23. Taylor Series

c. Constructing New Taylor Series from Old Ones

In the previous chapter, we saw that new power series could be constructed from old ones by applying the operations of substitution, addition and subtraction, multiplication by a polynomial, differentiation and integration. Since Taylor series are power series, we can apply these operations to old Taylor series to produce new Taylor series. In the previous chapter, the examples all started from the geometric series for the function \(\dfrac{1}{1-x}\). We now look at examples involving exponentials, sines and cosines.

The hyperbolic cosine is defined by \(\cosh x=\dfrac{e^x+e^{-x}}{2}\). Find its Maclaurin series.

We start with the exponential series \[ e^x=\sum_{n=0}^\infty \dfrac{x^n}{n!} =1+x+\dfrac{x^2}{2}+\dfrac{x^3}{3!}+\cdots \] We substitute \(x\rightarrow-x\): \[ e^{-x}=\sum_{n=0}^\infty \dfrac{(-x)^n}{n!} =1-x+\dfrac{x^2}{2}-\dfrac{x^3}{3!}+\cdots \] We add these two series and divide by \(2\): \[\begin{aligned} \cosh x &=\dfrac{e^x+e^{-x}}{2} =\sum_{n=0}^\infty \left(\dfrac{1+(-1)^n}{2}\right)\dfrac{x^n}{n!} \\ &=\dfrac{1}{2}\left(1+x+\dfrac{x^2}{2}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\cdots\right. \\ &\quad\;\;+\left.1-x+\dfrac{x^2}{2}-\dfrac{x^3}{3!}+\dfrac{x^4}{4!}-\cdots\right) \\ &=1+\dfrac{x^2}{2}+\dfrac{x^4}{4!}+\cdots \end{aligned}\] Notice that the odd power terms cancelled out and the even power terms added up and cancelled the \(\dfrac{1}{2}\). To account for only having even terms, we let \(n=2k\). Thus \[ \cosh x=\sum_{k=0}^\infty \dfrac{x^{2k}}{(2k)!}=1+ \dfrac{1}{2}x^2+\dfrac{1}{4!}x^4\cdots \]

Notice that the series for \(\cosh x\) is the same as the series for \(\cos x\) except there are no minus signs. This is one of the reasons the name hyperbolic cosine is used.

Find the Taylor series for \(\sin^2x\) centered at \(x=0\).

We use the identity \(\sin^2x=\dfrac{1-\cos 2x}{2}\). We start with the series for \(\cos x\): \[ \cos x=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!}x^{2n} =1-\dfrac{x^2}{2}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\cdots \] We substitute \(x\rightarrow 2x\): \[\begin{aligned} \cos 2x &=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!}(2x)^{2n} =1-\dfrac{(2x)^2}{2}+\dfrac{(2x)^4}{4!}-\dfrac{(2x)^6}{6!}+\cdots \\ &=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!}2^{2n}x^{2n} =1-\dfrac{2^2x^2}{2}+\dfrac{2^4x^4}{4!}-\dfrac{2^6x^6}{6!}+\cdots \end{aligned}\] We subtract from \(1\) and divide by \(2\). First in expanded form: \[\begin{aligned} 1-\cos 2x &=1-\left(1-\dfrac{2^2x^2}{2}+\dfrac{2^4x^4}{4!}-\dfrac{2^6x^6}{6!}+\cdots\right) \\ &=\dfrac{2^2x^2}{2}-\dfrac{2^4x^4}{4!}+\dfrac{2^6x^6}{6!}-\cdots \end{aligned}\] \[ \sin^2x=\dfrac{1-\cos 2x}{2}=\dfrac{2x^2}{2}-\dfrac{2^3x^4}{4!}+ \dfrac{2^5x^6}{6!}+\cdots \] Now in summation form: (Notice how we separate the \(n=0\) term from the rest of the sum.) \[\begin{aligned} 1-\cos 2x &=1-\left(1+\sum_{n=1}^\infty \dfrac{(-1)^n}{(2n)!}2^{2n}x^{2n}\right) \\ \end{aligned}\] \[\begin{aligned} \sin^2x &=\dfrac{1-\cos 2x}{2} =\sum_{n=1}^\infty \dfrac{(-1)^{n-1}}{(2n)!}2^{2n-1}x^{2n} \end{aligned}\]