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Topic 2/3
15 Flashcards in this deck.
A power series is an infinite series of the form:
$$ \sum_{n=0}^{\infty} a_n (x - c)^n = a_0 + a_1 (x - c) + a_2 (x - c)^2 + a_3 (x - c)^3 + \cdots $$Here, $a_n$ represents the coefficients of the series, and $c$ is the center of the series. The variable $x$ is the variable of the function being represented.
Power series are used to represent functions in a neighborhood around the center point $c$, allowing for the approximation of functions that may be difficult to express in standard algebraic forms.
For a power series, two critical parameters determine where the series converges:
The radius of convergence can be found using the Ratio Test or the Root Test. These tests determine the values of $x$ for which the series converges.
For example, applying the Ratio Test to the series $\sum_{n=0}^{\infty} a_n (x - c)^n$ involves evaluating:
$$ L = \lim_{n \to \infty} \left| \frac{a_{n+1} (x - c)^{n+1}}{a_n (x - c)^n} \right| = |x - c| \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $$If $L < 1$, the series converges absolutely; if $L > 1$, it diverges. Solving $|x - c| < \frac{1}{L}$ gives the radius of convergence.
Maclaurin and Taylor series are specific types of power series centered at $c = 0$ and $c = a$, respectively.
Maclaurin Series is a Taylor series expansion around $c = 0$. It is expressed as:
$$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n $$Where $f^{(n)}(0)$ is the $n^{th}$ derivative of $f$ evaluated at 0.
Taylor Series generalizes the Maclaurin series by expanding around any point $c = a$:
$$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n $$Taylor series are essential for approximating functions near a specific point, enabling the simplification of complex functions for analysis and computation.
To derive a power series representation of a function, we expand the function into its Taylor or Maclaurin series by calculating its derivatives.
For instance, consider the exponential function $f(x) = e^x$. Its derivatives are all equal to $e^x$, and evaluated at $0$ are $f^{(n)}(0) = 1$ for all $n$. Thus, the Maclaurin series for $e^x$ is:
$$ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$This series converges for all real numbers $x$, with a radius of convergence $R = \infty$.
Power series can be manipulated through various operations:
For example, differentiating the power series for $e^x$ term-by-term gives the same series, maintaining $e^x$:
$$ \frac{d}{dx} e^x = \frac{d}{dx} \left( \sum_{n=0}^{\infty} \frac{x^n}{n!} \right ) = \sum_{n=1}^{\infty} \frac{n x^{n-1}}{n!} = \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x $$Power series have widespread applications in various fields:
For example, in solving the differential equation $y'' - y = 0$, we assume a power series solution and determine the coefficients by substituting back into the equation and equating like terms.
When approximating functions with power series, it is crucial to estimate the error or remainder to understand the approximation's accuracy.
The remainder term in the Taylor series provides an estimate of the error after truncating the series:
$$ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x - a)^{n+1} $$Here, $c$ is a value between $a$ and $x$. By bounding the remainder, we can ensure the approximation meets desired accuracy levels.
Let's explore several functions and their power series representations:
And $\cos(x)$ as:
$$ \cos(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots $$These representations are instrumental in computations and theoretical analyses where exact forms are challenging to handle.
Power series can be converted or transformed to represent different functions. For instance, multiplying two power series corresponds to the convolution of their coefficients.
Consider two power series:
$$ f(x) = \sum_{n=0}^{\infty} a_n x^n \quad \text{and} \quad g(x) = \sum_{n=0}^{\infty} b_n x^n $$The product $f(x) \times g(x)$ is given by:
$$ f(x) \times g(x) = \sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} a_k b_{n-k} \right ) x^n $$This convolution allows for the combination of two power series into a single series, representing the product function.
While primarily discussed in the context of real functions in Calculus BC, power series also play a pivotal role in complex analysis. In the complex plane, power series can represent analytic functions, which are functions that are differentiable at every point within their radius of convergence.
The convergence behavior of power series in the complex plane is more intricate due to the two-dimensional nature of complex numbers. However, the fundamental principles remain similar, with the radius of convergence defining a disk within which the series converges absolutely.
This extension to complex functions opens avenues into advanced mathematical concepts such as Laurent series and residue calculus, further emphasizing the versatility of power series.
To find a power series representation of a given function around a point $c$, follow these steps:
Example: Find the Maclaurin series for $f(x) = \frac{1}{1 - x}$.
Thus, $|x| < 1$, so $R = 1$.
Not all functions have straightforward power series expansions using elementary functions. In such cases, alternative methods like manipulating known series or integrating known expansions can be employed.
Example: Find the power series for $f(x) = \arctan(x)$. Start with the known series for $\frac{1}{1 + x^2}$:
$$ \frac{1}{1 + x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n} \quad \text{for}\ |x| < 1 $$Integrate term-by-term to get:
$$ \arctan(x) = \int \frac{1}{1 + x^2} dx = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n + 1} + C $$Setting $x = 0$ to find $C$, we have $\arctan(0) = 0$, so $C = 0$. Therefore:
$$ \arctan(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n + 1} \quad \text{for}\ |x| < 1 $$This method showcases the flexibility in deriving power series for more complex functions.
Beyond the Ratio and Root Tests, other convergence tests can be applied to power series:
These tests ensure rigorous assessment of where the power series converges, which is crucial for accurate function representation and approximation.
Sometimes, one might need to express a power series in terms of another variable or shift the center of the series. This involves substitution and manipulation of the variable within the series terms.
Example: Given the Maclaurin series for $e^x$, find the series for $e^{2x}$.
Original series:
$$ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $$Substitute $2x$ for $x$:
$$ e^{2x} = \sum_{n=0}^{\infty} \frac{(2x)^n}{n!} = \sum_{n=0}^{\infty} \frac{2^n x^n}{n!} $$Thus, the power series for $e^{2x}$ has coefficients $a_n = \frac{2^n}{n!}$.
Mastery of these techniques enhances problem-solving efficiency and accuracy in calculus, particularly in advanced topics such as infinite series and their applications.
While power series are versatile, representing certain functions can present challenges:
Addressing these challenges involves a solid understanding of convergence properties, efficient computation techniques, and strategic selection of expansion points.
Aspect | Power Series | Taylor Series | Maclaurin Series |
Definition | An infinite series of the form $\sum_{n=0}^{\infty} a_n (x - c)^n$ | A power series centered at $c = a$, representing a function near $a$ | A Taylor series centered at $c = 0$ |
Center | Any point $c$ | Specific point $c = a$ | Fixed at $c = 0$ |
Usage | General function representation around any point | Function expansion at a chosen point for approximation | Particular case of Taylor series for ease of computation |
Examples | $\sum_{n=0}^{\infty} a_n (x - c)^n$ | $\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n$ | $\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$ |
Convergence | Depends on $a_n$ and $c$; requires determining radius | Determined by testing around point $a$ | Determined by testing around point $0$ |
Flexibility | Can represent any analytic function within radius | Focused around specific expansion point | Limited to expansions around zero, but simplifies computations |
Tip 1: Always start by identifying the center of your power series expansion. Whether it's a Maclaurin (center at 0) or a Taylor series (center at $a$), this determines your approach.
Tip 2: Memorize common power series expansions like those for $e^x$, $\sin(x)$, and $\ln(1+x)$ to save time during exams.
Tip 3: Use the Ratio Test efficiently by simplifying the limit expression early to quickly find the radius of convergence.
Power series aren't just theoretical tools—they have practical applications in technology and science. For instance, the algorithms behind smartphone cameras use power series to process images efficiently. Additionally, NASA employs power series in trajectory calculations for space missions, ensuring precise navigation through the complexities of space travel.
Moreover, the discovery of the natural exponential function's power series by mathematician Brook Taylor in the 18th century revolutionized mathematical analysis, paving the way for modern calculus and engineering innovations.
Mistake 1: Forgetting to determine the radius of convergence, leading to incorrect application of the series beyond its valid interval.
Incorrect: Using the Maclaurin series for $e^x$ without checking convergence.
Correct: Noting that the Maclaurin series for $e^x$ converges for all real numbers $x$.
Mistake 2: Miscalculating derivatives when forming Taylor series, resulting in incorrect coefficients.
Incorrect: Wrongly computing $f''(a)$ for a Taylor series expansion.
Correct: Carefully differentiating the function and evaluating each derivative at the center point $a$.