149}{2250} approx 5 0.0661 5.0661 ]

Hence, the value of ( f(130) ) using the Taylor series expansion is approximately 5.0661.

Conclusion

This detailed approach using the Taylor series expansion provides a precise method to approximate the value of the cube root of 130. By leveraging this technique, we can solve similar problems with high accuracy.

Related Keywords

- Taylor Series

- Approximation

- Value Calculation