Question

Compute the coefficients for the Taylor series for the following functions about the given point \(a\), and then use the first four terms of the series to approximate the given number. $$f(x)=\sqrt[4]{x} \text { with } a=16 ; \text { approximate } \sqrt[4]{13}$$.

Short answer

Question: Using the first four terms of the Taylor series, approximate the value of \(\sqrt[4]{13}\). Answer: The approximate value of \(\sqrt[4]{13}\) is 1.5444.

Step-by-step solution

Taylor series formula

The Taylor series for a function \(f(x)\) about the point \(a\) is given by: $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$$ where the \(f^{(n)}(a)\) denotes the \(n\)-th derivative of the function \(f(x)\) evaluated at the point \(a\).

Function and its derivatives

To form the Taylor series, we need to find the \(n\)-th derivatives of the given function, \(f(x) = \sqrt[4]{x}\), evaluated at the point \(a=16\). First, let's compute the first few derivatives of \(f(x)\): 1. \(f(x) = x^{\frac{1}{4}}\) 2. \(f'(x) = \frac{1}{4} x^{-\frac{3}{4}}\) 3. \(f''(x) = -\frac{3}{16} x^{-\frac{7}{4}}\) 4. \(f'''(x) = \frac{21}{64} x^{-\frac{11}{4}}\)

Evaluate derivatives at \(a=16\)

Next, we find the value of the first few derivatives at \(a=16\): 1. \(f(16) = 16^{\frac{1}{4}} = 2\) 2. \(f'(16) = \frac{1}{4}16^{-\frac{3}{4}} = \frac{1}{8}\) 3. \(f''(16) = -\frac{3}{16}16^{-\frac{7}{4}} = -\frac{3}{128}\) 4. \(f'''(16) = \frac{21}{64}16^{-\frac{11}{4}} = \frac{21}{2048}\)

Forming the Taylor series

Now we plug these values into the Taylor series formula to create the first several terms: $$f(x) \approx 2 + \frac{1}{8}(x-16) - \frac{3}{128}(x-16)^2 + \frac{21}{2048}(x-16)^3$$

Approximating \(\sqrt[4]{13}\)

Finally, we can use the Taylor series to approximate \(\sqrt[4]{13}\): $$f(13) \approx 2 + \frac{1}{8}(13-16) - \frac{3}{128}(13-16)^2 + \frac{21}{2048}(13-16)^3 = 2 - \frac{3}{8} - \frac{9}{128} + \frac{63}{2048}$$ $$f(13) \approx 1.54443359375$$ So, by using the first four terms of the Taylor series, we have approximated \(\sqrt[4]{13}\) to be approximately 1.5444.

Key concepts

Derivative Evaluation

When working with Taylor series, understanding the role of derivatives is crucial. A derivative measures how a function changes as its input changes. For a Taylor series, we require derivatives because they help articulate the behavior of a function near a certain point. In the context of this problem, the function in question is \(f(x) = \sqrt[4]{x}\) and we are interested in its behavior around \(x = 16\). By calculating the derivatives of this function and evaluating them at the annular point (\(a = 16\)), we catch a glimpse of the function's slope, curvature, and changes in curvature—all of which inform us on how to properly expand a Taylor series.Determining these derivatives is achieved through standard differentiation rules. For example:

• The first derivative, \(f'(x)\), tells us how the function's value changes with a small change in \(x\); here it turned out to be \(\frac{1}{8}\) at \(x = 16\).
• Further derivatives (like \(f''(x)\) and \(f'''(x)\)) provide even more insight, refining our approximation through additional series terms.

These derivatives are then plugged into the Taylor series formula to polish our approximation.

Function Approximation

Taylor series is a powerful tool for approximating complex functions with polynomials. The main idea is to express a function as an infinite sum of its derivatives at a certain point, making computation easier near that point. This is especially useful for evaluating functions at points where direct computation might be tough or cumbersome—for example, attempting to find \(\sqrt[4]{13}\).Using a Taylor series, we achieve a local polynomial approximation that significantly simplifies computation:

• The first term is the function value at the chosen point \(a\), which sets the stage for the approximation.
• Subsequent terms refine the estimate by accounting for the function's behavior, as revealed by its derivatives.

This approximated polynomial closely mirrors the original function over the interval where the calculation is relevant. For our exercise, only the first four terms are used, offering a balance between computational ease and result accuracy.Each additional term brings the series closer to the function, improving approximation precision.

Polynomial Expansion

Polynomial expansion within the Taylor series context is an elegant method of describing complex functions via simpler polynomial terms. When we expand a function like \(\sqrt[4]{x}\) around a point (here, \(a = 16\)), we are breaking down the function into a sum that includes powers of \((x - a)\).This expansion makes it possible to effectively sum an infinite series when evaluating functions values at nearby points, like \(\sqrt[4]{13}\). Here are the vital parts:

• The zeroth term \(f(a)\) represents the simplest polynomial: a constant.
• Higher-order terms involve powers of deviations \((x-a)\) and included increased influence from higher-order derivatives.
• Each term in the expansion progressively refines the approximation, resembling the original function more closely.

By retaining only a few terms (in our scenario, the first four), we strike a balance—gaining a precise approximation without unnecessary computational complexity. This approach makes handling otherwise unwieldy or inaccessible calculations straightforward and accessible, ultimately enabling efficient function evaluation, like approximating \(\sqrt[4]{13}\) with a polynomial instead of directly navigating complex function values.