Question

a. Find the first four nonzero terms of the Taylor series centered at 0 for the given finction. b. Use the first four terms of the series to approximate the given quantity. $$f(x)=\sqrt[4]{1+x} ; \text { approximate } \sqrt[4]{1.12}$$

Short answer

Question: Approximate the value of $$\sqrt[4]{1.12}$$ using the first four nonzero terms of the Taylor series centered at 0 for the function $$f(x) = \sqrt[4]{1+x}$$. Answer: The approximation of $$\sqrt[4]{1.12}$$ using the first four terms of the Taylor series is 1.028846.

Step-by-step solution

Finding the derivatives of the function

The given function is $$f(x) = (1+x)^{\frac{1}{4}}$$. We will find the first four derivatives of the function and evaluate them at 0. First derivative: $$f'(x) =\frac{1}{4}(1+x)^{-\frac{3}{4}}$$ At x=0, $$f'(0) = \frac{1}{4}$$. Second derivative: $$f''(x) =-\frac{3}{16}(1+x)^{-\frac{7}{4}}$$ At x=0, $$f''(0) =-\frac{3}{16}$$. Third derivative: $$f'''(x) =\frac{21}{64}(1+x)^{-\frac{11}{4}}$$ At x=0, $$f'''(0) =\frac{21}{64}$$. Fourth derivative: $$f^{(4)}(x) =-\frac{429}{256}(1+x)^{-\frac{15}{4}}$$ At x=0, $$f^{(4)}(0) =-\frac{429}{256}$$.

Finding the Taylor series

Now that we have found the first four derivatives and their values at 0, we can find the Taylor series of the given function $$f(x) = (1+x)^{\frac{1}{4}}$$ centered at 0. The formula for the Taylor series is given by: $$f(x) = f(0) + f'(0)(x) + \frac{1}{2!}f''(0)(x)^2 + \frac{1}{3!}f'''(0)(x)^3 + \cdots$$ Plug in the values of the derivatives: $$f(x) = 1 + \frac{1}{4}x -\frac{3}{16 \times 2!}x^2 +\frac{21}{64 \times 3!}x^3 + \cdots$$ Simplify the series: $$f(x) = 1 + \frac{1}{4}x -\frac{3}{32}x^2 +\frac{7}{64}x^3 + \cdots$$

Approximating the value of the given expression

We are asked to approximate the value of $$\sqrt[4]{1.12}$$. To do this, we can plug in $$x = 0.12$$ into the first 4 terms of the Taylor series: $$f(0.12) \approx 1 + \frac{1}{4}(0.12) - \frac{3}{32}(0.12)^2 + \frac{7}{64}(0.12)^3$$ Calculate the approximation: $$f(0.12) \approx 1 + 0.03 - 0.00135 + 0.000196$$ $$f(0.12) \approx 1.028846$$ Thus, the approximation of $$\sqrt[4]{1.12}$$ using the first four terms of the Taylor series is approximately 1.028846.

Key concepts

Derivatives

To start using the Taylor series, it's essential to understand derivatives. They are a fundamental tool in calculus that measure how a function changes as its input changes. For each order of the derivative, you are computing a higher level of change.

• The first derivative, noted as \(f'(x)\), gives the rate of change of the function itself.
• The second derivative, \(f''(x)\), tells us how the rate of change is changing. It's often connected to concepts like concavity and acceleration in physics.
• Higher order derivatives, like the third or fourth, provide even more insight into the behavior of the function over an interval.

By evaluating these derivatives at a particular point (usually 0, for simplicity in series expansion), we can gather the necessary information to construct the Taylor series for that function.

Function approximation

The essence of Taylor series lies in its power of function approximation. A Taylor series is used to approximate functions that may be difficult to compute directly.

• It takes a complex function and breaks it down into a series of polynomial terms, which are much simpler to work with.
• The approximation becomes more accurate as more terms are added. For many practical purposes, only a few terms are needed.
• This process is particularly helpful in scenarios where we need quick, efficient calculations without needing the exact answer.

In our example, the Taylor series approximation is used to estimate \(\sqrt[4]{1.12}\).

Series expansion

The Taylor series is a specific type of series expansion, which expresses a function as a sum of terms calculated from the values of its derivatives at a single point.
The general formula for a Taylor series centered at point \(a\) is:\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \dots\]

• Each term of the series has a derivative, evaluated at the center of the series, multiplied by a power of \((x-a)\).
• The factorial in the denominator helps to balance the growth of the series, ensuring convergence.
• In practice, we often center it around \(a = 0\), resulting in a Maclaurin series, a special case of the Taylor series.

For \(1+x\), the expansion becomes simpler and focuses on how functions behave for small \(x\), a common theme in series expansions.

Mathematical approximation

Mathematical approximation involves using models to get values that are close to the exact answer. With the Taylor series, the aim is to find an approximation that is as close to the real function as possible.

• A good approximation maintains high accuracy with as few terms as possible. Thus, efficiency is key.
• For example, using the first four non-zero terms of a Taylor series might provide a sufficient approximation for numerical computations, as we saw with the approximation of \(\sqrt[4]{1.12}\).
• It's significant in areas like physics and engineering, where exact answers are often less important than getting a result that’s "close enough" for practical purposes.

While the exact function gives a precise output, the series offers a close estimation that suffices in many scientific and engineering contexts.