Tag: Series

Prove some properties of the Bessel functions of the first kind of orders zero and one

by RoRi

April 22, 2016

We define the Bessel functions of the first kind of orders zero and one by

$J_0 (x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(n!)^2 2^{2n}}, \qquad J_1 (x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{n!(n+1)! 2^{2n+1}}.$

Prove that both $J_0(x)$ and $J_1(x)$ converge for all $x \in \mathbb{R}$ .
Prove that $J'_0(x) = -J_1(x)$ .
If we define two new functions
$j_0(x) = xJ_0(x), \qquad j_1 (x) = x J_1(x)$

prove that $j_0(x) = j_1'(x)$ .

Proof. For the order zero Bessel function of the first kind $J_0$ we have $a_n = \frac{x^{2n}}{(n!)^2 2^{2n}}$ and so using the ratio test we have
$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{x^{2n+2}}{(n+1)!^2 2^{2n+2}} \right) \left( \frac{(n!)^2 2^{2n}}{x^{2n}} \right) \\[9pt] &= \lim_{n \to \infty} \frac{x^2}{4 (n+1)^2} \\[9pt] &= 0. \end{align*}$

Hence, $J_0 (x)$ converges for all $x \in \mathbb{R}$ .

For $J_1(x)$ we have $a_n = \frac{x^{2n+1}}{n!(n+1)! 2^{2n+1}}$ and so

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{x^{2n+3}}{(n+1)! (n+2)! 2^{2n+3}} \right) \left( \frac{n!(n+1)! 2^{2n+1}}{x^{2n+1}} \right) \\[9pt] &= \lim_{n \to \infty} \frac{x^2}{4(n+1)(n+2)} \\[9pt] &= 0. \end{align*}$

Hence, $J_1(x)$ converges for all $x \in \mathbb{R}. \qquad \blacksquare$
Proof. We compute the derivative of $J_0(x)$ directly,
$\begin{align*} && J_0(x) &= \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(n!)^2 2^{2n}} \\[9pt] \implies && J_0'(x) &= \sum_{n=1}^{\infty} (-1)^n \frac{2nx^{2n-1}}{(n!)^2 2^{2n}} \\[9pt] &&&= \sum_{n=1}^{\infty} (-1)^n \frac{x^{2n-1}}{(n-1)!n! 2^{2n-1}} \\[9pt] &&&= \sum_{n=0}^{\infty} (-1)^{n+1} \frac{x^{2(n+1)-1}}{(n+1-1)!(n+1)! 2^{2(n+1)-1}} \\[9pt] &&&= - \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{n! (n+1)! 2^{2n+1}} \\[9pt] &&&= - J_1(x). \qquad \blacksquare \end{align*}$
Proof. First, we have
$j_0 (x) = x J_0(x) = x \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(n!)^2 2^{2n}} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(n!)^2 2^{2n}}.$

On the other hand we have,

$j_1 (x) = xJ_1(x) = x \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{n!(n+1)! 2^{2n+1}} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+2}}{n!(n+1)! 2^{2n+1}}.$

Therefore,

$\begin{align*} j_1'(x) &= \sum_{n=0}^{\infty} (-1)^n \frac{(2n+2)x^{2n+1}}{n! (n+1)! 2^{2n+1}} \\[9pt] &= \sum_{n=0}^{\infty} (-1)^n \frac{2(n+1)x^{2n+1}}{n! (n+1)! 2^{2n+1}} \\[9pt] &= \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(n!)^2 2^{2n}}. \end{align*}$

Hence, we indeed have $j_0(x) = j_1'(x). \qquad \blacksquare$

Compute the sum of the series ∑ (x-1)ⁿ / (n+2)!

by RoRi

March 31, 2016

Find all $x \in \mathbb{R}$ such that the series

$\sum_{n=0}^{\infty} \frac{(x-1)^n}{(n+2)!}$

converges and compute the sum.

First, we apply the ratio test,

$\begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left( \frac{|x-1|^{n+1}}{(n+3)!} \right) \left( \frac{(n+2)!}{|x-1|^n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{|x-1|}{n+3} \\[9pt] &= 0 \end{align*}$

for all $x$ . Therefore, the series converges for all $x \in \mathbb{R}$ . Next, we compute the sum for $x \neq 1$ ,

$\begin{align*} \sum_{n=0}^{\infty} \frac{(x-1)^n}{(n+2)!} &= \sum_{n=2}^{\infty} \frac{(x-1)^{n-2}}{n!} \\[9pt] &= \frac{1}{(x-1)^2} \sum_{n=2}^{\infty} \frac{(x-1)^n}{n!} \\[9pt] &= \frac{1}{(x-1)^2} \left( \sum_{n=0}^{\infty} \frac{(x-1)^n}{n!} - 1 - x + 1 \right) \\[9pt] &= \frac{1}{(x-1)^2} \left( e^{x-1} - x \right). \end{align*}$

In the case $x = 1$ then all of the terms except the $n = 0$ term are 0, and that term is $\frac{1}{2}$ ; hence, the sum is $\frac{1}{2}$ .

Compute the sum of the series ∑ xⁿ / (n+3)!

by RoRi

March 31, 2016

Find all $x \in \mathbb{R}$ such that the series

$\sum_{n=0}^{\infty} \frac{x^n}{(n+3)!}$

converges and compute the sum.

First, we apply the ratio test

$\begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left( \frac{|x|^{n+1}}{(n+4)!} \right) \left( \frac{(n+3)!}{|x|^n} \right)\\[9pt] &= \lim_{n \to \infty} \frac{|x|}{n+4} \\[9pt] &= 0 \end{align*}$

for all $x$ . Therefore this converges for all $x \in \mathbb{R}$ . Furthermore, we compute the sum, for $x \neq 0$ (since we’ll want to divide by $x$ in here),

$\begin{align*} \sum_{n=0}^{\infty} \frac{x^n}{(n+3)!} &= \sum_{n=3}^{\infty} \frac{x^{n-3}}{n!} \\[9pt] &= \frac{1}{x^3} \sum_{n=3}^{\infty} \frac{x^n}{n!} \\[9pt] &= \frac{1}{x^3} \left( \sum_{n=0}^{\infty} \frac{x^n}{n!} - 1 - x - \frac{x^2}{2} \right) \\[9pt] &= \frac{1}{x^3} \left( e^x - 1 - x - \frac{x^2}{2} \right). \end{align*}$

If $x = 0$ then the sum is 0.

Compute the sum of the series ∑ ((-1)ⁿ x²ⁿ) / n!

by RoRi

March 31, 2016

Find all $x \in \mathbb{R}$ such that the series

$\sum_{n=0}^{\infty} \frac{(-1)^n x^{3n}}{n!}$

converges and compute the sum.

By the ratio test we have

$\begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left( \frac{x^{3n+3}}{(n+1)!} \right) \left( \frac{n!}{x^{3n}} \right) \\[9pt] &= \lim_{n \to \infty} \frac{x^3}{n+1} \\[9pt] &= 0 \end{align*}$

for all $x$ . Thus, the series converges for all $x \in \mathbb{R}$ . Furthermore, we compute the sum from the series expansion of $e^x$ ,

$\sum_{n=0}^{\infty} \frac{(-1)(x^3)^n}{n!} = e^{-x^3}.$

Compute the sum of the series ∑ (((-1)ⁿ) / (2n+1)) (x/2)²ⁿ*

by RoRi

March 31, 2016

Find all $x \in \mathbb{R}$ such that the series

$\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \left( \frac{x}{2} \right)^{2n}$

converges and compute the sum.

First, we have

$\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \left( \frac{x}{2} \right)^{2n} = \sum_{n=0}^{\infty} \frac{(-1)^n \left( \frac{x^2}{4} \right)^n}{2n+1}.$

Then, applying the ratio test we have

$\begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left( \frac{ \left( \frac{x^2}{4} \right)^{n+1}}{2n+3} \right) \left( \frac{2n+1}{\left( \frac{x^2}{4} \right)^n} \right) \\[9pt] &= \left( \frac{x^2}{4} \right) \left( \frac{2n+1}{2n+3} \right) \\[9pt] &= \frac{x^2}{4}. \end{align*}$

Thus, the series converges for all $x$ with $x^2 < 4$ which implies $|x| < 2$ . Furthermore, if $x = \pm 2$ , then the series converges as the alternating harmonic series. Then, we compute the sum for $|x| < 2$ ,

$\begin{align*} \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \left( \frac{x}{2} \right)^{2n} &= \left( \frac{2}{x} \right) \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \left( \frac{x}{2} \right)^{2n+1} \\[9pt] &= \left(\frac{2}{x} \right) \arctan \left( \frac{x}{2} \right). \end{align*}$

Compute the sum of the series ∑ (2ⁿ xⁿ) / n

by RoRi

March 31, 2016

Find all $x \in \mathbb{R}$ such that the series

$\sum_{n=1}^{\infty} \frac{2^n x^n}{n}$

converges and compute the sum.

First, we apply the ratio test,

$\begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left( \frac{(2|x|)^{n+1}}{n+1} \right) \left( \frac{n}{(2|x|)^n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{|2x|n}{n+1} \\[9pt] &= |2x|. \end{align*}$

Thus, the series converges for $|x| < \frac{1}{2}$ . For the boundary, if $x = -\frac{1}{2}$ then this is the alternating harmonic series which converges. If $x = \frac{1}{2}$ then it is the harmonic series, which diverges. Therefore, the series converges for all $x \in \left[-\frac{1}{2}, \frac{1}{2} \right)$ . Then, we compute,

$\begin{align*} && P(x) &= \sum_{n=1}^{\infty} \frac{(2x)^n}{n} \\[9pt] \implies && P'(x) &= \sum_{n=1}^{\infty} (2x)^{n-1} \\[9pt] &&&= \sum_{n=0}^{\infty} (2x)^n \\[9pt] &&&= \frac{2}{1-2x} \\[9pt] \implies && P(x) &= \int_0^x \frac{2}{1-2t} \, dt \\[9pt] &&&= -\log(1-2x). \end{align*}$

Compute the sum of the series ∑ (-2)ⁿ ((n+2) / (n+1)) xⁿ

by RoRi

March 31, 2016

Find all $x \in \mathbb{R}$ such that the series

$\sum_{n=0}^{\infty} (-2)^n \frac{n+2}{n+1} x^n$

converges and compute the sum.

First, we apply the ratio test,

$\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(-2)^{n+1}(n+3)x^{n+1}}{n+2} \right) \left( \frac{n+1}{(-2)^n (n+2) x^n} \right) \\[9pt] &= \lim_{n \to \infty} \frac{-2x (n+1)(n+3)}{(n+2)^2} \\[9pt] &= -2x. \end{align*}$

Thus, the series converges if $|-2x| < 1$ which implies $|x| < \frac{1}{2}$ . For the boundary, $|x| = \frac{1}{2}$ we have in the case $x = \frac{1}{2}$ ,

$\sum_{n=0}^{\infty} (-2)^n \frac{n+2}{n+1} x^n = \sum_{n=0}^{\infty} (-1)^n \frac{n+2}{n+1}$

which diverges since the terms do not go to 0. Similarly, if $x = -\frac{1}{2}$ we get $\sum \frac{n+2}{n+1}$ which also diverges since the terms do not go to 0. Therefore, the given series converges absolutely on the interval $|x| < \frac{1}{2}$ . Then, to compute the sum for $|x| < \frac{1}{2}$ we have

$\begin{align*} \sum_{n=0}^{\infty} \frac{(-2x)^n (n+2)}{n+1} &= \sum_{n=0}^{\infty} \frac{(-2x)^n (n+1+1)}{n+1} \\[9pt] &= \sum_{n=0}^{\infty} \frac{(-2x)^n (n+1)}{n+1} + \sum_{n=0}^{\infty} \frac{(-2x)^n}{n+1} &(\text{Abs. Conv. Series}) \\[9pt] &= \sum_{n=0}^{\infty} (-2x)^n + \frac{1}{2x} \sum_{n=0}^{\infty} \frac{(-2x)^{n+1}}{n+1} \\[9pt] &= \frac{1}{1+2x} + \frac{\log (1+2x)}{2x}. \end{align*}$

Compute the sum of the series ∑ (-1)ⁿ nxⁿ

by RoRi

March 31, 2016

Find all $x \in \mathbb{R}$ such that the series

$\sum_{n=0}^{\infty} (-1)^n nx^n$

converges and compute the sum.

First, we write

$\sum_{n=0}^{\infty} (-1)^n n x^n = x \cdot \sum_{n=0}^{\infty} (-1)^n nx^{n-1}.$

Then, from Theorem 11.9 we know we can differentiate term-by-term and we have

$\begin{align*} x \cdot \sum_{n=0}^{\infty} (-1)^n nx^{n-1} &= x \cdot D_x \left( \sum_{n=0}^{\infty} (-1)^n x^n \right) \\[9pt] &= x \cdot D_x \left( \sum_{n=0}^{\infty} (-x)^n \right) \\[9pt] &= x \cdot D_x \left( \frac{1}{1+x} \right) &(|x| < 1)\\[9pt] &= x \cdot \left( \frac{-1}{(1+x)^2} \right)\\[9pt] &= \frac{-x}{(1+x)^2}. \end{align*}$

Compute the sum of the series ∑ nxⁿ

by RoRi

March 31, 2016

Find all $x \in \mathbb{R}$ such that the series

$\sum_{n=0}^{\infty} nx^n$

converges and compute the sum.

First, we write

$\sum_{n=0}^{\infty} nx^n = x \cdot \sum_{n=0}^{\infty} nx^{n-1}.$

By Theorem 11.9 we know that we can differentiate a power series term-by-term, so we have

$\begin{align*} x \cdot \sum_{n=0}^{\infty} nx^{n-1} &= x \cdot D_x \left( \sum_{n=0}^{\infty} x^n \right) \\[9pt] &= x \cdot D_x \left( \frac{1}{1-x} \right) &(\text{Geometric series})\\[9pt] &= x \cdot \left( \frac{1}{(1-x)^2} \right) \\[9pt] &= \frac{x}{(1-x)^2}, \end{align*}$

valid for $|x| < 1$ (since this was where the geometric series was valid).

Compute the sum of the series ∑ xⁿ / 3ⁿ⁺¹

by RoRi

March 31, 2016

Find all $x \in \mathbb{R}$ such that the series

$\sum_{n=0}^{\infty} \frac{x^n}{3^{n+1}}$

converges and compute the sum.

From the geometric series expansion we have,

$\begin{align*} \sum_{n=0}^{\infty} \frac{x^n}{3^{n+1}} &= \frac{1}{3} \cdot \sum_{n=0}^{\infty} \left( \frac{x}{3} \right)^n \\[9pt] &= \frac{1}{3} \left( \frac{1}{1-\frac{x}{3}} \right) \\[9pt] &= \frac{1}{3-x}. \end{align*}$

This is valid for $|x| < 3$ .