5.3 Analytic functions represented by uniformly convergent series.

Let \(\sum\limits_{n=0}^\infty f_n(z)\) be a series such that (i) it converges uniformly along a contour \(C\), (ii) \(f_n (z)\) is analytic throughout \(C\) and its interior.

Then \(\sum\limits_{n=0}^\infty f_n(z)\) converges, and the sum of the series is an analytic function throughout \(C\) and its interior.

For let \(a\) be any point inside \(C\); on \(C\), let \(\sum\limits_{n=0}^\infty f_n(z)=\Phi(z)\).

Then \[\begin{align*} \frac{1}{2\pi i}\!\int_C \frac{\Phi(z)}{z-a} dz &= \frac{1}{2\pi i}\!\int_C \left\{\sum_{n=0}^\infty \:\! f_n(z)\right\}\frac{dz}{z-a}\\ \\&=\sum_{n=0}^\infty \left\{\frac{1}{2\pi i}\!\int_C \frac{f_n(z)}{z-a} dz \right\}, \end{align*}\] by §4.7.[1] But this last series, by §5.21, is \(\sum\limits_{n=0}^\infty f_n(a)\); the series under consideration therefore converges at all points inside \(C\); let its sum inside \(C\) (as well as on \(C\)) be called \(\Phi(z)\). Then the function is analytic if it has a unique differential coefficient at all points inside \(C\).

[1]Since \(\left|\, z - a \, \right|^{-1}\) is bounded when \(a\) is fixed and \(z\) is on \(C\), the uniformity of the convergence of \(\sum\limits_{n=0}^\infty \left. f_n(z) \middle/ (z-a) \right.\) follows from that of \(\sum\limits_{n=0}^\infty f_n(z)\).  ↩

But if \(a\) and \(a + h\) be inside \(C\), \[\frac{\Phi(a+h)-\Phi(a)}{h} = \frac{1}{2\pi i}\!\int_C \frac{\Phi(z)\,dz}{(z-a)(z-a-h)},\] and hence, as in §5.22, \(\lim\limits_{h \rightarrow 0} \left[\left\{\Phi(a+h)-\Phi(a)\right\}h^{-1}\right]\) exists and is equal to \[ \frac{1}{2\pi i}\!\int_C \frac{\Phi(z)\,dz}{(z-a)^2};\] and therefore \(\Phi(z)\) is analytic inside \(C\). Further, by transforming the last integral in the same way as we transformed the first one, we see that \(\Phi'(a)=\sum\limits_{n=0}^\infty f'_n (a)\), so that \(\sum\limits_{n=0}^\infty f_n (a)\) may be ‘differentiated term by term.’

If a series of analytic functions converges only at points of a curve which is not closed nothing can be inferred as to the convergence of the derived series.[2]

[2]This might have been anticipated as the main theorem of this section deals with uniformity of convergence over a two-dimensional region.  ↩

Thus \(\sum\limits_{n=1}^\infty (-1)^n \dfrac{\cos nx}{n^2}\) converges uniformly for real values of \(x\) (§3.34). But the derived series \(\sum\limits_{n=1}^\infty (-1)^{n-1} \dfrac{\sin nx}{n}\) converges non-uniformly near \(x=(2m+1)\,\pi\), (\(m\) any integer); and the derived series of this, viz. \(\sum\limits_{n=1}^\infty (-1)^{n-1} \cos nx\), does not converge at all.

Corollary. By §3.7, the sum of a power series is analytic inside its circle of convergence.

5.31 Analytic functions represented by integrals.

Let \(f(t, z)\) satisfy the following conditions when \(t\) lies on a certain path of integration \([a, b]\) and \(z\) is any point of a region \(S\):

  1. \(f\) and \(\dfrac{\partial f}{\partial z}\) are continuous functions of \(t\).
  2. \(f\) is an analytic function of \(z\).
  3. The continuity of \(\dfrac{\partial f}{\partial z}\) qua[3] function of \(z\) is uniform with respect to the variable \(t\).
[3]Editor’s Note: In this usage, the word qua means treated as … ↩
[4]Editor’s Note: What was done in §4.2 (and §4.44 (II) corollary) was for \(f(x, \alpha)\) where \(x\) and \(\alpha\) were real – the complex derivate had not yet been defined. However, the proofs in those sections may be re-interpreted without any changes to apply for complex \(\alpha\). ↩

Then \(\int_a^b f(t,z)\, dt\) is an analytic function of \(z\). For, by §4.2, it has the unique derivate[4] \[\int_a^b\! \frac{\partial f(t,z)}{\partial z}dt.\]

5.32 Analytic functions represented by infinite integrals.

From §4.44 (II) corollary, it follows that \(\int_a^\infty \! f(t,z)\, dt\) is an analytic function of \(z\) at all points of a region \(S\) if

  1. the integral converges,
  2. \(f(t, z)\) is an analytic function of \(z\) when \(t\) is on the path of integration and \(z\) is on \(S\),
  3. \(\dfrac{\partial f(t,z)}{\partial z}\)is a continuous function of both variables,
  4. \(\displaystyle \int_a^\infty \!\frac{\partial f(t,z)}{\partial z}dt\) converges uniformly throughout \(S\).

For if these conditions are satisfied, \(\int_a^\infty f(t, z) \,dt\) has the unique derivate \[\int_a^\infty\! \frac{\partial f(t,z)}{\partial z}dt.\]

A case of very great importance is afforded by the integral \(\int_0^\infty \! e^{-t\:\!z} f(t) \, dt\), where \(f(t)\) is continuous and \(\left|\, f(t)\, \right| < Ke^{r\:\!t}\) where \(K\), \(r\) are independent of \(t\); it is obvious from the conditions stated that the integral is an analytic function of \(z\) when \(\mathfrak{Re}(z) \geq r_1 > r\). [Condition (iv) is satisfied, by §4.431 (I), since \(\int_0^\infty t e^{(r-r_1)\:\!t}dt\) converges.]

5.4 Taylor’s Theorem.[5]

[5]The formal expansion was first published by Dr Brook Taylor (1715) in his Methodus Incrementorum.  ↩

Consider a function \(f(z)\), which is analytic in the neighbourhood of a point \(z = a\). Let \(C\) be a circle with \(a\) as centre in the \(z\)-plane, which does not have any singular point of the function \(f(z)\) on or inside it; so that \(f(z)\) is analytic at all points on and inside \(C\). Let \(z = a + h\) be any point inside the circle \(C\). Then, by §5.21, we have[6] \[f(a+h)=\frac{1}{2\pi i}\! \int_C \frac{f(z)\, dz}{z-a-h}\] \[=\frac{1}{2\pi i}\! \int_C \! f(z)\, dz\left\{\!\frac{1}{z-a}\!+\!\frac{h}{(z-a)^2}\!+\!\cdots \!+\!\frac{h^n}{(z-a)^{n+1}}\!+\!\frac{h^{n+1}}{(z-a)^{n+1}(z-a-h)}\!\right\}\] \[=f(a)+hf'\!(a)+\!\frac{h^2}{2!}f''\!(a)+\! \cdots \!+\!\frac{h^n}{n!}f^{(n)}(a)+\!\frac{1}{2\pi i}\!\int_C \frac{h^{n+1}f(z)\,dz}{(z-a)^{n+1}(z-a-h)}. \]

[6]Editor’s Note: For those unfamiliar with this expansion, it follows from the geometric series formula, \[\begin{align*}\sum_{m=0}^n r^m &= \frac{1-r^{n+1}}{1-r} \\ \frac{1}{1-r} &= \sum_{m=0}^n r^m + \frac{r^{n+1}}{1-r}. \end{align*}\] Hence, if \(a-b=c\), \[\begin{align*}\frac{1}{c} &= \frac{1}{a \left(1- \vphantom{g} b \middle/ a \right)} \\ &= \frac{1}{a}\left(\sum_{m=0}^n \left(\frac{b}{a}\right)^m + \frac{b^{n+1}}{a^n c}\right) .\end{align*}\] Whittaker and Watson make regular use of this expansion (see §5.6). ↩
But when \(z\) is on \(C\), the modulus of \(\dfrac{f(z)}{z-a-h}\) is continuous, and so, by §3.61 cor. (ii), will not exceed some finite number \(M\).

Therefore, by §4.62, \[\left|\,\frac{1}{2\pi i}\!\int_C \frac{h^{n+1}f(z)\,dz}{(z-a)^{n+1}(z-a-h)}\,\right| \leq \frac{M \cdot 2\pi R}{2\pi}\left( \frac{\left|\,h\,\right|}{R}\right)^{n+1},\] where \(R\) is the radius of the circle \(C\), so that \(2\pi R\) is the length of the path of integration in the last integral, and \(R = \left|\,z - a\,\right|\) for points \(z\) on the circumference of \(C\).

The right-hand side of the last inequality tends to zero as \(n \rightarrow \infty\). We have therefore \[f(a + h)=f(a) + hf'(a)+\frac{h^2}{2!}f''(a) + \cdots +\frac{h^n}{n!}f^{(n)}(a) + \cdots, \] which we can write \[f(z) = f(a) + (z-a)f'(a) + \frac{(z-a)^2}{2!} f''(a) + \cdots + \frac{(z-a)^n}{n!}f^{(n)}(a)+ \cdots .\]

This result is known as Taylor’s Theorem; and the proof given is due to Cauchy. It follows that the radius of convergence of a power series is always at least so large as only just to exclude from the interior of the circle of convergence the nearest singularity of the function represented by the series. And by §5.3 corollary, it follows that the radius of convergence is not larger than the number just specified. Hence the radius of convergence is just such as to exclude from the interior of the circle that singularity of the function which is nearest to \(a\).

At this stage we may introduce some terms which will be frequently used.

If \(f(a) = 0\), the function \(f(z)\) is said to have a zero at the point \(z = a\). If at such a point \(f'(a)\) is different from zero, the zero of \(f(a)\) is said to be simple; if, however, \(\,f'(a)\), \(\, f''(a), \dots ,\, f^{(n-1)}(a)\) are all zero, so that the Taylor’s expansion of \(f(z)\) at \(z = a\) begins with a term in \((z - a)^n\), then the function \(f(z)\) is said to have a zero of the nth order at the point \(z = a\).

Example 1. Find the function \(f(z)\), which is analytic throughout the circle \(C\) and its interior, whose centre is at the origin and whose radius is unity, and has the value \[\frac{\alpha-\cos \theta}{\alpha^2-2\alpha \cos \theta + 1}+i\frac{\sin \theta}{\alpha^2-2\alpha \cos \theta + 1}\] (where \(\alpha > 1\) and \(\theta\) is the vectorial angle) at points on the circumference of \(C\).

[We have \[\begin{align*} f^{(n)}(0) &= \frac{n!}{2\pi i}\! \int_C \frac{f(z)\,dz}{z^{n+1}}\\ \\ &=\frac{n!}{2\pi i}\! \int_0^{2\pi} \! e^{-ni\theta} i\,d\theta\left\{\frac{\alpha-\cos \theta+i\sin \theta}{\alpha^2-2\alpha \cos \theta + 1}\right\}, \; \text{(putting}\, z=e^{i\theta}\, \text{)}\\ \\ &=\frac{n!}{2\pi i}\! \int_0^{2\pi} \!\frac{e^{-ni\theta}\, d\theta}{\alpha-e^{-i\theta}}=\frac{n!}{2\pi i}\! \int_C \frac{dz}{z^n(\alpha-z)}=\left[\frac{d^n}{dz^n} \frac{1}{\alpha-z}\right]_{z=0}\\ \\ &=\frac{n!}{\alpha^{n+1}}. \end{align*}\] Therefore by Maclaurin’s Theorem,[7] \[f(z)=\sum_{n=0}^\infty \frac{z^n}{\alpha^{n+1}},\] or \(f(z) = (a - z)^{-1}\) for all points within the circle.

This example raises the interesting question, Will it still be convenient to define \(f(z)\) as \((a - z)^{-1}\) at points outside the circle? This will be discussed in §5.51.]

[7]The result \(f(z) =f(0) +z\,f'(0) +\dfrac{z^2}{2} f''(0) + \cdots\), obtained by putting \(a =0\) in Taylor’s Theorem, is usually called Maclaurin’s Theorem; it was discovered by Stirling (1717) and published by Maclaurin (1742) in his Fluxions.  ↩

Example 2. Prove that the arithmetic mean of all values of \(z^{-n} \sum\limits_{v=0}^\infty a_\nu z^{\nu}\), for points \(z\) on the circumference of the circle \(\left|\, z\,\right| = 1\), is \(a_n\), if \(\sum a_\nu z^\nu\) is analytic throughout the circle and its interior.

[Let \(\sum\limits_{\nu=0}^\infty a_\nu z^\nu = f(z)\), so that \(a_\nu=\dfrac{f^{(n)}(0)}{n!}\). Then, writing \(z = e^{i\theta}\), and calling \(C\) the circle \(\left|\, z\,\right| = 1\),

\(\displaystyle \frac{1}{2 \pi}\!\int_0^{2\pi} \! \frac{f(z)\,d\theta}{z^n} = \frac{1}{2 \pi i}\!\int_C \frac{f(z)\,dz}{z^{n+1}}=\frac{f^{(n)}(0)}{n!}=a_n.\:\! \)]

Example 3. Let \(f(z) = z^r\); then \(f(z + h)\) is an analytic function of \(h\) when \(\left|\, h\,\right| < \left|\, z\, \right|\;\!\) for all values of \(r\); and so \[(z + h)^r =z^r + rz^{r-1} h + \frac{r(r-1)}{2}z^{r-2}h^2 + \cdots,\] this series converging when \(\left|\, h\,\right| < \left|\, z\, \right|\). This is the binomial theorem.

Example 4. Prove that if \(h\) is a positive constant, and \((1+2zh+h^2)^{-\frac{1}{2}}\) is expanded in the form

\(\quad\) \[ \begin{equation}1 + h P_1(z) + h^2 P_2(z) + h^3 P_3(z) + \cdots , \end{equation} \] \(\qquad\) (A) \(\quad\)

(where \(P_n(z)\) is easily seen to be a polynomial of degree \(n\) in \(z\)), then this series converges so long as \(z\) is in the interior of an ellipse whose foci are the points \(z=1\) and \(z= -1\), and whose semi-major axis is \(\frac{1}{2}(h+h^{-1}) \).

Let the series be first regarded as a function of \(h\). It is a power series in \(h\), and therefore converges so long as the point \(h\) lies within a circle in the \(h\)-plane. The centre of this circle is the point \(h=0\), and its circumference will be such as to pass through that singularity of \((1+2zh+h^2)^{-\frac{1}{2}}\) which is nearest to \(h = 0\).

But \[1+2zh+h^2=\left\{h-z+(z^2-1)^{\frac{1}{2}}\right\}\left\{h-z-(z^2-1)^{\frac{1}{2}}\right\},\] so the singularities of \((1+2zh+h^2)^{-\frac{1}{2}}\) are the points \(h=z-(z^2-1)^{\frac{1}{2}}\) and \(h=z+(z^2-1)^{\frac{1}{2}}\). [These singularities are branch points (see §5.7 ).]

[8]Editor’s Note: See “Eccentric Angle.” at MathWorld. ↩

Thus the series (A) converges so long as \(\left|\, h \,\right|\) is less than both \[\left|\, z-(z^2-1)^{\frac{1}{2}} \:\!\right| \quad \text{and} \quad \left|\, z+(z^2-1)^{\frac{1}{2}} \:\!\right|\,. \] Draw an ellipse in the \(z\)-plane passing through the point \(z\) and having its foci at \(\pm 1\). Let \(\alpha\) be its semi-major axis, and \(\theta\) the eccentric angle[8] of \(z\) on it.

Then \[z=\alpha \cos \theta + i\,(\alpha^2-1)^{\frac{1}{2}} \sin \theta,\] which gives \[z \pm (z^2 - 1)^{\frac{1}{2}} =\left\{\alpha \pm (\alpha^2-1)^{\frac{1}{2}}\right\}\left(\cos \theta \pm i \sin \theta \right),\] so \[\left|\,z \pm (z^2 -1)^{\frac{1}{2}}\,\right| = \alpha \pm (\alpha^2 -1)^{\frac{1}{2}}.\]

Thus the series (A) converges so long as \(h\) is less than the smaller of the numbers \(\alpha +(\alpha^2 -1)^{\frac{1}{2}}\) and \(\alpha-(\alpha^2 -1)^{\frac{1}{2}}\), i.e. so long as \(h\) is less than \(\alpha-(\alpha^2 -1)^{\frac{1}{2}}\). But \(h =\alpha-(\alpha^2 -1)^{\frac{1}{2}}\) when \(\alpha =\frac{1}{2}(h+h^{-1}) \).[9]

Therefore the series (A) converges so long as \(z\) is within an ellipse whose foci are \(1\) and \(-1\), and whose semi-major axis is \(\frac{1}{2}(h+h^{-1}) \).

[9]Editor’s Note: It might be helpful to observe, \(h < \alpha-(\alpha^2 -1)^{\frac{1}{2}}\) whenever \(1 \leq \alpha < \frac{1}{2}(h+h^{-1}) \). ↩

5.41 Forms of the remainder in Taylor’s series.

Let \(f(x)\) be a real function of a real variable; and let it have continuous differential coefficients of the first \(n\) orders when \(a \leq x \leq a+h\).

If \(0 \leq t \leq 1\), we have \[\frac{d}{dt} \left\{\sum_{m=1}^{n-1} \frac{h^m}{m!} (1-t)^m f^{(m)}(a+th)\right\}= \frac{h^n (1-t)^{n-1}}{(n-1)!} f^{(n)}(a+th)-hf'(a+th).\]

Integrating this between the limits 0 and 1, we have \[f(a + h) =f(a) +\sum_{m=1}^{n-1}\frac{h^m}{m!} f^{(m)}(a)+ \int_0^1 \! \frac{h^n(1-t)^{n-1}}{(n-1)!} f^{(n)}(a+th)\, dt.\]

Let \[R_n=\frac{h^n}{(n-1)!}\!\int _0^1 \! (1-t)^{n-1} f^{(n)}(a+th)\,dt;\] and let \(p\) be a positive integer such that \(p \leq n\).

Then \[R_n=\frac{h^n}{(n-1)!}\!\int _0^1 \! (1-t)^{p-1}(1-t)^{n-p} f^{(n)}(a+th)\,dt.\]

Let \(U\), \(L\) be the upper and lower bounds of \((1-t)^{n-p} f^{(n)}(a+th)\). Then \[\int_0^1 \!L(1-t)^{p-1}\,dt < \int_0^1 \! (1-t)^{p-1}(1-t)^{n-p} f^{(n)}(a+th)\,dt < \int_0^1 \!U(1-t)^{p-1}\,dt.\]

Since \((1-t)^{n-p} f^{(n)}(a+th)\) is a continuous function it passes through all values between \(U\) and \(L\), and hence we can find \(\theta \) such that \(0 \leq \theta \leq 1 \), and \[\int _0^1 \! (1-t)^{n-1} f^{(n)}(a+th)\,dt = p^{-1}(1-\theta)^{n-p}f^{(n)}(a+\theta h).\] Therefore \[R_n = \frac{h^n}{(n-1)! p}(1-\theta)^{n-p}f^{(n)}(a+ \theta h). \]

Writing \(p = n\),we get \(\displaystyle R_n = \frac{h^n}{n!}f^{(n)}(a+\theta h)\), which is Lagrange’s form for the remainder; and writing \(p = 1\), we get \(\displaystyle R_n = \frac{h^n}{(n-1)!}(1-\theta)^{n-1}f^{(n)}(a+\theta h)\), which is Cauchy’s form for the remainder.

Taking \(n=1\) in this result, we get \[f(a+h)-f(a) = hf'(a+\theta h)\] if \(f'(x)\) is continuous when \(a \leq x \leq a+h\); this result is usually known as the First Mean Value Theorem (see also §4.14).

Darboux gave in 1876 (Journal de Math. (3) ii. p. 291) a form for the remainder in Taylor’s Series, which is applicable to complex variables and resembles the above form given by Lagrange for the case of real variables.