Calculus of power series

-

Functions defined by power series

If the power series \(\sum_{n=0}^{\infty}a_nx^n\) has the radius of convergence \(r > 0\), then it defines a function \(f(x) = \sum_{n=0}^{\infty}a_nx^n\) on open interval \((-r, r)\).



Theorem (Continuous power series)

Let \(\sum_{n=0}^{\infty}a_nx^n\) be a power series with its radius of convergence \(r > 0\). Then, the function \(f(x) = \sum_{n=0}^{\infty}a_nx^n\) is continuous on \((-r, r)\).

Proof. It suffices to show that \(f(x)\) is continuous on open interval \(I = (-s, s)\) where \(s\) is an arbitrary real number such that \(0 < s < r\). Let \(t = \frac{r + s}{2}\). Then \(s < t < r\) and the power series \(\sum_{n=0}^{\infty}a_nx^n\) converges absolutely at \(x = t\).

Step 1. For the partial sum \(f_n(x) = \sum_{k=0}^{n}a_kx^k\), we show the following:

  • (*) For any \(\varepsilon > 0\), there exists an \(N\in\mathbb{N}_0\) such that, for all \(n\in\mathbb{N}_0\) and all \(x\in I\), if \(n \geq N\), then \(|f(x) - f_n(x)| < \varepsilon\).
(In other words, \(\{f_n(x)\}\) uniformly converges to \(f(x)\).) Take an arbitrary \(\varepsilon > 0\). Since the series \(\sum_{n=0}^{\infty}|a_n|t^n\) has a sum, \(\lim_{n\to\infty}|a_n|t^n = 0\). In particular, the sequence \(\{|a_n|t^n\}\) is bounded so that there exists \(M>0\) such that \(|a_n|t^n < M\) for all \(n\). Also, since \(0 < \frac{s}{t} < 1\), the geometric series \(\sum_{n=0}^{\infty}\left(\frac{s}{t}\right)^n\) converges. Therefore, we can find a sufficiently large \(N\) such that \(\sum_{k=n+1}^{\infty}\left(\frac{s}{t}\right)^k < \frac{\varepsilon}{M}\) for all \(n \geq N\).
Noting that, for all \(n\geq N\) and \(x \in (-s, s)\), \(|x| < s\), we have
\[\begin{eqnarray*} |f(x) - f_n(x)| & = & \left|\sum_{k=n+1}^{\infty}a_kx^k\right| \leq \sum_{k=n+1}^{\infty}|a_k||x|^k\\ & = & \sum_{k=n+1}^{\infty}|a_k|t^k\left(\frac{|x|}{t}\right)^k < \sum_{k=n+1}^{\infty}|a_k|t^k\left(\frac{s}{t}\right)^k\\ & < & \sum_{k=n+1}^{\infty}M\left(\frac{s}{t}\right)^k < \varepsilon. \end{eqnarray*}\]
Thus, (*) holds. This means that the function \(f(x)\) can be approximated by a polynomial function \(f_n(x)\) to arbitrary precision. In other words, \(f(x) = \lim_{n\to\infty}f_n(x)\).

Step 2. Next, consider a sequence of functions \(\{f_n(x)\}\) that are continuous on interval \(I\) and satisfy the property (*) in Step 1. Then, we show that its limit \(f(x) = \lim_{n\to\infty}f_n(x)\) is also continuous on \(I\).
Take an arbitrary \(a \in I\). Let \(\varepsilon > 0\). By (*), for all \(x \in I\), there exists some \(n\) such that \(|f(x) - f_n(x)| < \frac{\varepsilon}{3}\); in particular, \(|f(a) - f_n(a)| < \frac{\varepsilon}{3}\). Furthermore, since \(f_n(x)\) is continuous, there exists some \(\delta > 0\) such that, if \(|x - a| < \delta\), then \(|f_n(x)  - f_n(a)| < \frac{\varepsilon}{3}\). Combining these together, we have
\[\begin{eqnarray*} |f(x) - f(a)| & \leq & |f(x) - f_n(x)| + |f_n(x) - f_n(a)| + |f_n(a) - f(a)|\\ & < & \frac{\varepsilon}{3} +\frac{\varepsilon}{3} +\frac{\varepsilon}{3} = \varepsilon. \end{eqnarray*}\]
Now,
\[\sum_{n=0}^{\infty}a_nx^n = \lim_{n\to\infty}\sum_{k=0}^{n}a_kx^k\]
and polynomial functions (i.e., partial sums) are continuous (everywhere). Therefore, \(f(x)\) is continuous on \(I\). ■

Term-wise integration and term-wise differentiation

Given the power series \(f(x) = \sum_{n=0}^{\infty}a_nx^n\), consider the following power series:
\[\begin{eqnarray*} F(x) &=& \sum_{n=0}^{\infty}\frac{a_n}{n+1}x^{n+1},\\ g(x) &=& \sum_{n=1}^{\infty}na_nx^{n-1}. \end{eqnarray*}\]
If \(f(x)\) is a finite sum (i.e., polynomial function; \(\exists N\in\mathbb{N} ~ \forall n \geq N ~ (a_n = 0)\)), then \(F(x)\) and \(g(x)\) are the primitive function and derivative of \(f(x)\), respectively. As we show below, this holds true even when \(f(x)\) is not a polynomial. That is, if \(f(x)\) has the radius of convergence \(r > 0\), then \(F(x)\) is the primitive function of \(f(x)\) and \(g(x)\) is the derivative of \(f(x)\) on the open interval \((-r, r)\). This means that we can obtain the primitive function and derivative of a function defined by a power series by term-wise (term-by-term) integration and term-wise (term-by-term) differentiation of \(f(x)\), respectively.

Theorem (Term-wise integration of power series)

Suppose the power series \(f(x) = \sum_{n=0}^{\infty}a_nx^n\) has a radius of convergence \(r>0\). Then, the following holds on \((-r, r)\):
\[\int_0^xf(t)dt = \sum_{n=0}^{\infty}\frac{a_n}{n+1}x^{n+1}.\]
Proof. Consider the sequence \(\{f_n(x)\}\) of polynomial functions given by the partial sums \(f_n(x) = \sum_{k=0}^{n}a_kx^k\). For \(x\in (-r, r)\), take an \(s\in\mathbb{R}\) such that \(|x| < s < r\). Then, on the interval \(I = (-s, s)\), we have \(f(x) = \lim_{n\to\infty}f_n(x)\) (c.f., the property (*) in the proof of the above Theorem (Continuous power series)). For any \(\varepsilon > 0\), there exists an \(N\in \mathbb{N}_0\) such that, for all \(n\geq N\), \(|f(x) - f_n(x)| < \varepsilon\). Thus, if \(n \geq N\), then
\[\left|\int_0^xf(t)dt - \int_0^xf_n(t)dt\right| \leq \int_0^x|f(t) - f_n(t)|dt < \varepsilon|x|.\]
Therefore,
\[\lim_{n\to\infty}\int_0^xf_n(t)dt =\int_0^xf(t)dt.\]
Since
\[\int_0^x\left(\sum_{k=0}^{n}a_kt^k\right) dt = \sum_{k=0}^{n}\left(\int_0^xa_kt^k dt\right) = \sum_{k=0}^{n}\frac{a_k}{k+1}x^{k+1},\]
we have
\[\int_0^xf(t)dt = \sum_{n=0}^{\infty}\frac{a_n}{n+1}x^{n+1}.\]
■

Theorem (Term-wise differentiation of power series)

Given the power series \(f(x) = \sum_{n=0}^{\infty}a_nx^n\), consider the power series \(g(x) = \sum_{n=1}^{\infty}na_nx^{n-1}\). The following hold:
  1. The radius of convergence of \(g(x)\) is equal to that of \(f(x)\).
  2. If \(f(x)\) has a radius of convergence \(r > 0\), then \(f(x)\) is differentiable on the open interval \((-r, r)\) and \(f'(x) = g(x)\).
Proof. 
  1. Let \(r\) and \(r'\) be the radii of convergence of \(f(x)\) and \(g(x)\), respectively. First, we show that \(r \leq r'\). This is trivial if \(r = 0\). Suppose \(r > 0\). It suffices to show that \(g(x)\) converges absolutely for all \(x\) such that \(|x| < r\). Let \(s\) be such that \(|x| < s < r\). Since the series \(\sum_{n=0}^{\infty}|a_n|s^n\) has a sum, there exists \(M\) such that \(|a_n|s^n < M\) for all \(n\). Let \(t = \frac{|x|}{s}\), then \(0 < t < 1\) and \[\sum_{n=1}^{\infty}|na_nx^{n-1}| = \sum_{n=1}^{\infty}\frac{n}{s}|a_ns^n|t^{n-1} < M\sum_{n=1}^{\infty}\frac{n}{s}t^{n-1}.\tag{Eq:gbound}\] But \[\lim_{n\to\infty}\frac{\frac{n+1}{s}t^n}{\frac{n}{s}t^{n-1}} = \lim_{n\to\infty}\frac{(n+1)t}{n} = t < 1,\] so by D'Alembert's criterion, the right-hand side of (Eq:gbound) has a sum, and hence \(g(x)\) converges absolutely. Next, we show that \(r' \leq r\). This is trivial if \(r' = 0\). Suppose \(r' > 0\). It suffices to show that \(f(x)\) converges absolutely for all \(x \in (-r', r)\). But this is true because for all \(n\geq 1\), \[|a_nx^n| \leq |x||na_nx^{n-1}| < r'|na_nx^{n-1}|.\] That is, \(r'g(x)\) is a dominating series of \(f(x)\). Therefore, we have shown that \(r = r'\).
  2. By Part 1, \(g(x)\) is a power series with the radius of convergence \(r > 0\). By term-wise integration, for all \(x \in (-r, r)\), we have \[\int_0^xg(t)dt = \sum_{n=1}^{\infty}a_nx^n = f(x) - a_0.\] The left-hand side is differentiable with respect to \(x\). Thus, \(f(x)\) is also differentiable. By differentiating both sides, we have \(f'(x) = g(x)\) (c.f. the Fundamental Theorem of Calculus).
■

Corollary 1

Suppose the power series \(f(x) = \sum_{n=0}^{\infty}a_nx^n\) has a radius of convergence \(r > 0\). Then, the function \(f(x)\) is of class \(C^\infty\) on the open interval \((-r, r)\), and its derivative \(f'(x)\) is the power series \(\sum_{n=1}^{\infty}na_nx^{n-1}\) with the radius of convergence \(r\).
Proof. Exercise. ■

Corollary 2

Suppose the power series \(f(x) = \sum_{n=0}^{\infty}a_nx^n\) has a radius of convergence \(r > 0\). For each \(k = 0, 1, 2, \cdots\), we have
\[a_k = \frac{f^{(k)}(0)}{k!}\tag{Eq:maccof}\]
where \(f^{(k)}(x)\) is the \(k\)-th derivative of \(f(x)\).
Proof. For \(k = 0\), (Eq:maccof) is trivial. By repeatedly term-wise differentiating \(f(x)\), we can show by mathematical induction (exercise!) that
\[f^{(k)}(x) = \sum_{n=k}^{\infty}n(n-1)\cdots(n-k+1)a_nx^{n-k}.\]
Substituting \(x = 0\) to this, we have
\[f^{(k)}(0) = k!a_k.\]
■



Comments

Post a Comment

Popular posts from this blog

Applications of multiple integrals

-
Image
We can use multiple integrals to compute areas and volumes of various shapes. Area of a planar region Definition (Area) Let \(D\) be a bounded closed region in \(\mathbb{R}^2\). \(D\) is said to have an area if the multiple integral of the constant function 1 over \(D\), \(\iint_Ddxdy\), exists. Its value is denoted by \(\mu(D)\): \[\mu(D) = \iint_Ddxdy.\] Example . Let us calculate the area of the disk \(D = \{(x,y)\mid x^2 + y^2 \leq a^2\}\). Using the polar coordinates, \(x = r\cos\theta, y = r\sin\theta\), \(dxdy = rdrd\theta\), and the ranges of \(r\) and \(\theta\) are \([0, a]\) and \([0, 2\pi]\), respectively. Thus, \[\begin{eqnarray*} \mu(D) &=& \iint_Ddxdy\\ &=&\int_0^a\left(\int_0^{2\pi}rd\theta\right)dr\\ &=&2\pi\int_0^a rdr\\ &=&2\pi\left[\frac{r^2}{2}\right]_0^a = \pi a^2. \end{eqnarray*}\] □ Volume of a solid figure Definition (Volume) Let \(V\) be a solid figure in the \((x,y,z)\) space \(\mathbb{R}^3\). \(V\) is...
Read more

Birth process

-
Image
  Defining the birth process  Consider a colony of bacteria that never dies. We study the following process known as  the birth process , also known as  the Yule process . The colony starts with \(n_0\) cells at time \(t = 0\). Assume that the probability that any individual cell divides in the time interval \((t, t + \delta t)\) is proportional to \(\delta t\) for small \(\delta t\). Further assume that each cell division is independent of others. Let \(\lambda\) be the  birth rate.  The probability of a cell division for a population of \(n\) cells during \(\delta t\) is \(\lambda n \delta t\). We assume that the probability that two or more births take place in the time interval \(\delta t\) is \(o(\delta t)\). That is, it can be ignored. Consequently, the probability that no cell divides during \(\delta t\) is \(1 - \lambda n \delta t - o(\delta t)\). Note that this process is an example of the Markov chain with states \({n_0}, {n_0 + 1}, {n_0 + 2}...
Read more

Improper multiple integrals

-
Image
Consider integrating a function \(f(x,y)\) over a region \(D\) which may not be bounded or closed. In the case of a univariate function, this corresponds to the improper integral where we took the limits of the endpoints of a closed interval. In the case of multiple integrals, we adopt the notion of a "sequence of regions." Consider a sequence of regions \(\{K_n\}\) where each \(K_n\) is a subset of \(\mathbb{R}^2\) that satisfies the following conditions: (a) \(K_1 \subset K_2\)\(\subset \cdots \subset\) \(K_n \subset K_{n+1} \subset \cdots\). (b) For all \(n\in \mathbb{N}\), \(K_n \subset D\). (c) For all \(n \in\mathbb{N}\), \(K_n\) is bounded and closed. (d) For any bounded closed set \(F\) that is included in \(D\) (i.e., \(F \subset D\)), if \(n\) is sufficiently large, then \(F \subset K_n\).  In other words: for all bounded closed \(F \subset D\), there exists some \(N\in \mathbb{N}\) such that, for all \(n\in \mathbb{N}\), if \(n \geq N\) then \(F \subset K_...
Read more