Professor Baker's Math Class - April 11, 2024: Calculus 2, Section 11.9
Welcome back to Calc 2! Today's focus is on Section 11.9, where we delve into representing functions as power series. This is a powerful technique with many applications, and we'll explore some key examples.
Expressing Functions as Power Series
Our main goal is to express a given function, $f(x)$, as a power series of the form:
$$f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + ...$$where $c_n$ are coefficients and $a$ is the center of the series. A crucial aspect of working with power series is determining the interval of convergence, which is the range of $x$ values for which the series converges.
Example 1: Expressing 1/(1 + x²) as a Power Series
Let's consider the function $f(x) = \frac{1}{1 + x^2}$. We can rewrite this in a form that resembles a geometric series:
$$\frac{1}{1 + x^2} = \frac{1}{1 - (-x^2)}$$Recall the formula for the sum of an infinite geometric series: $\frac{1}{1-r} = \sum_{n=0}^{\infty} r^n$, which converges when $|r| < 1$.
In this case, $r = -x^2$. Therefore, we have:
$$\frac{1}{1 + x^2} = \sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^n x^{2n} = 1 - x^2 + x^4 - x^6 + ...$$To find the interval of convergence, we need $|-x^2| < 1$, which simplifies to $|x^2| < 1$, and further to $|x| < 1$. Thus, the interval of convergence is $(-1, 1)$.
Example 2: Power Series Representation of x³/(x + 2)
To find the power series for $f(x) = \frac{x^3}{x + 2}$, we manipulate the expression to fit the geometric series form. First, factor out a 2 from the denominator:
$$ \frac{x^3}{x+2} = \frac{x^3}{2(\frac{x}{2}+1)} = \frac{x^3}{2(1 + \frac{x}{2})} = \frac{x^3}{2} \cdot \frac{1}{1 - (-\frac{x}{2})} $$ Now, use the geometric series formula with $r = -\frac{x}{2}$: $$ \frac{x^3}{2} \cdot \frac{1}{1 - (-\frac{x}{2})} = \frac{x^3}{2} \sum_{n=0}^{\infty} (-\frac{x}{2})^n = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+3}}{2^{n+1}} $$Example 3: Power Series for ln(1 + x)
We can find the power series for $ln(1 + x)$ by integrating the power series representation of its derivative, which is $\frac{1}{1+x}$:
$$\ln(1+x) = \int \frac{1}{1+x} dx = \int \sum_{n=0}^{\infty} (-1)^n x^n dx = \sum_{n=0}^{\infty} \int (-1)^n x^n dx$$ Integrating term by term, we get: $$\ln(1+x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1} + C$$ Setting $x = 0$, we find that $C = \ln(1+0) = 0$. So, the power series representation for $ln(1+x)$ is: $$\ln(1+x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...$$Important Theorem
If the power series $\sum c_n(x-a)^n$ has a radius of convergence $R > 0$, then the function $f$ defined by $f(x) = \sum c_n(x-a)^n$ is differentiable and continuous on the interval $(a-R, a+R)$. Furthermore, we can differentiate and integrate the power series term by term within this interval.
Keep practicing, and you'll master these techniques in no time! See you in the next lecture!