/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Use power series operations to f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use power series operations to find the Taylor series at \(x=0\) for the functions in Exercises \(11-28 .\) $$x \ln (1+2 x)$$

Short Answer

Expert verified
The Taylor series is \(2x^2 - 2x^3 + \frac{8x^4}{3} - \cdots\).

Step by step solution

01

Recall the Taylor Series Formula

The Taylor series for a function \(f(x)\) centered at \(x = 0\) is given by \(\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\), where \(f^{(n)}(0)\) is the \(n\)-th derivative of the function evaluated at \(x=0\).
02

Write the Series for the Inner Function

For the function \(\ln(1+2x)\), the Taylor series at \(x=0\) is \[\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} (2x)^n = 2x - \frac{(2x)^2}{2} + \frac{(2x)^3}{3} - \cdots\].
03

Multiply by \(x\)

Since the function is \(x \ln(1+2x)\), multiply each term in the series by \(x\): \[x \cdot \left(2x - \frac{(2x)^2}{2} + \frac{(2x)^3}{3} - \cdots\right).\]
04

Simplify the Series Terms

Simplify the series: \[2x^2 - \frac{4x^3}{2} + \frac{8x^4}{3} - \cdots\].
05

Find Terms of the Series

Term by term, simplify: \[2x^2 - 2x^3 + \frac{8x^4}{3} - \cdots\]. Continue this process to obtain more terms if needed.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Power Series
A power series is an infinite series of the form \( \sum_{n=0}^{\infty} a_n x^n \), where \( a_n \) is a constant coefficient for each term, and \( x \) is a variable raised to the power \( n \). This type of series is foundational in calculus and analysis because it allows us to express functions as infinite sums. Power series are widely used because they simplify difficult functions and make calculations more manageable.

In the context of the exercise, a power series represents the function \( \ln(1+2x) \). Each term in the power series for \( \ln(1+2x) \) is formulated using powers of \( 2x \). By expanding this function in terms of power series, we can easily manipulate it to find the Taylor series, which is power series approximation at specific points.
  • Make sure to check the radius of convergence for your power series using the formula or ratio test to ensure it converges, especially when substituting values into \( x \).
  • A power series can be differentiated or integrated term by term within its interval of convergence, making it an adaptable tool.
Derivatives
Derivatives represent how a function changes as its input changes. They are at the heart of finding Taylor series because each term of the Taylor series is based on the value of the function's derivative at a certain point. The general formula for a Taylor series includes the term \( \frac{f^{(n)}(0)}{n!} x^n \), where \( f^{(n)}(0) \) is the \( n \)-th derivative evaluated at \( x = 0 \). This derivative calculation determines the coefficient for each term in the series.

In the original step-by-step solution, the series \( \ln(1+2x) \) is built upon derivatives of the function evaluated at \( x=0 \). This is why calculating derivatives correctly is crucial. By understanding derivatives and their role in a Taylor series, you grasp how an infinite series can approximate functions through incremental, small changes in function values.
  • Remember that your first derivative indicates slope, while higher derivatives involve curvature and changes in curvature.
  • Being able to calculate and use derivatives effectively helps open up a wide array of mathematical concepts and applications.
Series Expansion
Series expansion refers to expressing a function as an infinite sum of terms using a series, such as a Taylor series. This expansion is highly useful, as it allows complicated functions to be represented in simpler, more computationally friendly terms. In a Taylor series, each term \( \frac{f^{(n)}(0)}{n!} x^n \) represents an incremental approximation.

In the exercise, the series expansion derives from multiplying each term of the power series \( \ln(1+2x) \) by \( x \), yielding \( x \cdot \left(2x - \frac{(2x)^2}{2} + \frac{(2x)^3}{3} - \cdots\right) \). This step illustrates how functions can be built incrementally from basic polynomials.
  • Each additional term in the series provides a better approximation for the original function over its interval of convergence.
  • After expanding, simplifying each term can make the calculations easier and clearer, so try to simplify halfway to make sense of calculations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that \(\sum_{n=2}^{\infty}\left((\ln n)^{q} / n^{p}\right)\) converges for \(-\infty < q <\infty\) and \(p >1.\) (Hint: Limit Comparison with \(\sum_{n=2}^{\infty} 1 / n^{r}\) for \(1 < r < p.)\)

In the alternating harmonic series, suppose the goal is to arrange the terms to get a new series that converges to \(-1 / 2 .\) Start the new arrangement with the first negative term, which is \(-1 / 2 .\) Whenever you have a sum that is less than or equal to \(-1 / 2,\) start introducing positive terms, taken in order, until the new total is greater than \(-1 / 2 .\) Then add negative terms until the total is less than or equal to \(-1 / 2\) again. Continue this process until your partial sums have been above the target at least three times and finish at or below it. If \(s_{n}\) is the sum of the first \(n\) terms of your new series, plot the points \(\left(n, s_{n}\right)\) to illustrate how the sums are behaving.

Determine how many terms should be used to estimate the sum of the entire series with an error of less than \(0.001 .\) $$ \sum_{n=1}^{\infty}(-1)^{n} \frac{1}{\ln (\ln (n+2))} $$

The (second) second derivative test Use the equation $$f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}\left(c_{2}\right)}{2}(x-a)^{2}$$ to establish the following test. \begin{equation} \begin{array}{l}{\text { Let } f \text { have continuous first and second derivatives and }} \\ {\text { suppose that } f^{\prime}(a)=0 . \text { Then }} \\ {\text { a. } f \text { has a local maximum at } a \text { if } f^{\prime \prime} \leq 0 \text { throughout an interval }} \\ {\text { whose interior contains } a ;} \\ {\text { b. } f \text { has a local minimum at } a \text { if } f^{\prime \prime} \geq 0 \text { throughout an interval }} \\\ {\text { whose interior contains a. }}\end{array} \end{equation}

The limit \(L\) of an alternating series that satisfies the conditions of Theorem 15 lies between the values of any two consecutive partial sums. This suggests using the average $$\frac{s_{n}+s_{n+1}}{2}=s_{n}+\frac{1}{2}(-1)^{n+2} a_{n+1}$$ to estimate \(L .\) Compute $$s_{20}+\frac{1}{2} \cdot \frac{1}{21}$$ as an approximation to the sum of the alternating harmonic series. The exact sum is \(\ln 2=0.69314718 \ldots .\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.