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11.2 Properties of Power Series (part 2)

last time:

\[ \frac{1}{1-x} = \sum_{k=0}^{\infty} x^k \quad \text{converges } |x| < 1 \]

reuse it for similar expressions

\[ \frac{1}{1+x^2} = \frac{1}{1-(-x^2)} \]

change \( x \) to \( -x^2 \)

\[ = \sum_{k=0}^{\infty} (-x^2)^k = \sum_{k=0}^{\infty} (-1)^k x^{2k} \]

converges if

\( |-x^2| < 1 \)

\( |x^2| < 1 \)

today: generate more power series by differentiating or integrating

"model series"

\[ \frac{1}{1-x} = \sum_{k=0}^{\infty} x^k \]
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example

\( g(x) = \frac{1}{(1+x^2)^2} \) find power series representation

re-use \( \frac{1}{1-x} = \sum_{k=0}^{\infty} x^k \), we have \( \frac{1}{1+x^2} = \sum_{k=0}^{\infty} (-1)^k x^{2k} \)

notice \[ \frac{d}{dx} \left( \frac{1}{1+x^2} \right) = \frac{-2x}{(1+x^2)^2} = -2x \cdot \left( \frac{1}{(1+x^2)^2} \right) \]

\( \leftarrow g(x) \) power series?

we get:

\[ \frac{1}{(1+x^2)^2} = -\frac{1}{2x} \frac{d}{dx} \left( \frac{1}{1+x^2} \right) \]

\( \leftarrow \) we have power series

\[ \frac{d}{dx} \left( \frac{1}{1+x^2} \right) = \frac{d}{dx} \left( \sum_{k=0}^{\infty} (-1)^k x^{2k} \right) = \frac{d}{dx} (1 - x^2 + x^4 - x^6 + x^8 - \dots) \]
\[ = -2x + 4x^3 - 6x^5 + 8x^7 - 10x^9 + 12x^{11} - \dots \]
\[ -\frac{1}{2x} \frac{d}{dx} \left( \frac{1}{1+x^2} \right) = -\frac{1}{2x} (-2x + 4x^3 - 6x^5 + 8x^7 - 10x^9 + 12x^{11} - \dots) \]
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\[ = 1 - 2x^2 + 3x^4 - 4x^6 + 5x^8 - 6x^{10} + 7x^{12} - \dots \]

this is the power series of \(\frac{1}{(1+x^2)^2}\)

let's put into summation notation

patterns: alternating
coefficients go up by 1
powers are even

\[ = 1 - 2x^2 + 3x^4 - 4x^6 + 5x^8 - 6x^{10} + 7x^{12} - \dots \]

choose to start at \(k = 1\)

k=1k=2k=3k=4k=5k=6
\[ = \sum_{k=1}^{\infty} (-1)^{k-1} \cdot k \cdot x^{2k-2} \]

2k-2 (mistake in lecture)

differentiation / integration does NOT change the radius of convergence of the "model series"

So, this series is based on

\[ \frac{1}{1-x} = \sum_{k=0}^{\infty} x^k \quad |x| < 1 \]

so the power series we got still requires \(|x| < 1\), but the end behaviors can change (\(x=1, x=-1\))

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example

\(\ln \sqrt{16-x^2}\)

model series: \(\frac{1}{1-x} = \sum_{k=0}^{\infty} x^k \quad |x| < 1\)

\(\ln (16-x^2)^{1/2} = \frac{1}{2} \ln (16-x^2)\)

what happens if we differentiate \(\frac{1}{2} \ln (16-x^2)\)?

\[ \frac{d}{dx} \left[ \frac{1}{2} \ln (16-x^2) \right] = \frac{1}{2} \cdot \frac{-2x}{16-x^2} = -x \cdot \left( \frac{1}{16-x^2} \right) \]

use \(\frac{1}{1-x} = \sum_{k=0}^{\infty} x^k\)

\[ \frac{1}{16-x^2} = \frac{1}{16(1-\frac{x^2}{16})} = \frac{1}{16} \cdot \left( \frac{1}{1-(\frac{x}{4})^2} \right) \]

change \(x\) to \((\frac{x}{4})^2\)

\[ = \frac{1}{16} \sum_{k=0}^{\infty} \left[ \left( \frac{x}{4} \right)^2 \right]^k = \frac{1}{16} \sum_{k=0}^{\infty} \frac{x^{2k}}{4^{2k}} \]

what we want

\[ -x \cdot \frac{1}{16-x^2} = -x \cdot \frac{1}{16} \sum_{k=0}^{\infty} \frac{x^{2k}}{4^{2k}} = \frac{d}{dx} \left[ \frac{1}{2} \ln (16-x^2) \right] \]

from two lines above

\[ \text{so } \frac{1}{2} \ln (16-x^2) = \ln \sqrt{16-x^2} = \int \frac{-x}{16} \sum_{k=0}^{\infty} \frac{x^{2k}}{4^{2k}} dx \]
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\[ = \int -\frac{x}{16} \left( 1 + \frac{x^2}{4^2} + \frac{x^4}{4^4} + \frac{x^6}{4^6} + \frac{x^8}{4^8} + \dots \right) dx \]

Note: \( 16 = 4^2 \)

\[ = \int \left( -\frac{x}{4^2} - \frac{x^3}{4^4} - \frac{x^5}{4^6} - \frac{x^7}{4^8} - \frac{x^9}{4^{10}} - \dots \right) dx \]
\[ = -\frac{x^2}{2 \cdot 4^2} - \frac{x^4}{4 \cdot 4^4} - \frac{x^6}{6 \cdot 4^6} - \frac{x^8}{8 \cdot 4^8} - \frac{x^{10}}{10 \cdot 4^{10}} - \frac{x^{12}}{12 \cdot 4^{12}} - \dots + C \]

\( = \ln \sqrt{16 - x^2} \)

What is C?

At \( x = 0 \), \( \ln \sqrt{16 - x^2} = -\frac{x^2}{2 \cdot 4^2} - \frac{x^4}{4 \cdot 4^4} - \dots + C \)

\[ \ln \sqrt{16} = 0 + C \]

\( \ln 4 = C \)

So, \( \ln \sqrt{16 - x^2} = \ln 4 - \left( \frac{x^2}{2 \cdot 4^2} + \frac{x^4}{4 \cdot 4^4} + \frac{x^6}{6 \cdot 4^6} + \frac{x^8}{8 \cdot 4^8} + \dots \right) \)

\( C \)

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\[ \ln 4 - \left( \frac{x^2}{2 \cdot 4^2} + \frac{x^4}{4 \cdot 4^4} + \frac{x^6}{6 \cdot 4^6} + \frac{x^8}{8 \cdot 4^8} + \dots \right) \]
\( k = 1 \)\( k = 2 \)\( k = 3 \)\( k = 4 \)choose to start at \( k = 1 \)
\[ = \ln 4 - \sum_{k=1}^{\infty} \frac{x^{2k}}{2k \cdot 4^{2k}} \]
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Example: Power Series Representation

What function is represented by \[ \sum_{k=0}^{\infty} \frac{(x-2)^k}{3^{2k}} \]?

Again, use the geometric series formula:

\[ \frac{1}{1-x} = \sum_{k=0}^{\infty} x^k \]

We can rewrite the given series as:

\[ \sum_{k=0}^{\infty} \left( \frac{x-2}{3^2} \right)^k \]

Change \( x \) to \( \frac{x-2}{3^2} \) in the geometric series formula.

So,

\[ \sum_{k=0}^{\infty} \left( \frac{x-2}{3^2} \right)^k = \frac{1}{1 - \left( \frac{x-2}{9} \right)} = \frac{9}{9 - (x-2)} = \frac{9}{11 - x} \]
\[ \frac{9}{11 - x} \]