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11.3 Taylor Series

Taylor series:

\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!} (x-a)^k \]
\[ = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!} (x-a)^2 + \frac{f'''(a)}{3!} (x-a)^3 + \dots \]

a: center (where the Taylor series is built)

When \( a = 0 \) we call the resulting Taylor series the Maclaurin Series.

Maclaurin series of \( f(x) \):

\[ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!} x^k \]
\[ = f(0) + f'(0)x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \dots \]
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Example: Maclaurin series of \( f(x) = \sin x \)

\( f(x) = \sin x \)

\( f'(x) = \cos x \)

\( f''(x) = -\sin x \)

\( f'''(x) = -\cos x \)

\( f^{(4)}(x) = \sin x \)

\( f(0) = 0 \)

\( f'(0) = 1 \)

\( f''(0) = 0 \)

\( f'''(0) = -1 \)

\( f^{(4)}(0) = 0 \)

all even derivs are 0

all odd derivs are \( \pm 1 \)

pattern repeats after 4 derivatives

Maclaurin series of \( \sin x \)

\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \dots \]
\[ = x - \frac{1}{3!} x^3 + \frac{1}{5!} x^5 - \frac{1}{7!} x^7 + \dots \]

pack into summation

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

near \( x = 0 \), \( \sin x \approx x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots \)

\( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)

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Interval of Convergence for Sine Series

Interval of convergence of \(\sin x = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}\)

Ratio Test:

\[ \lim_{k \to \infty} \left| \frac{(-1)^{k+1} \frac{x^{2(k+1)+1}}{(2(k+1)+1)!}}{(-1)^k \frac{x^{2k+1}}{(2k+1)!}} \right| \]
\[ = \lim_{k \to \infty} \left| \frac{x^{2k+3}}{(2k+3)!} \cdot \frac{(2k+1)!}{x^{2k+1}} \right| \]
\[ = \lim_{k \to \infty} \left| x^2 \cdot \frac{(2k+1)(2k)(2k-1)(2k-2) \dots (1)}{(2k+3)(2k+2)(2k+1)(2k)(2k-1)(2k-2) \dots (1)} \right| = 0 \]

So converge on \(-\infty < x < \infty\) or \((-\infty, \infty)\)

Radius of convergence \(\infty\)

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We can re-use \(\sin(x) = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!}\) for anything that resembles \(\sin(x)\).

For example, \(\sin(2x) = \sum_{k=0}^{\infty} (-1)^k \frac{(2x)^{2k+1}}{(2k+1)!}\)

Example: Composite Function

\(f(x) = x^3 \sin\left(\frac{x^2}{3}\right)\) converges on \((-\infty, \infty)\)

\[ \sin x = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k+1}}{(2k+1)!} \]
\[ \sin\left(\frac{x^2}{3}\right) = \sum_{k=0}^{\infty} (-1)^k \frac{1}{(2k+1)!} \left(\frac{x^2}{3}\right)^{2k+1} \]
\[ = \sum_{k=0}^{\infty} (-1)^k \frac{1}{(2k+1)!} \frac{x^{4k+2}}{3^{2k+1}} \text{ converges on } (-\infty, \infty) \]

Bump up by 3:

\[ x^3 \sin\left(\frac{x^2}{3}\right) = x^3 \sum_{k=0}^{\infty} (-1)^k \frac{1}{(2k+1)!} \frac{x^{4k+2}}{3^{2k+1}} = \sum_{k=0}^{\infty} (-1)^k \frac{1}{(2k+1)!} \frac{x^{4k+5}}{3^{2k+1}} \]
\[ = \frac{x^5}{3} - \frac{x^9}{3! \cdot 3^3} + \frac{x^{13}}{5! \cdot 3^5} - \dots \]
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Common Maclaurin Series

a = 0 only

\[ \frac{1}{1-x} = 1 + x + x^2 + \dots + x^k + \dots = \sum_{k=0}^{\infty} x^k, \quad \text{for } |x| < 1 \]

\[ \frac{1}{1+x} = 1 - x + x^2 - \dots + (-1)^k x^k + \dots = \sum_{k=0}^{\infty} (-1)^k x^k, \quad \text{for } |x| < 1 \]

\[ e^x = 1 + x + \frac{x^2}{2!} + \dots + \frac{x^k}{k!} + \dots = \sum_{k=0}^{\infty} \frac{x^k}{k!}, \quad \text{for } |x| < \infty \]

\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots + \frac{(-1)^k x^{2k+1}}{(2k+1)!} + \dots = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{(2k+1)!}, \quad \text{for } |x| < \infty \]

\[ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots + \frac{(-1)^k x^{2k}}{(2k)!} + \dots = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(2k)!}, \quad \text{for } |x| < \infty \]

\[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots + \frac{(-1)^{k+1} x^k}{k} + \dots = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} x^k}{k}, \quad \text{for } -1 < x \le 1 \]

\[ -\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots + \frac{x^k}{k} + \dots = \sum_{k=1}^{\infty} \frac{x^k}{k}, \quad \text{for } -1 \le x < 1 \]

\[ \tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots + \frac{(-1)^k x^{2k+1}}{2k+1} + \dots = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{2k+1}, \quad \text{for } |x| \le 1 \]

\[ \sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \dots + \frac{x^{2k+1}}{(2k+1)!} + \dots = \sum_{k=0}^{\infty} \frac{x^{2k+1}}{(2k+1)!}, \quad \text{for } |x| < \infty \]

\[ \cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots + \frac{x^{2k}}{(2k)!} + \dots = \sum_{k=0}^{\infty} \frac{x^{2k}}{(2k)!}, \quad \text{for } |x| < \infty \]

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Example

Find the Maclaurin series for \( f(x) = \frac{e^x - 1}{5x} \).

From Table: \[ e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} \quad (-\infty, \infty) \]

\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \]

Subtracting 1 from the series:

\[ e^x - 1 = (1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots) - 1 \]

\[ = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots = \sum_{k=1}^{\infty} \frac{x^k}{k!} \]

Dividing by \( 5x \):

\[ \frac{e^x - 1}{5x} = \frac{1}{5x} \left( x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \right) = \frac{1}{5x} \sum_{k=1}^{\infty} \frac{x^k}{k!} \]

\[ = \frac{1}{5} + \frac{x}{5 \cdot 2!} + \frac{x^2}{5 \cdot 3!} + \frac{x^3}{5 \cdot 4!} + \dots = \sum_{k=1}^{\infty} \frac{x^{k-1}}{5 \cdot k!} \]

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The Table we saw is a list of Maclaurin series → \( a=0 \) only

ANY other \( a \), find Taylor series by differentiating

Example: \( f(x) = \sin x \), \( a = \frac{\pi}{6} \)

\( f(x) = \sin x \)

\( f'(x) = \cos x \)

\( f''(x) = -\sin x \)

\( f'''(x) = -\cos x \)

\( f^{(4)}(x) = \sin x \)

cycle repeats

\( f(\frac{\pi}{6}) = \frac{1}{2} \)

\( f'(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \)

\( f''(\frac{\pi}{6}) = -\frac{1}{2} \)

\( f'''(\frac{\pi}{6}) = -\frac{\sqrt{3}}{2} \)

\( f^{(4)}(\frac{\pi}{6}) = \frac{1}{2} \)

General Taylor Series Formula:

\[ \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!} (x-a)^k \]

Near \( x = \frac{\pi}{6} \), \( \sin x \) can be approximated as:

\[ \sin x = \frac{1}{2} + \frac{\sqrt{3}}{2}(x - \frac{\pi}{6}) - \frac{1/2}{2!}(x - \frac{\pi}{6})^2 - \frac{\sqrt{3}/2}{3!}(x - \frac{\pi}{6})^3 + \frac{1/2}{4!}(x - \frac{\pi}{6})^4 + \dots \]
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In practice, we chop it off after some \( k \), then look at the error.

Remainder of \( k^{th} \)-order Taylor polynomial is

\[ R_k = \frac{f^{(k+1)}(c)}{(k+1)!} (x-a)^{k+1} \]

For \( \sin x \) with \( a = \frac{\pi}{6} \):

\[ R_k = \frac{\pm \cos(c)}{(k+1)!} (x - \frac{\pi}{6})^{k+1} \quad \text{or} \quad \frac{\pm \sin(c)}{(k+1)!} (x - \frac{\pi}{6})^{k+1} \]

How do we know as \( k \to \infty \), the Taylor series = true function?

→ If \( \lim_{k \to \infty} R_k = 0 \)

\( R_k = \frac{\pm \cos(c)}{(k+1)!} (x - \frac{\pi}{6})^{k+1} \)

  • Numerator \( \pm \cos(c) \) is between -1 and 1
  • Denominator \( (k+1)! \) goes to \( \infty \) really quickly

It's clear that \( R_k \to 0 \) as \( k \to \infty \) (same for \( \pm \sin(c) \) on top)