In Section 4.4 we showed that functions defined by power series have derivatives of all orders (Theorem 4.4.7). In Section 6.5 we demonstrated that analytic functions also have derivatives of all orders (Corollary 6.5.10). It seems natural, therefore, that there would be some connection between analytic functions and power series. As you might guess, the connection exists via the Taylor and Maclaurin series of analytic functions.
is called the Taylor series for\(\boldsymbol f\) centered at\(\boldsymbol \alpha\) . When the center is \(\alpha =0\text{,}\) the series is called the Maclaurin series for\(\boldsymbol f\text{.}\)
\(\dfrac{1}{z-z_0}=\dfrac{1}{(z-\alpha)-(z_0-\alpha)} = \dfrac{1}{z-\alpha }\left(\dfrac{1}{1-\frac{z_0-\alpha}{z-\alpha}}\right)\text{.}\) The result now follows from Corollary 4.3.3 if in it we replace \(z\) with \(\left(\frac{z_0-\alpha}{z-\alpha}\right)\text{.}\) We leave verification of the details as an exercise.
Suppose that \(f\) is analytic in a domain \(G\) and that \(D_{R}(\alpha)\) is any disk contained in \(G\text{.}\) Then the Taylor series for \(f\) converges to \(f(z)\) for all \(z\) in \(D_{R}(\alpha)\text{;}\) that is,
\begin{equation}
f(z)=\sum_{k=0}^{\infty}\frac{f\,^{(k)}(\alpha)}{k!}(z-\alpha)^k\, \text{ for all } z \in D_{R}(\alpha)\text{.}\tag{7.2.1}
\end{equation}
Furthermore, for any \(r\text{,}\)\(0\lt r\lt R\text{,}\) the convergence is uniform on the closed subdisk \(\overline{D}_r(\alpha) =\{z:|z-\alpha| \le r\}\text{.}\)
If we can establish Equation (7.2.1), the uniform convergence on \(\overline{D}_r(\alpha)\) for \(0\lt r\lt R\) will follow immediately from Theorem 7.1.5 by equating the \(c_k\) of that theorem with \(\frac{f\,^{(k)}(\alpha)}{k!}\text{.}\)
Let \(z_0 \in D_{R}(\alpha)\) and let \(r\) designate the distance between \(z_0\) and \(\alpha\) so that \(|z_0-\alpha | =r\text{.}\) Note that \(0 \le r\lt R\) because \(z_0\) belongs to the open disk \(D_{R}(\alpha)\text{.}\) We choose \(\rho\) such that \(0 \le r\lt \rho \lt R\text{,}\) and let \(C=C_{\rho }^{+}(\alpha)\) be the positively oriented circle centered at \(\alpha\) with radius \(\rho\) as shown in Figure 7.2.4.
The summation on the right-hand side of this last expression is the first \(n+1\) terms of the Taylor series. Verification of Equation (7.2.1) relies on our ability to show that we can make the remainder term, \(E_n(z_0)\text{,}\) as small as we please by making \(n\) sufficiently large. We will use the ML inequality (Theorem 6.2.19) to get a bound for \(|E_n(z_0)|\text{.}\) According to the constructions shown in Figure 7.2.4, we have
\begin{equation}
|z_0-a| = r \text{ and } |z-\alpha| =\rho\text{.}\tag{7.2.4}
\end{equation}
Because \(0 \le r \lt \rho \lt R\text{,}\) the fraction \(\frac{r}{\rho}\) is less than 1, so \((\frac{r}{\rho })^{n+1}\) (and hence the right side of Equation (7.2.7)) goes to zero as \(n\) goes to infinity. Thus, for any \(\varepsilon>0\text{,}\) we can find an integer \(N_{\varepsilon}\) such that \(|E_n(z_0)| \lt \varepsilon\) for \(n \ge N_{\varepsilon}\text{,}\) and this fact completes the proof.
A singular point of a function is a point at which the function fails to be analytic. You will see in Section 7.4 that singular points of a function can be classified according to how badly the function behaves at those points. Loosely speaking, a nonremovable singular point of a function has the property that it is impossible to redefine the value of the function at that point so as to make it analytic there. For example, the function \(f(z) =\frac{1}{1-z}\) has a nonremovable singularity at \(z=1\text{.}\) We give a formal definition of this concept in Section 7.4, but with this language we can nuance Taylor’s theorem a bit.
Suppose that \(f\) is analytic in the domain \(G\) that contains the point \(\alpha\text{.}\) Let \(z_0\) be a nonremovable singular point of minimum distance to the point \(\alpha\text{.}\) If \(|z_0-\alpha| =R\text{,}\) then
i.
the Taylor series \(\sum\limits_{k=0}^{\infty}\frac{f\,^{(k)}(\alpha)}{k!}(z-\alpha)^k\) converges to \(f(z)\) on all of \(D_{R}(\alpha)\text{,}\) and
if \(|z_1-\alpha| =S>R\text{,}\) the Taylor series \(\sum\limits_{k=0}^{\infty}\frac{f\,^{(k)}(\alpha)}{k!}(z_1-\alpha)^k\) does not converge to \(f(z_1)\text{.}\)
Taylor’s theorem gives us part (i) immediately. To establish part (ii), we note that if \(|z_0-\alpha|=R\text{,}\) then \(z_0 \in D_{S}(\alpha)\) whenever \(S>R\text{.}\) If for some \(z_1\text{,}\) with \(|z_1-\alpha| =S>R\text{,}\) the Taylor series converged to \(f(z_1)\text{,}\) then according to Theorem 4.4.7, the radius of convergence of the series \(\sum\limits_{k=0}^{\infty}\frac{f\,^{(k)}(\alpha)}{k!}(z-\alpha)^k\) would be at least equal to \(S\text{.}\) We could then make \(f\) differentiable at \(z_0\) by redefining \(f(z_0)\) to equal the value of the series at \(z_0\text{,}\) thus contradicting the fact that \(z_0\) is a nonremovable singular point.
In Example 4.4.9 we established this identity with the use of Theorem 4.4.7. We now do so via Theorem 7.2.3. If \(f(z) =\frac{1}{(1-z)^2}\text{,}\) then a standard induction argument (which we leave as an exercise) will show that \(f\,^{(n)}(z) =\frac{(n+1)!}{(1-z)^{n+2}}\) for \(z \in D_1(0)\text{.}\) Thus \(f\,^{(n)}(0)=(n+1) !\text{,}\) and Taylor’s theorem gives
If we let \(z^2\) take the role of \(z\) in Equation (7.2.9), we get that \(\frac{1}{1-z^2}= \sum\limits_{n=0}^{\infty}(z^2)^n=\sum\limits_{n=0}^{\infty}z^{2n}\) for \(z^2 \in D_1(0)\text{.}\) But \(z^2 \in D_1(0)\) iff \(z \in D_1(0)\text{.}\) Letting \(-z^2\) take the role of \(z\) in Equation (7.2.9) gives the second part of Equations (7.2.8).
For many students, it makes sense that the first series in Equations (7.2.10) converges only on the interval \((-1,1)\) because \(\frac{1}{1-x^2}\) is undefined at the points \(x=\pm 1\text{.}\) It seems unclear as to why this should also be the case for the series representing \(\frac{1}{1+x^2}\text{,}\) since the real-valued function \(f(x) =\frac{1 }{1+x^2}\) is defined everywhere. The explanation, of course, comes from the complex domain. The complex function \(f(z) =\frac{1}{1+z^2}\) is not defined everywhere. In fact, the singularities of \(f\) are at the points \(\pm i\text{,}\) and the distance between them and the point \(\alpha =0\) equals 1. According to Corollary 7.2.5, therefore, Equations (7.2.8) are valid only for \(z \in D_1(0)\text{,}\) so Equations (7.2.10) are valid only for \(x \in (-1,1)\text{.}\)
Alas, there is a potential fly in this ointment: Corollary 7.2.5 applies to Taylor series. To form the Taylor series of a function, we must compute its derivatives. We didn’t get the series in Equations (7.2.8) by computing derivatives, so how do we know that they are indeed the Taylor series centered at \(\alpha=0\text{?}\) Perhaps the Taylor series would give completely different expressions from those given by Equations (7.2.8). Fortunately, Theorem 7.2.9 removes this possibility.
Computing derivatives for \(f(z)\) would be an onerous task. Fortunately, we can make use of the trigonometric identity
\begin{equation*}
\sin^3 z = \frac{3}{4}\sin z - \frac{1}{4}\sin 3z\text{.}
\end{equation*}
Recall that the series for \(\sin z\) (valid for all \(z\)) is \(\sin z= \sum\limits_{n=0}^{\infty}(-1)^n\frac{z^{2n+1}}{(2n+1)!}\text{.}\) Using the identity for \(\sin ^3z\text{,}\) we obtain
In the preceding argument we used some obvious results of power series representations that we haven’t yet formally stated. The requisite results are part of Theorem 7.2.11.
We leave the details of establishing Equations (7.2.11) and (7.2.12) for you to do as an exercise. To establish Equation (7.2.13), we note that the function \(h(z) =f(z)g(z)\) is analytic in \(D_r(\alpha)\text{.}\) Thus, for \(z \in D_r(\alpha)\text{,}\)
By mathematical induction, we can generalize the preceding pattern to the \(n\) th derivative, giving Leibniz’s formula for the derivative of a product of functions:
We let \(f(z) = g(z) = \frac{1}{1-z}=\sum\limits_{n=0}^{\infty}z^n\text{,}\) for \(z \in D_1(0)\text{.}\) In terms of Theorem 7.2.11, we have \(a_n=b_n=1\text{,}\) for all \(n\text{,}\) and thus Equation (7.2.13) gives
\(\frac{1-z}{z-2} = \frac{z-1}{1-(z-1)} = (z-1)\left[\frac{1}{1-(z-1)}\right]\text{.}\) Expand the expression in brackets by replacing \(z\) with \(z-1\) in the geometric series (valid, therefore, for \(|z-1|\lt 1\)), then multiply by the \((z-1)\) term.
\(f(z) = \frac{1}{1-z} = \frac{1}{1-i}\left[\frac{1}{1-\frac{z-i}{1-i}}\right]\text{.}\) Expand the expression in brackets by replacing \(z\) with \(\frac{z-i}{1-i}\) in the geometric series (valid, therefore, for \(|\frac{z-i}{1-i}|\lt 1\text{,}\) or \(|z-i|\lt \sqrt{2}\)). Explain.
Let \(f(z) =(1+z)^{\beta }=\exp[\beta\mathrm{Log}(1+z)]\) be the principal branch of \((1+z)^{\beta }\text{,}\) where \(\beta\) is a fixed complex number. Establish the validity for \(z \in D_1(0)\) of the binomial expansion
Let \(f(z) =\sum\limits_{n=0}^{\infty}c_nz^n=1+z+2z^2+3z^3+5z^4+8z^5+13z^6+\cdots\text{,}\) where the coefficients \(c_n\) are the Fibonacci numbers defined by \(c_0=1\text{,}\)\(c_1=1\text{,}\) and \(c_n=c_{n-1}+c_{n-2}\text{,}\) for \(n \ge 2\text{.}\)
Observe that \(1+zf(z)+z^2f(z) = 1 + \sum\limits_{n=0}^{\infty}c_nz^{n+1} +
\sum\limits_{n=0}^{\infty}c_nz^{n+2}\text{.}\) Reindex and write this as \(1+\sum\limits_{n=1}^{\infty}c_{n-1}z^n +
\sum\limits_{n=2}^{\infty}c_{n-2}z^n = 1 + z + \sum\limits_{n=2}^{\infty}(c_{n-1}+c_{n-2})z^n\text{.}\) Now use the relation \(c_n=c_{n-1}+c_{n-2}\) for \(n \ge 2\) to conclude \(1+zf(z)+z^2f(z) = f(z)\text{,}\) then solve for \(f(z)\text{.}\)
We used Lemma 7.2.2 in establishing Identity (7.2.2). However, Lemma 7.2.2 is valid provided \(z \ne z_0\) and \(z \ne\alpha\text{.}\) Explain why these conditions are indeed the case in Identity (7.2.2).
The point \(z\) is on the circle \(C_{\rho }(\alpha)\) with center \(\alpha\text{,}\) so \(z \ne \alpha\text{.}\) Also, \(z_0\) is in the interior of this circle, so again \(z\ne z_0\text{.}\)
Let \(f\) be defined in a domain that contains the origin. The function \(f\) is said to be even if \(f(-z) =f(z)\text{,}\) and it is called odd if \(f(-z) = -f(z)\text{.}\)
By definition, \(f(-z) = -f(z)\text{,}\) so using the chain rule, we see that \(f\,'(z) = \frac{d}{dz}f(z) = -\frac{d}{dz}f(-z) = -f\,'(-z)(-1) = f\,'(-z)\text{.}\) But this means that \(f\,'\) is an even function.
If \(f\) is even, then by part b \(f\,'\) is odd, so \(f\,'(0) = -f\,'(-0) = -f\,'(0)\text{.}\) Of course, this implies \(f'(0) = 0\text{.}\) Similarly, from part a \(f\,''\) is even, so \(f\,'''(0)=0\text{.}\) An induction argument gives \(f\,^{(2n-1)}(0)=0\) for all positive integers \(n\text{.}\) Show the details.
Obviously, the radius of convergence of this series equals 1 (ratio test). However, the distance between 0 and the nearest singularity of \(f\) equals \(\frac{1}{2}\text{.}\) Explain why this condition does not contradict Corollary 7.2.5.
The point \(z=\frac{1}{2}\) is a removable singularity, since \(f\) may be redefined at \(\frac{1}{2}\) to be analytic. State what \(f\) should equal at that point.
Explain why the function \(f\) is an example of a function that, although differentiable everywhere on the real line, is not expressible as a Taylor series about 0 that is valid for any interval \((-\varepsilon, \varepsilon)\text{,}\) no matter how small \(\varepsilon\) is. Hint.
\(\text{First, evaluate the Taylor series representation for } \, f(x) \, \text{ when } \, x \ne 0.\)
\(\text{Then show that the series does not equal } \, f(x)\text{.}\)
