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Chapter 9: Problem 26

Find the \(n\) th Taylor polynomial centered at \(c .\) $$ f(x)=\frac{2}{x^{2}}, \quad n=4, \quad c=2 $$

Short Answer

Expert verified
The \(4\)th degree Taylor polynomial for the function \(f(x) = \frac{2}{x^2}\) centered at \(c = 2\) is \(P_{4}(x) = 0.5 - 0.5(x - 2) + 0.375(x - 2)^2 - 0.125(x - 2)^3 + 0.625(x - 2)^4\).

Step by step solution

01

Compute the function derivatives at \(c = 2\)

First, compute the first four derivatives of \(f(x) = \frac{2}{x^2}\): \[f^{'}(x)=-\frac{4}{x^{3}}\] \[f^{''}(x)=\frac{12}{x^{4}}\] \[f^{'''}(x)=-\frac{48}{x^{5}}\] \[f^{''''}(x)=\frac{240}{x^{6}}\] Then, evaluate these at \(c=2\): \[f'(2)=-\frac{1}{2}\] \[f''(2)=\frac{3}{4}\] \[f'''(2)=\(-1.5\)\] \[f''''(2)=\frac{15}{4}\]
02

Find Taylor polynomial coefficients

Calculate the Taylor series coefficients by dividing the evaluated derivatives by factorial of the corresponding number. The \(n\) th coefficient is given by \(\frac{f^{(n)}(c)}{n!}\). So, \[a_{0}=\frac{f(2)}{0!}=0.5\] \[a_{1}=\frac{f'(2)}{1!}=-0.5\] \[a_{2}=\frac{f''(2)}{2!}=0.375\] \[a_{3}=\frac{f'''(2)}{3!}=-0.125\] \[a_{4}=\frac{f''''(2)}{4!}=0.625\]
03

Assemble Taylor Polynomial

Finally, construct the \(4\)th degree Taylor polynomial for \(f(x)\) centered at \(c=2\), \( P_{4}(x) = a_0 + a_1 (x - c) + a_2 (x - c)^2 + a_3 (x - c)^3 + a_4 (x - c)^4. Therefore, the Taylor polynomial is, \(P_{4}(x) = 0.5 - 0.5(x - 2) + 0.375(x - 2)^2 - 0.125(x - 2)^3 + 0.625(x - 2)^4\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus
Calculus is an extensive branch of mathematics that deals with change and motion. Its two main types are differential calculus and integral calculus. Differential calculus concerns the rate at which quantities change, while integral calculus is about accumulation of quantities and the tending of values toward a limit.

In the context of Taylor polynomials, calculus helps us understand how a function behaves near a certain point and allows us to create polynomial approximations that closely match the function's behavior around that point. This is particularly useful in many fields like physics, engineering, and economics where exact solutions are difficult to find or unnecessary.
Derivatives
Within calculus, derivatives are a fundamental tool. They measure how a function's output changes as its input changes. The derivative of a function at a certain point is a way of encoding the slope of the function at that point.

To find a Taylor polynomial like in our example, derivatives are essential. We calculate the derivatives of the function at the center point鈥攖he point around which we're building the polynomial. These derivatives tell us how the function is acting in the vicinity of the center point, information we then encode into the polynomial to replicate the function's behavior.
Series Expansion
Series expansion is a way of expressing a function as an infinite sum of terms calculated from the values of its derivatives at a single point. The Taylor series is a type of series expansion that represents a function as a sum of polynomial terms.

The Taylor series is particularly powerful because it offers a way to approximate complex functions with simpler polynomial functions. In our example, we are looking at a series expansion of the function \(f(x) = \frac{2}{x^2}\) around the point \(x = 2\), which allows us to approximate the function near this point using a polynomial.
Factorial
A factorial, denoted with an exclamation point \(n!\), is the product of all positive integers less than or equal to \(n\). It's a concept from combinatorics, but it pops up in calculus, especially in series expansions and Taylor polynomials.

When we construct Taylor polynomials, factorials appear in the denominator of the coefficients. They are used to scale the derivatives appropriately, ensuring that the polynomial mimics the curvature of the original function at various degrees. Factorials grow very fast, and this rapid increase is a key factor why Taylor polynomials provide good approximations even with a relatively low degree.

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Most popular questions from this chapter

Prove, using the definition of the limit of a sequence, that \(\lim _{n \rightarrow \infty} \frac{1}{n^{3}}=0\)

Let \(f(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n+1}}{(2 n+1) !}\) and \(g(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n) !} .\) (a) Find the intervals of convergence of \(f\) and \(g\). (b) Show that \(f^{\prime}(x)=g(x)\). (c) Show that \(g^{\prime}(x)=-f(x)\). (d) Identify the functions \(f\) and \(g\).

\mathrm{\\{} B e s s e l ~ F u n c t i o n ~ T h e ~ B e s s e l ~ f u n c t i o n ~ o f ~ o r d e r ~ 1 is \(J_{1}(x)=x \sum_{k=0}^{\infty} \frac{(-1)^{k} x^{2 k}}{2^{2 k+1} k !(k+1) !}\) (a) Show that the series converges for all \(x\). (b) Show that the series is a solution of the differential equation \(x^{2} J_{1}^{\prime \prime}+x J_{1}^{\prime}+\left(x^{2}-1\right) J_{1}=0 .\) (c) Use a graphing utility to graph the polynomial composed of the first four terms of \(J_{1}\). (d) Show that \(J_{0}^{\prime}(x)=-J_{1}(x)\).

Find all values of \(x\) for which the series converges. For these values of \(x\), write the sum of the series as a function of \(x\). $$ \sum_{n=0}^{\infty}(-1)^{n} x^{2 n} $$

Let \(f(n)\) be the sum of the first \(n\) terms of the sequence 0,1 , \(1,2,2,3,3,4, \ldots\), where the \(n\) th term is given by \(a_{n}=\left\\{\begin{array}{cc}n / 2, & \text { if } n \text { is even } \\\ (n-1) / 2, & \text { if } n \text { is odd }\end{array}\right.\) Show that if \(x\) and \(y\) are positive integers and \(x>y\) then \(x y=f(x+y)-f(x-y)\)
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