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In Exercises \(21-28\) , find the Taylor series generated by \(f\) at \(x=a\) $$ f(x)=2 x^{3}+x^{2}+3 x-8, \quad a=1 $$

Short Answer

Expert verified
The Taylor series is \( -2 + 11(x-1) + 7(x-1)^2 + 2(x-1)^3 \).

Step by step solution

01

Write down the formula for the Taylor series

The Taylor series for a function \( f(x) \) centered at \( x = a \) is given by \( \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n \). We will use this formula to find the Taylor series for \( f(x) \) around \( x = a = 1 \).
02

Evaluate the function \( f(x) \) at \( x = a \)

Substitute \( x = 1 \) into the function: \[ f(1) = 2(1)^3 + (1)^2 + 3(1) - 8 = 2 + 1 + 3 - 8 = -2 \]. So, \( f(1) = -2 \).
03

Calculate the first derivative \( f'(x) \) and evaluate at \( x = a \)

Find the first derivative: \( f'(x) = 6x^2 + 2x + 3 \). Substitute \( x = 1 \) to get \( f'(1) = 6(1)^2 + 2(1) + 3 = 6 + 2 + 3 = 11 \). Thus, \( f'(1) = 11 \).
04

Calculate the second derivative \( f''(x) \) and evaluate at \( x = a \)

Find the second derivative: \( f''(x) = 12x + 2 \). Calculate \( f''(1) = 12(1) + 2 = 14 \). Hence, \( f''(1) = 14 \).
05

Calculate the third derivative \( f'''(x) \) and evaluate at \( x = a \)

The third derivative is \( f'''(x) = 12 \) (since it is a constant). So, \( f'''(1) = 12 \).
06

Substitute into the Taylor series formula and compute terms

The Taylor series expansion of \( f(x) \) around \( x = 1 \) is: \[ f(x) \approx f(1) + f'(1)(x-1) + \frac{f''(1)}{2!} (x-1)^2 + \frac{f'''(1)}{3!} (x-1)^3 \].Substituting the values:\[ -2 + 11(x-1) + \frac{14}{2} (x-1)^2 + \frac{12}{6} (x-1)^3 = -2 + 11(x-1) + 7(x-1)^2 + 2(x-1)^3 \].

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Key Concepts

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

Polynomial Functions
Polynomial functions are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. In a single-variable polynomial, the function is typically expressed as a sequence of terms like \(a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0\), where each term has a coefficient \(a_i\) and a non-negative integer power of \(x\), known as the degree. The highest exponent in the polynomial dictates the degree of the polynomial. Here are some important points about polynomial functions:

  • The degree of a polynomial gives us an idea about the shape and end behavior of its graph.
  • Polynomial functions are smooth and continuous, meaning there are no sharp turns or holes in their graphs.
  • The coefficients of a polynomial can be any real number, and these numbers impact the position and orientation of the graph.
In the given exercise, the function \(f(x) = 2x^3 + x^2 + 3x - 8\) is a cubic polynomial, as the highest exponent is 3. Understanding the behavior of such a function requires looking at its derivatives and how they influence its series expansion.
Derivatives
Derivatives represent the rate of change of a function concerning its variable. For a polynomial function, each term of the function is differentiated according to the power rule. Here's how derivatives influence the behavior of polynomial functions:

  • The first derivative, \(f'(x)\), provides information about the slope of the tangent to the curve at any point and helps in identifying increasing or decreasing behavior.
  • The second derivative, \(f''(x)\), informs us about the concavity of the graph, indicating whether the graph is curving upwards or downwards at a given point.
  • Higher-order derivatives become progressively more straightforward for polynomials due to reducing powers and eventually reaching constant values.
In the Taylor series, derivatives are utilized to estimate how a function behaves around a particular point. For the exercise given, we computed several derivatives of the polynomial, \(f'(x) = 6x^2 + 2x + 3\), \(f''(x) = 12x + 2\), and \(f'''(x) = 12\). These calculations are crucial to finding the coefficients of the Taylor series expansion.
Series Expansion
Series expansion is a method to represent functions as infinite sums of terms calculated from the function's derivatives at a single point. The Taylor series is a type of series expansion centered on a point \(x = a\), allowing approximation of functions around that point using derivatives.

This is especially useful as it transforms complex functions into manageable polynomial forms that approximate the original function within a certain range. The Taylor series formula is:

\[\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n\]

Here is how each component fits into the picture:
  • \(f^{(n)}(a)\) are the derivatives of the function evaluated at the point \(a\).
  • \((x-a)^n\) represents the power series about the center \(x = a\).
  • \(n!\) is the factorial of \(n\), used to normalize each term of the series.
Returning to the exercise, the Taylor series derived for \(f(x)\) around \(x = 1\) was obtained by substituting the calculated derivatives into this formula. Thus, approximating \(f(x)\) as \(-2 + 11(x-1) + 7(x-1)^2 + 2(x-1)^3\). This polynomial form allows us to approximate \(f(x)\) near \(x = 1\) effectively.

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Most popular questions from this chapter

The Cauchy condensation test says: Let \(\left\\{a_{n}\right\\}\) be a nonincreasing sequence \(\left(a_{n} \geq a_{n+1} \text { for all } n\right)\) of positive terms that converges to \(0 .\) Then \(\sum a_{n}\) converges if and only if \(\sum 2^{n} a_{2 n}\) converges. For example, \(\sum(1 / n)\) diverges because \(\Sigma 2^{n} \cdot\left(1 / 2^{n}\right)=\sum 1\) diverges. Show why the test works.

Euler's constant Graphs like those in Figure 11.8 suggest that as \(n\) increases there is little change in the difference between the sum $$1+\frac{1}{2}+\cdots+\frac{1}{n}$$ and the integral $$\ln n=\int_{1}^{n} \frac{1}{x} d x$$ To explore this idea, carry out the following steps. a. By taking \(f(x)=1 / x\) in the proof of Theorem 9 , show that $$\ln (n+1) \leq 1+\frac{1}{2}+\cdots+\frac{1}{n} \leq 1+\ln n$$ or $$0<\ln (n+1)-\ln n \leq 1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \leq 1$$ Thus, the sequence $$ a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n $$ is bounded from below and from above. b. Show that $$ \frac{1}{n+1}<\int_{n}^{n+1} \frac{1}{x} d x=\ln (n+1)-\ln n $$ and use this result to show that the sequence \(\left\\{a_{n}\right\\}\) in part (a) is decreasing. since a decreasing sequence that is bounded from below converges (Exercise 107 in Section 11.1\()\) , the numbers \(a_{n}\) defined in part (a) converge: $$1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \rightarrow \gamma$$ The number \(\gamma,\) whose value is \(0.5772 \ldots,\) is called Euler's constant. In contrast to other special numbers like \(\pi\) and \(e,\) no other expression with a simple law of formulation has ever been found for \(\gamma .\)

Use a CAS to perform the following steps for the sequences in Exercises \(129-140 .\) a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=\left(1+\frac{0.5}{n}\right)^{n} $$

Use a CAS to perform the following steps for the sequences in Exercises \(129-140 .\) a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=\frac{n^{41}}{19^{n}} $$

In Exercises \(97-100,\) determine if the sequence is nondecreasing and if it is bounded from above. $$ a_{n}=\frac{2^{n} 3^{n}}{n !} $$

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