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

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

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

  1. Prove that both J_0(x) and J_1(x) converge for all x \in \mathbb{R}.
  2. Prove that J'_0(x) = -J_1(x).
  3. 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).


  1. 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

  2. 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*}

  3. 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 ∑ (-2)n ((n+2) / (n+1)) xn

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*}