24. Applications of Taylor Series

d.1. Numeric Series

When you first learned about numerical series, the only series you could sum were the geometric series and the telescoping series. There are now many other series you can sum by using the known series for \(\dfrac{1}{1-x}\), \(e^x\), \(\sin x\), \(\cos x\), \(\ln(1+x)\) and \(\arctan x\) and series derived from these.

Compute \(\displaystyle \sum_{n=0}^\infty (-1)^n\dfrac{2^{2n+1}}{(2n+1)!}\).

This series is alternating and has odd factorials in the denominator. This should tip you off that you should look at the series for \(\sin x\): \[ \sin x=\sum_{n=0}^\infty (-1)^n\dfrac{x^{2n+1}}{(2n+1)!} \] Turn this around and plug in \(x=2\): \[ \sum_{n=0}^\infty (-1)^n\dfrac{2^{2n+1}}{(2n+1)!}=\sin 2 \]

For each series, identify its sum.

1. \(\displaystyle \sum_{n=1}^\infty \dfrac{(-1)^{n+1}}{n}\)
\(\sin 1\)
\(\ln 2\)
\(e^{-1}\)
\(\cos 2\)
\(e\)

A Incorrect. \[ \sin x=\sum_{n=0}^\infty \dfrac{(-1)^nx^{2n+1}}{(2n+1)!} \]

B Great! Substitute \(x=1\) into \[ \ln(1+x)=\sum_{n=1}^\infty \dfrac{(-1)^{n+1}x^n}{n} \] to get \[ \ln 2=\sum_{n=1}^\infty \dfrac{(-1)^{n+1}}{n} \]

C Not quite. \[ e^x=\sum_{n=0}^\infty \dfrac{x^n}{n!} \]

D Nope. \[ \cos x=\sum_{n=0}^\infty \dfrac{(-1)^nx^{2n}}{(2n)!} \]

E Try again. \[ e^x=\sum_{n=0}^\infty \dfrac{x^n}{n!} \]

2. \(\displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n2^{2n}}{(2n)!}\)
\(\sin 1\)
\(\ln 2\)
\(e^{-1}\)
\(\cos 2\)
\(e\)

A Incorrect. \[ \sin x=\sum_{n=0}^\infty \dfrac{(-1)^nx^{2n+1}}{(2n+1)!} \]

B Sorry. \[ \ln(1+x)=\sum_{n=1}^\infty \dfrac{(-1)^{n+1}x^n}{n} \]

C Not quite. \[ e^x=\sum_{n=0}^\infty \dfrac{x^n}{n!} \]

D Correct! Set \(x=2\) in: \[ \cos x=\sum_{n=0}^\infty \dfrac{(-1)^nx^{2n}}{(2n)!} \] producing \[ \cos 2=\sum_{n=0}^\infty \dfrac{(-1)^n2^{2n}}{(2n)!} \]

E Try again. \[ e^x=\sum_{n=0}^\infty \dfrac{x^n}{n!} \]

3. \(\displaystyle \sum_{n=0}^\infty \dfrac{1}{n!}\)
\(\sin 1\)
\(\ln 2\)
\(e^{-1}\)
\(\cos 2\)
\(e\)

A Incorrect. \[ \sin x=\sum_{n=0}^\infty \dfrac{(-1)^nx^{2n+1}}{(2n+1)!} \]

B Sorry. \[ \ln(1+x)=\sum_{n=1}^\infty \dfrac{(-1)^{n+1}x^n}{n} \]

C Not quite. \[ e^x=\sum_{n=0}^\infty \dfrac{x^n}{n!} \]

D Nope. \[ \cos x=\sum_{n=0}^\infty \dfrac{(-1)^nx^{2n}}{(2n)!} \]

That one is easy. Set \(x=1\) in: \[ De^x=\sum_{n=0}^\infty \dfrac{x^n}{n!} \] giving \[ e^1=\sum_{n=0}^\infty \dfrac{1}{n!} \]

4. \(\displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!}\)
\(\sin 1\)
\(\ln 2\)
\(e^{-1}\)
\(\cos 2\)
\(e\)

A Super. That \(x=1\) in: \[ \sin x=\sum_{n=0}^\infty \dfrac{(-1)^nx^{2n+1}}{(2n+1)!} \] giving \[ \sin 1=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!} \]

B Sorry. \[ \ln(1+x)=\sum_{n=1}^\infty \dfrac{(-1)^{n+1}x^n}{n} \]

C Not quite. \[ e^x=\sum_{n=0}^\infty \dfrac{x^n}{n!} \]

D Nope. \[ \cos x=\sum_{n=0}^\infty \dfrac{(-1)^nx^{2n}}{(2n)!} \]

E Try again. \[ e^x=\sum_{n=0}^\infty \dfrac{x^n}{n!} \]

5. \(\displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{n!}\)
\(\sin 1\)
\(\ln 2\)
\(e^{-1}\)
\(\cos 2\)
\(e\)

A Incorrect. \[ \sin x=\sum_{n=0}^\infty \dfrac{(-1)^nx^{2n+1}}{(2n+1)!} \]

B Sorry. \[ \ln(1+x)=\sum_{n=1}^\infty \dfrac{(-1)^{n+1}x^n}{n} \]

C Correct. Just slightly tricky. Set \(x=-1\) in \[ e^x=\sum_{n=0}^\infty \dfrac{x^n}{n!} \] to give \[ e^{-1}=\sum_{n=0}^\infty \dfrac{(-1)^n}{n!} \]

D Nope. \[ \cos x=\sum_{n=0}^\infty \dfrac{(-1)^nx^{2n}}{(2n)!} \]

E Try again. \[ e^x=\sum_{n=0}^\infty \dfrac{x^n}{n!} \]

Sometimes you need to do a little more work to find the appropriate series.

Compute \(\displaystyle \sum_{n=1}^\infty n\left(\dfrac{1}{2}\right)^n\).

If the series did not have the \(n\) coefficient, it would look like the geometric series (evaluated at \(x=\dfrac{1}{2})\): \[ \sum_{n=0}^\infty x^n=\dfrac{1}{1-x} \] To get an \(n\) factor, we differentiate: \[ \sum_{n=0}^\infty nx^{n-1}=\dfrac{1}{(1-x)^2} \] But now the power of \(x\) is off. So we multiply by \(x\): \[ \sum_{n=0}^\infty nx^n=\dfrac{x}{(1-x)^2} \] Also notice we can drop the \(n=0\) term because the coefficient is \(n\). Finally we plug in \(x=\dfrac{1}{2}\): \[ \sum_{n=1}^\infty n\left(\dfrac{1}{2}\right)^n =\dfrac{\left(\dfrac{1}{2}\right)}{\left[1-\left(\dfrac{1}{2}\right)\right]^2} =\dfrac{2}{(2-1)^2}=2 \]

Compute \(\displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n4}{(2n+1)}\).

\(\displaystyle \arctan x=\sum_{n=0}^\infty (-1)^n\dfrac{x^{2n+1}}{(2n+1)}\)

\(\displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n4}{(2n+1)} =4\arctan 1=\pi\)

This series is alternating and has odd numbers in the denominator without factorials. This looks like the series for \(\arctan x\): \[ \arctan x=\sum_{n=0}^\infty (-1)^n\dfrac{x^{2n+1}}{(2n+1)} \] Turn this around and plug in \(x=1\): \[ \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)} =\arctan 1 \] Finally to get the series we want, we multiply by 4 and remember that \(\arctan 1=\dfrac{\pi}{4}\): \[ \sum_{n=0}^\infty \dfrac{(-1)^n4}{(2n+1)} =4\arctan 1=\pi \]

Series like these can, in fact, be used to find numerical approximations for the fundamental constants as done on the next page.

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Supported in part by NSF Grant #1123255