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11.2 Properties of Power Series (part 1)

power series:
\[ \sum_{k=0}^{\infty} c_k (x-a)^k \]

\( a \): center

\( c_k \): coefficients of \( k^{\text{th}} \)-order term

convergence in general depends on \( x \)

the interval on which the power series converges \( \rightarrow \) interval of convergence
(already saw this in Ratio Test section)

for example,

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

\( c_k = 1, \quad a = 0 \)

converges if \( |x| < 1 \)

\( -1 < x < 1 \)

Sketch:

A number line showing an interval from -1 to 1 centered at 0, with parentheses at the endpoints.

this is the radius of convergence (center to one end)

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example

\[ \sum_{k=1}^{\infty} \frac{(-1)^k k}{4^k} (x+3)^k \]

\( c_k = \frac{(-1)^k k}{4^k} \)

\( (x-a)^k \rightarrow (x+3)^k \) so \( a = -3 \) (center of interval of convergence)

find interval of convergence

Ratio Test is good for this

\[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \]
\[ = \lim_{k \to \infty} \left| \frac{\frac{(-1)^{k+1} (k+1)}{4^{k+1}} (x+3)^{k+1}}{\frac{(-1)^k (k)}{4^k} (x+3)^k} \right| = \lim_{k \to \infty} \left| \frac{(-1)^{k+1} (k+1)}{4^{k+1}} \cdot \frac{4^k}{(-1)^k (k)} (x+3) \right| \]
\[ = \lim_{k \to \infty} \left| (-1) \frac{(k+1)}{k} \cdot \frac{1}{4} \cdot (x+3) \right| = \left| \frac{x+3}{4} \right| < 1 \]

ratio < 1 converges

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\[ |x+3| < 4 \]\[ -4 < x+3 < 4 \]\[ -7 < x < 1 \]

this is NOT it! At ends the ratio is 1 (inconclusive)

\[ \sum_{k=1}^{\infty} \frac{(-1)^k k}{4^k} (x+3)^k \]

at \( x = -7 \)

\[ \sum_{k=1}^{\infty} \frac{(-1)^k k}{4^k} (4^k) = \sum_{k=1}^{\infty} (-1)^k k \]\[ = \sum_{k=1}^{\infty} k \]

diverges at \( x = -7 \)

at \( x = 1 \)

\[ \sum_{k=1}^{\infty} \frac{(-1)^k k}{4^k} (4^k) = \sum_{k=1}^{\infty} (-1)^k k \]

diverges at \( x = 1 \)

so interval of convergence is \( -7 < x < 1 \) or \( (-7, 1) \)

radius of conv. \( R = 4 \)

Number line showing the interval from -7 to 1, centered at -3, with open parentheses at the endpoints.
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example

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

Note: \( c_k = k^k \) and \( a = 0 \)

Ratio Test

\[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = \lim_{k \to \infty} \left| \frac{(k+1)^{k+1} \cdot x^{k+1}}{k^k \cdot x^k} \right| < 1 \]\[ = \lim_{k \to \infty} \left| \left( \frac{k+1}{k} \right)^k \cdot (k+1) \cdot x \right| = \infty \text{ unless } x = 0 \]

Note: \( \left( \frac{k+1}{k} \right)^k = (1 + \frac{1}{k})^k \to e \) and \( (k+1) \to \infty \)

ratio \( < 1 \) only if \( x = 0 \)

the only value of \( x \) for which the series converges is \( x = 0 \)

interval of convergence: \( x = 0 \)

radius of conv: \( R = 0 \)

Number line with a single solid point at the origin labeled 0.

series when \( x = 0 \)

\[ \sum_{k=1}^{\infty} k^k \cdot (0)^k = \sum_{k=1}^{\infty} 0 = 0 + 0 + 0 + 0 + 0 + \dots \]
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Summation Notation and Intervals of Convergence

Summation notation is useful when finding intervals of convergence. You may need to find it yourself.

Example

Express the following series in summation notation:

\[ x - \frac{x^3}{4} + \frac{x^5}{9} - \frac{x^7}{16} + \dots \]

Identifying Patterns

  • Numerator: \( x \) to powers of odd numbers, alternating
  • Denominator: something squared
\[ = \frac{x^1}{1^2} - \frac{x^3}{2^2} + \frac{x^5}{3^2} - \frac{x^7}{4^2} + \frac{x^9}{5^2} - \frac{x^{11}}{6^2} + \dots \]
k=1k=2k=3k=4k=5k=6

Decide a starting \( k \). Here, \( k = 1 \).

\[ = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{x^{2k-1}}{k^2} \]
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Power Series as Functions

We know:

\[ \sum_{k=0}^{\infty} ar^k = a + ar^1 + ar^2 + ar^3 + \dots = \frac{a}{1-r} \text{ if } |r| < 1 \]

Treat it as a function: \( r = x \), \( a = 1 \)

\[ \frac{a}{1-r} = \frac{1}{1-x} = \sum_{k=0}^{\infty} x^k = 1 + x + x^2 + x^3 + x^4 + \dots \text{ if } |x| < 1 \]

Taylor series of \( \frac{1}{1-x} \) (without taking derivatives!)

Modifying the Series

We can modify it to find Taylor series of similar functions:

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

Now replace \( x \) in \( \frac{1}{1-x} = \sum_{k=0}^{\infty} x^k \) with \( (-x) \).

\[ = 1 - x + x^2 - x^3 + x^4 - x^5 + \dots \]\[ \text{converges if } |-x| < 1 \]\[ |x| < 1 \]
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Taylor Series Expansion

Find the Taylor series for:

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

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

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

\[ = x^2 (1 - x^2 + x^4 - x^6 + x^8 - \dots) \]\[ = x^2 - x^4 + x^6 - x^8 + x^{10} - \dots \]

Interval of Convergence

\[ |-x^2| < 1 \]\[ x^2 < 1 \]\[ -1 < x < 1 \]
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Manipulating the Geometric Series

The "1" in \( \frac{1}{1-x} = \sum_{k=0}^{\infty} x^k \) is important.

For \( \frac{1}{2-x} \), we must rearrange so the denominator starts with 1.

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

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

\( |x| < 1 \)

Interval of Convergence

We changed \( x \) to \( \frac{x}{2} \):

\[ |x| < 1 \implies |\frac{x}{2}| < 1 \]\[ |x| < 2 \]\[ -2 < x < 2 \]