/*! 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 15 Find a power series representati... [FREE SOLUTION] | 91Ó°ÊÓ

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Find a power series representation for the function and determine the radius of convergence. $$ f(x)=\ln (5-x) $$

Short Answer

Expert verified
The power series for \( f(x) = \ln(5-x) \) is \( \ln(5) - \sum_{n=1}^{\infty} \frac{x^n}{5^n n} \) with radius of convergence 5.

Step by step solution

01

Define the Function's Expansion

The function given is \( f(x) = \ln(5-x) \). To find a power series representation, we need to rewrite the argument as \( \ln(a) \) where \( a = 5 \), and express \( (5-x) \) in terms of \((1-u)\) such that \( u = \frac{x}{5} \). Therefore, \( 5-x = 5(1 - \frac{x}{5}) \).
02

Use Taylor Series for \( \ln(1-u) \)

Utilize the Taylor series for \( \ln(1-u) \), which is \(- \sum_{n=1}^{\infty} \frac{u^n}{n} \). Here \( u = \frac{x}{5} \), thus:\[\ln(1 - \frac{x}{5}) = - \sum_{n=1}^{\infty} \frac{(\frac{x}{5})^n}{n}\]
03

Transform to Match \( \ln(5-x) \)

The function \( \ln(5-x) \) can be expressed using the expansion for \( \ln(1-u) \):\[f(x) = \ln(5) + \ln(1 - \frac{x}{5}) = \ln(5) - \sum_{n=1}^{\infty} \frac{(\frac{x}{5})^n}{n}\]Expand to:\[\ln(5) - \sum_{n=1}^{\infty} \frac{x^n}{5^n n}\]
04

Determine Radius of Convergence

The power series \(- \sum_{n=1}^{\infty} \frac{x^n}{5^n n}\) is based on the geometric series with \( u = \frac{x}{5} \).The radius of convergence \( R \) is determined from the condition \(|u| < 1\), which gives us \( |\frac{x}{5}| < 1 \) or \(|x| < 5\). Therefore, the radius of convergence \( R = 5 \).

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Key Concepts

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

Taylor Series
The Taylor series is a powerful tool in calculus, used to approximate intricate functions with infinite sums of polynomial terms. It provides a way to express functions that might be difficult to handle directly, by representing them as an infinite sum based on the function's derivatives at a given point.
The general formula for the Taylor series of a function \( f(x) \) around a point \( a \) is:
  • \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \)
This infinite series provides an exact representation of many types of functions if all orders of derivatives exist at the point \( a \). For the function \( \ln(5-x) \), Taylor series help in transforming the function into a power series representation, allowing us to analyze its behavior and characteristics.
Finding this series involves calculating derivatives, which means it can capture the essence of a function accurately using simpler, polynomial terms that are easier to manage.
Radius of Convergence
The radius of convergence is a crucial concept in the analysis of power series. It represents the interval within which a power series converges to the actual function it represents. Outside this interval, the series may not converge, leading to inaccurate results.
For a power series of the form \( \sum_{n=0}^{\infty} c_n(x-a)^n \), determining convergence involves using the ratio test, and the radius of convergence \( R \) can be derived from the condition:
  • \( \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| = \frac{1}{R} \)
For the exercise, the series for \( \ln(5-x) \) was transformed to match a standard form by manipulation, leading to a power series expansion around a point \( a=5 \):
  • \( - \sum_{n=1}^{\infty} \frac{x^n}{5^n n} \)
By substituting \( u = \frac{x}{5} \), and knowing the general rule that the series converges where \( |u| < 1 \), we derive that \( |x| < 5 \). Hence, the radius of convergence for the series is \( R = 5 \). This ensures convergence within the interval \(-5 < x < 5\).
Logarithmic Functions
Logarithmic functions, represented as \( \ln(x) \) for natural logarithms, are the inverses of exponential functions. They play a vital role in various mathematical contexts because they can transform multiplicative processes into additive ones, making complex calculations more manageable.
In dealing with differential calculus and series, logarithmic functions help simplify expressions, especially for functions like \( \ln(5-x) \). By knowing the series expansion of \( \ln(1-u) \), derived as:
  • \( \ln(1-u) = - \sum_{n=1}^{\infty} \frac{u^n}{n} \)
the function \( \ln(5-x) \) could be expressed in a power series format, greatly simplifying the task of finding the power series representation.
This conversion not only provides easier methods for approximation but also assists in identifying convergence characteristics through the manipulation of the logarithmic terms into a series form. By visualizing these transformations, students can better grasp difficult concepts related to the relationship between exponential and logarithmic operations in calculus.

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Most popular questions from this chapter

After injection of a dose \(D\) of insulin, the concentration of insulin in a patient's system decays exponentially and so it can be written as \(D e^{-a t}\) , where \(t\) represents time in hours and \(a\) is a positive constant. (a) If a dose \(D\) is injected every \(T\) hours, write an expression for the sum of the residual concentrations just before the \((n+1)\) st injection. (b) Determine the limiting pre-injection concentration. (c) If the concentration of insulin must always remain at or above a critical value \(C,\) determine a minimal dosage \(D\) in terms of \(C, a,\) and \(T\)

(a) A sequence \(\left\\{a_{n}\right\\}\) is defined recursively by the equation \(a_{n}=\frac{1}{2}\left(a_{n-1}+a_{n-2}\right)\) for \(n \geqslant 3,\) where \(a_{1}\) and \(a_{2}\) can be any real numbers. Experiment with various values of \(a_{1}\) and \(a_{2}\) and use your calculator to guess the limit of the sequence. (b) Find \(\lim _{n \rightarrow \infty} a_{n}\) in terms of \(a_{1}\) and \(a_{2}\) by expressing \(a_{n+1}-a_{n}\) in terms of \(a_{2}-a_{1}\) and summing a series.

(a) Show that the function defined by $$f(x)=\left\\{\begin{array}{ll}{e^{-1 / x^{2}}} & {\text { if } x \neq 0} \\\ {0} & {\text { if } x=0}\end{array}\right.$$ is not equal to its Maclaurin series. (b) Graph the function in part (a) and comment on its behavior near the origin.

Show that if \(a_{n}>0\) and \(\Sigma a_{n}\) is convergent, then \(\Sigma \ln \left(1+a_{n}\right)\) is convergent.

Show that if \(\lim _{n \rightarrow \infty} \sqrt[n]{\left|c_{n}\right|}=c,\) where \(c \neq 0,\) then the radius of convergence of the power series \(\Sigma c_{n} x^{n}\) is \(R=1 / c\).

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