Taylor Theorem and Power Series
Introduction to Taylor Theorem
The Taylor theorem is a fundamental result in calculus that provides a way to approximate a function by a polynomial. This polynomial approximation becomes increasingly accurate as we include more terms. Essentially, it allows us to express a function as an infinite sum of terms, calculated from the values of its derivatives at a single point.
This theorem is incredibly powerful because it can be used to approximate complex functions with simpler polynomial expressions, which are much easier to work with for analysis, computation, and understanding the local behavior of the function. It forms the basis for many numerical methods and approximations used in science and engineering.
The Taylor Polynomial
Let $f(x)$ be a function that is $n$ times differentiable at a point $x = a$. The Taylor polynomial of degree $n$ for $f(x)$ centered at $a$ is given by:
$P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \dots + \frac{f^{(n)}(a)}{n!}(x-a)^n$
This can be written more compactly using summation notation:
$P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k$
Here, $f^{(k)}(a)$ denotes the $k$-th derivative of $f$ evaluated at $a$, and $k!$ is the factorial of $k$. We define $f^{(0)}(a) = f(a)$ and $0! = 1$.
Maclaurin Series: A Special Case
A Maclaurin series is a Taylor series centered at $a=0$. If a function $f(x)$ has derivatives of all orders at $x=0$, its Maclaurin series is:
$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}x^k$
The Taylor Theorem with Remainder
While the Taylor polynomial approximates the function, it's important to understand the error introduced by this approximation. The Taylor theorem with remainder quantifies this error. If $f(x)$ has derivatives up to order $n+1$ on an interval containing $a$ and $x$, then there exists a number $c$ between $a$ and $x$ such that:
$f(x) = P_n(x) + R_n(x)$
Where $R_n(x)$ is the remainder term. The most common form of the remainder is the Lagrange form:
$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$
This form tells us that the error in approximating $f(x)$ by $P_n(x)$ is similar in form to the next term in the Taylor series, but evaluated at some unknown point $c$ between $a$ and $x$.
Example: Approximating $e^x$
Let's find the Taylor polynomial of degree 3 for $f(x) = e^x$ centered at $a=0$ (Maclaurin polynomial).
First, find the derivatives of $f(x) = e^x$:
- $f(x) = e^x \implies f(0) = e^0 = 1$
- $f'(x) = e^x \implies f'(0) = e^0 = 1$
- $f''(x) = e^x \implies f''(0) = e^0 = 1$
- $f'''(x) = e^x \implies f'''(0) = e^0 = 1$
Now, plug these values into the Maclaurin polynomial formula:
$P_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3$
$P_3(x) = 1 + 1 \cdot x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3$
$P_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$
This polynomial is a good approximation of $e^x$ for values of $x$ close to 0. The remainder term $R_3(x)$ would be $\frac{f^{(4)}(c)}{4!}x^4 = \frac{e^c}{24}x^4$ for some $c$ between 0 and $x$. As $n$ increases, $P_n(x)$ gets closer to $e^x$.
Taylor Series
If the remainder term $R_n(x)$ approaches 0 as $n \to \infty$ for a given $x$, then the infinite Taylor series converges to the function $f(x)$ at that point:
$f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}(x-a)^k$
This infinite series is called the Taylor series of $f(x)$ centered at $a$. If $a=0$, it's the Maclaurin series.
Important Maclaurin Series
Memorizing these common Maclaurin series is crucial for quickly solving problems involving power series and Taylor expansions.
| Function $f(x)$ | Maclaurin Series | Interval of Convergence |
|---|---|---|
| $e^x$ | $1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots = \sum_{k=0}^{\infty} \frac{x^k}{k!}$ | $(-\infty, \infty)$ |
| $\sin(x)$ | $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}$ | $(-\infty, \infty)$ |
| $\cos(x)$ | $1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!}$ | $(-\infty, \infty)$ |
| $\frac{1}{1-x}$ | $1 + x + x^2 + x^3 + \dots = \sum_{k=0}^{\infty} x^k$ | $(-1, 1)$ |
| $\ln(1+x)$ | $x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{x^k}{k}$ | $(-1, 1]$ |
Power Series
A power series is an infinite series of the form:
$\sum_{k=0}^{\infty} c_k (x-a)^k = c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + \dots$
where $c_k$ are constants called the coefficients, and $a$ is the center of the series. If $a=0$, the series is $\sum_{k=0}^{\infty} c_k x^k$. A Taylor series is a specific type of power series where the coefficients are determined by the derivatives of a function $f(x)$ at the center $a$.
Radius and Interval of Convergence
For a given power series, there are three possibilities for the values of $x$ for which the series converges:
- The series converges only at the center $x=a$. In this case, the radius of convergence $R=0$.
- The series converges for all real numbers $x$. In this case, the radius of convergence $R=\infty$.
- There exists a positive number $R$ such that the series converges for $|x-a| < R$ and diverges for $|x-a| > R$. $R$ is called the radius of convergence. The interval of convergence is the set of $x$ values for which the series converges, which includes the interval $(a-R, a+R)$ and possibly the endpoints $a-R$ and $a+R$.
The radius of convergence $R$ can often be found using the Ratio Test or the Root Test.
Ratio Test for Power Series
Given a power series $\sum_{k=0}^{\infty} c_k (x-a)^k$, let $L = \lim_{k \to \infty} \left| \frac{c_{k+1}(x-a)^{k+1}}{c_k(x-a)^k} \right|$.
- If $L < 1$, the series converges absolutely.
- If $L > 1$, the series diverges.
- If $L = 1$, the test is inconclusive.
The radius of convergence $R$ is given by $R = \frac{1}{\lim_{k \to \infty} \left| \frac{c_{k+1}}{c_k} \right|}$, provided the limit exists.
Root Test for Power Series
Given a power series $\sum_{k=0}^{\infty} c_k (x-a)^k$, let $L = \lim_{k \to \infty} \sqrt[k]{|c_k (x-a)^k|} = |x-a| \lim_{k \to \infty} \sqrt[k]{|c_k|}$.
- If $L < 1$, the series converges absolutely.
- If $L > 1$, the series diverges.
- If $L = 1$, the test is inconclusive.
The radius of convergence $R$ is given by $R = \frac{1}{\lim_{k \to \infty} \sqrt[k]{|c_k|}}$, provided the limit exists.
Determining the Interval of Convergence
Once the radius of convergence $R$ is found, we need to check the endpoints of the interval $(a-R, a+R)$ separately by substituting $x = a-R$ and $x = a+R$ into the series and testing for convergence using standard series tests (like the p-series test, alternating series test, etc.).
Example: Interval of Convergence for $\sum_{k=1}^{\infty} \frac{x^k}{k}$
Here, $c_k = \frac{1}{k}$ and $a=0$. We use the Ratio Test:
$L = \lim_{k \to \infty} \left| \frac{x^{k+1}/(k+1)}{x^k/k} \right| = \lim_{k \to \infty} \left| \frac{x^{k+1}}{k+1} \cdot \frac{k}{x^k} \right| = \lim_{k \to \infty} \left| x \cdot \frac{k}{k+1} \right| = |x| \lim_{k \to \infty} \frac{k}{k+1} = |x| \cdot 1 = |x|$
For convergence, $L < 1$, so $|x| < 1$. The radius of convergence is $R=1$. The interval is $(-1, 1)$.
Now, check the endpoints:
- At $x=1$: The series becomes $\sum_{k=1}^{\infty} \frac{1^k}{k} = \sum_{k=1}^{\infty} \frac{1}{k}$. This is the harmonic series, which diverges.
- At $x=-1$: The series becomes $\sum_{k=1}^{\infty} \frac{(-1)^k}{k}$. This is the alternating harmonic series, which converges by the Alternating Series Test.
Therefore, the interval of convergence is $[-1, 1)$.
Operations on Power Series
Power series can be manipulated like polynomials within their interval of convergence. This means we can:
- Add/Subtract Series: Add or subtract corresponding coefficients for terms of the same power of $(x-a)$.
- Multiply Series: Use the Cauchy product for multiplication.
- Differentiate Series Term-by-Term: If $f(x) = \sum c_k (x-a)^k$ converges on $(a-R, a+R)$, then $f'(x) = \sum k c_k (x-a)^{k-1}$ converges on the same interval.
- Integrate Series Term-by-Term: If $f(x) = \sum c_k (x-a)^k$ converges on $(a-R, a+R)$, then $\int f(x) dx = \sum \frac{c_k}{k+1} (x-a)^{k+1} + C$ converges on the same interval.
Example: Finding the Series for $\frac{1}{(1-x)^2}$
We know the Maclaurin series for $\frac{1}{1-x}$ is $\sum_{k=0}^{\infty} x^k = 1 + x + x^2 + x^3 + \dots$ for $|x| < 1$. Notice that $\frac{1}{(1-x)^2}$ is the derivative of $\frac{1}{1-x}$ with respect to $x$. So, we can differentiate the series term-by-term:
$\frac{d}{dx} \left( \frac{1}{1-x} \right) = \frac{d}{dx} (1-x)^{-1} = -1 (1-x)^{-2} (-1) = \frac{1}{(1-x)^2}$
Differentiating the series:
$\frac{d}{dx} \left( \sum_{k=0}^{\infty} x^k \right) = \frac{d}{dx} (1 + x + x^2 + x^3 + \dots)$
$= 0 + 1 + 2x + 3x^2 + 4x^3 + \dots = \sum_{k=1}^{\infty} k x^{k-1}$
Let $j = k-1$. When $k=1, j=0$. When $k \to \infty, j \to \infty$. So $k = j+1$. The series becomes:
$\sum_{j=0}^{\infty} (j+1) x^j = 1 + 2x + 3x^2 + 4x^3 + \dots$
Thus, the Maclaurin series for $\frac{1}{(1-x)^2}$ is $\sum_{k=0}^{\infty} (k+1) x^k$, valid for $|x| < 1$.
Applications of Taylor and Power Series
Taylor and power series have wide-ranging applications:
- Approximation: Approximating functions, especially in numerical analysis and physics, where complex functions can be replaced by simpler polynomials.
- Solving Differential Equations: Finding series solutions to differential equations.
- Evaluating Limits: Simplifying indeterminate forms in limits.
- Calculating Integrals: Integrating functions that do not have elementary antiderivatives.
- Understanding Function Behavior: Analyzing the local behavior of functions, including their concavity and rate of change.
- Physics and Engineering: Used extensively in areas like quantum mechanics, signal processing, and control theory.