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Chapter 10: Problem 9

For Exercises \(1-9,\) find the first four nonzero terms of the Taylor series for the function about 0. $$\ln (5+2 x)$$

Short Answer

Expert verified
The first four nonzero terms are: \( \ln(5) + \frac{2}{5}x - \frac{2}{25}x^2 + \frac{2}{125}x^3 \).

Step by step solution

01

Understand Taylor Series

The Taylor series for a function \( f(x) \) about 0 is given by: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \] To find the first four nonzero terms, we need to compute the derivatives of \( \ln(5 + 2x) \) at \( x = 0 \).
02

Evaluate Function at Zero

The first term of the Taylor series is \( f(0) \). For \( f(x) = \ln(5 + 2x) \), at \( x = 0 \), we have: \[ f(0) = \ln(5) \].
03

Calculate First Derivative and Evaluate

Find the first derivative of \( \ln(5 + 2x) \) and evaluate at \( x = 0 \). \[ f'(x) = \frac{2}{5 + 2x} \] Evaluating at \( x = 0 \), we get: \[ f'(0) = \frac{2}{5} \].
04

Calculate Second Derivative and Evaluate

Next, find the second derivative and evaluate at \( x = 0 \). \[ f''(x) = -\frac{4}{(5 + 2x)^2} \] Therefore, \( f''(0) = -\frac{4}{25} \).
05

Calculate Third Derivative and Evaluate

Calculate the third derivative and evaluate at \( x = 0 \). \[ f'''(x) = \frac{16}{(5 + 2x)^3} \] At \( x = 0 \), \( f'''(0) = \frac{16}{125} \).
06

Write First Four Nonzero Terms

Substitute the computed derivatives into the Taylor series formula: \[ \ln(5 + 2x) \approx \ln(5) + \left(\frac{2}{5}\right)x + \left(-\frac{4}{25}\right)\frac{x^2}{2} + \left(\frac{16}{125}\right)\frac{x^3}{6} \] Simplifying, the first four nonzero terms are: \[ \ln(5) + \frac{2}{5}x - \frac{2}{25}x^2 + \frac{2}{125}x^3 \].

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Key Concepts

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

Derivatives
In mathematics, derivatives play a fundamental role. They measure how a function changes as its input changes. Think of it as the slope of a line that touches a point on a curve. In the context of Taylor series, derivatives help us determine the coefficients of the series. For a function like \( \ln(5 + 2x) \), its derivatives at a specific point (like \( x = 0 \)) are crucial for constructing the series.
  • The first derivative tells us the slope of the tangent line at the point.
  • Higher-order derivatives give the curvature and help define the shape of the polynomial.
This process involves calculating successive derivatives, such as:
  • First derivative: \( f'(x) = \frac{2}{5 + 2x} \)
  • Second derivative: \( f''(x) = -\frac{4}{(5 + 2x)^2} \)
  • Third derivative: \( f'''(x) = \frac{16}{(5 + 2x)^3} \)
By evaluating these at \( x = 0 \), we determine the coefficients that multiply each term in the Taylor series.
ln Function
The linear function, often written as \( \ln(x) \), represents the natural logarithm. This function is unique because it describes the time needed to reach a certain level of growth. In calculus, the ln function is often involved in problems related to growth or decay. For example, \( \ln(5+2x) \) implies a shift in the logarithmic function by a constant and a factor of 2. When working with Taylor series, we aim to approximate \( \ln \) functions using polynomials.
  • The natural logarithm \( \ln(5) \) is the starting point of our polynomial approximation.
  • The derivatives of \( \ln(5+2x) \) help us understand how this function behaves around \( x = 0 \).
This makes it possible to express \( \ln(5+2x) \) as a series of polynomial terms.
Polynomial Approximation
Polynomial approximation is a powerful technique in calculus. It involves representing a complex function with simpler polynomial terms. The Taylor series is one such method that helps us approximate functions around a specific point, often zero. By using the derivatives of a function, we construct a polynomial that closely follows the original function near the chosen point.
  • The first term \( \ln(5) \) is the constant term that represents the value of the function at \( x = 0 \).
  • Each subsequent term involves derivatives evaluated at zero, multiplied by \( x \) raised to increasing powers.
  • This gives us an expression: \( \ln(5+2x) \approx \ln(5) + \frac{2}{5}x - \frac{2}{25}x^2 + \frac{2}{125}x^3 \)
Through polynomial approximation, complex functions can be made more understandable and easier to work with, especially when calculating values around a particular point.

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Most popular questions from this chapter

Explain what is wrong with the statement. If \(f(x)=\ln (2+x),\) then the second-degree Taylor polynomial approximating \(f(x)\) near \(x=0\) has a negative constant term.

Use the third degree Taylor polynomial \(P_{3}(x)=4+2(x-1)-2(x-1)^{2}+(x-1)^{3} / 2\) of \(f(x)\) about \(x=1\) to find the given value, or explain why you can't. $$f^{\prime \prime \prime}(1)$$

The theory of relativity predicts that when an object moves at speeds close to the speed of light, the object appears heavier. The apparent, or relativistic, mass, \(m\) of the object when it is moving at speed \(v\) is given by the formula $$m=\frac{m_{0}}{\sqrt{1-v^{2} / c^{2}}}$$, where \(c\) is the speed of light and \(m_{0}\) is the mass of the object when it is at rest. (a) Use the formula for \(m\) to decide what values of \(v\) are possible. (b) Sketch a rough graph of \(m\) against \(v,\) labeling intercepts and asymptotes. (c) Write the first three nonzero terms of the Taylor series for \(m\) in terms of \(v\) (d) For what values of \(v\) do you expect the series to converge?

Give an example of: Two different functions having the same Taylor polynomial of degree 1 about \(x=0\)

(a) Find and multiply the Taylor polynomials of degree 1 near \(x=0\) for the two functions \(f(x)=\) \(1 /(1-x)\) and \(g(x)=1 /(1-2 x)\) (b) Find the Taylor polynomial of degree 2 near \(x=0\) for the function \(h(x)=f(x) g(x)\) (c) Is the product of the Taylor polynomials for \(f(x)\) and \(g(x)\) equal to the Taylor polynomial for the function \(h(x) ?\)
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