Functions and Series
- Page ID
- 219493
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The Geometric Power Series
Recall that
\( \displaystyle\sum_{n=0}^{\infty} a r^n = \frac{a}{1-r} \) for \( |r| < 1\)
Substituting x for r, we have
\( \displaystyle\sum_{n=0}^{\infty} a x^n = \frac{a}{1-x} \) for \( |x| < 1\)
We write
|
\( \frac{1}{1-x} = 1 + x + x^2 + x^3 + ... = \displaystyle\sum_{n=0}^{\infty} x^n \) |
Milking the Geometric Power Series
By using substitution, we can obtain power series expansions from the geometric series.
Example 1
Substituting x2 for x, we have
\( \frac{1}{1-x^2} = 1 + x^2 + x^4 + x^6 + ... = \displaystyle\sum_{n=0}^{\infty} x^{2n} \)
Example 2
Multiplying by x we have
\( \frac{x}{1-x} = x \displaystyle\sum_{n=0}^{\infty} x^{n} = \displaystyle\sum_{n=0}^{\infty} x^{n+1} \)
Example 3
Suppose we want to find the power series for
1
f(x) =
2x - 3
centered at x = 4. We rewrite the function as
1 1
=
2(x - 4) + 8 - 3 2(x - 4) + 5
\( = \frac{1}{5}\frac{1}{\frac{2}{5}(x-4)+1} =\frac{1}{5}\frac{1}{1 - (\frac{-2}{5}(x-4))} \)
\( = \frac{1}{5} \displaystyle\sum_{n=0}^{\infty} (\frac{-2}{5}(x - 4))^n = \frac{1}{5} \displaystyle\sum_{n=0}^{\infty} (\frac{(-1)^n 2^n}{5^n}(x - 4)^n \)
\( = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n 2^n}{5^{n+1}} (x - 4)^n \)
Example 4
Substituting -x for x, we have
\( \frac{1}{1+x} = \displaystyle\sum_{n=0}^{\infty} (-x)^n = \displaystyle\sum_{n=0}^{\infty} (-1)^n x^n \)
Example 5
Substituting x2 in for x in the previous example, we have
\( = \frac{1}{1+x^2} = \displaystyle\sum_{n=0}^{\infty} (-1)^n x^{2n} \)
Example 6
Taking the integral of the previous example, we have
\( tan^{-1}(x) = \int \frac{1}{1+x^2}dx = \int \displaystyle\sum_{n=0}^{\infty} (-1)^n x^{2n}dx \)
\( = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n + 1} + C \)
Exercise Find the power series that represents the following functions:
-
ln(1 + x)
-
tanh-1x
-
-(1 - x)-2
Integrating Impossible Functions
We can use power series to integrate functions where there are no standard techniques of integration available.
Example:
Use power series to find the integral
\( \int tan^{-1}(x^2)dx \)
Then use this integral to approximate
\( \int_0^1 tan^{-1}(x^2)dx \)
Solution:
Notice that this is a very difficult integral to solve. We resort to power series. First we use the series expansion from Example 6, replacing x with x2.
\( \int tan^{-1}(x^2)dx = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} (x^{2})^{2n+1} = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{4n+2} \)
Integrating we arrive at the solution
\( \int tan^{-1}(x^2)dx = \int \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} (x^{2})^{2n+1}dx = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)(4n+3)} x^{4n+3} + C \)
Now to solve the definite integral, notice that when we plug in 0 we get 0, hence the definite integral is
\( \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)(4n+3)}(1)^{4n+3} = \frac{1}{3} - \frac{1}{21} + \frac{1}{55} - \frac{1}{105} +\frac{1}{171} - \frac{1}{253} + ... \)
Using the first 5 terms to approximate this we get 0.300
Notice that the error is less than the next term (which comes from x23/253)
E < 1/253 = .004.