ANSWER on Question #84506 – Math – Differential Equations
QUESTION
Show that
1)
J 1 2 ( x ) = 2 π x ⋅ cos ( x ) J _ {\frac {1}{2}} (x) = \sqrt {\frac {2}{\pi x}} \cdot \cos (x) J 2 1 ( x ) = π x 2 ⋅ cos ( x )
2)
d d x ( x ⋅ J 1 ( x ) ) = x ⋅ J 0 ( x ) \frac {d}{d x} \big (x \cdot J _ {1} (x) \big) = x \cdot J _ {0} (x) d x d ( x ⋅ J 1 ( x ) ) = x ⋅ J 0 ( x )
where J x ( x ) J_{x}(x) J x ( x )
(More information: https://en.wikipedia.org/wiki/Bessel_function)
SOLUTION
2) To prove this formula, we will prove the general formulas:
{ J n − 1 ( x ) + J n + 1 ( x ) = 2 n x ⋅ J n ( x ) J n − 1 ( x ) − J n + 1 ( x ) = 2 ⋅ J n ′ ( x ) → d d x ( x n ⋅ J n ( x ) ) = x n ⋅ J n − 1 ( x ) \left\{ \begin{array}{l} J _ {n - 1} (x) + J _ {n + 1} (x) = \frac {2 n}{x} \cdot J _ {n} (x) \\ J _ {n - 1} (x) - J _ {n + 1} (x) = 2 \cdot J _ {n} ^ {\prime} (x) \end{array} \right. \to \frac {d}{d x} \big (x ^ {n} \cdot J _ {n} (x) \big) = x ^ {n} \cdot J _ {n - 1} (x) { J n − 1 ( x ) + J n + 1 ( x ) = x 2 n ⋅ J n ( x ) J n − 1 ( x ) − J n + 1 ( x ) = 2 ⋅ J n ′ ( x ) → d x d ( x n ⋅ J n ( x ) ) = x n ⋅ J n − 1 ( x )
For this we use the generating function:
e ( x 2 ) ( t − 1 t ) = ∑ n = − ∞ ∞ t n ⋅ J n ( x ) e ^ {\left(\frac {x}{2}\right) \left(t - \frac {1}{t}\right)} = \sum_ {n = - \infty} ^ {\infty} t ^ {n} \cdot J _ {n} (x) e ( 2 x ) ( t − t 1 ) = n = − ∞ ∑ ∞ t n ⋅ J n ( x )
(More information: https://en.wikipedia.org/wiki/Bessel_function)
2 FORMULA:
d d x ∣ e ( x 2 ) ( t − 1 t ) = ∑ n = − ∞ ∞ t n ⋅ J n ( x ) → \frac {d}{d x} \left| e ^ {\left(\frac {x}{2}\right) (t - \frac {1}{t})} = \sum_ {n = - \infty} ^ {\infty} t ^ {n} \cdot J _ {n} (x) \rightarrow \right. d x d ∣ ∣ e ( 2 x ) ( t − t 1 ) = n = − ∞ ∑ ∞ t n ⋅ J n ( x ) → e ( x 2 ) ( t − 1 t ) ⋅ 1 2 ( t − 1 t ) = ∑ n = − ∞ ∞ t n ⋅ J n ′ ( x ) → ( ∑ n = − ∞ ∞ t n ⋅ J n ( x ) ) ⋅ 1 2 ( t − 1 t ) = ∑ n = − ∞ ∞ t n ⋅ J n ′ ( x ) ∣ ⋅ ( 2 ) e ^ {\left(\frac {x}{2}\right) \left(t - \frac {1}{t}\right)} \cdot \frac {1}{2} \left(t - \frac {1}{t}\right) = \sum_ {n = - \infty} ^ {\infty} t ^ {n} \cdot J _ {n} ^ {\prime} (x) \rightarrow \left(\sum_ {n = - \infty} ^ {\infty} t ^ {n} \cdot J _ {n} (x)\right) \cdot \frac {1}{2} \left(t - \frac {1}{t}\right) = \sum_ {n = - \infty} ^ {\infty} t ^ {n} \cdot J _ {n} ^ {\prime} (x) \Bigg | \cdot (2) e ( 2 x ) ( t − t 1 ) ⋅ 2 1 ( t − t 1 ) = n = − ∞ ∑ ∞ t n ⋅ J n ′ ( x ) → ( n = − ∞ ∑ ∞ t n ⋅ J n ( x ) ) ⋅ 2 1 ( t − t 1 ) = n = − ∞ ∑ ∞ t n ⋅ J n ′ ( x ) ∣ ∣ ⋅ ( 2 ) ( ∑ n = − ∞ ∞ t n ⋅ J n ( x ) ) ⋅ ( t − 1 t ) = 2 ⋅ ∑ n = − ∞ ∞ t n ⋅ J n ′ ( x ) → \left(\sum_ {n = - \infty} ^ {\infty} t ^ {n} \cdot J _ {n} (x)\right) \cdot \left(t - \frac {1}{t}\right) = 2 \cdot \sum_ {n = - \infty} ^ {\infty} t ^ {n} \cdot J _ {n} ^ {\prime} (x) \rightarrow ( n = − ∞ ∑ ∞ t n ⋅ J n ( x ) ) ⋅ ( t − t 1 ) = 2 ⋅ n = − ∞ ∑ ∞ t n ⋅ J n ′ ( x ) → ∑ n = − ∞ ∞ t n + 1 ⋅ J n ( x ) − ∑ n = − ∞ ∞ t n − 1 ⋅ J n ( x ) = 2 ⋅ ∑ n = − ∞ ∞ t n ⋅ J n ′ ( x ) \sum_ {n = - \infty} ^ {\infty} t ^ {n + 1} \cdot J _ {n} (x) - \sum_ {n = - \infty} ^ {\infty} t ^ {n - 1} \cdot J _ {n} (x) = 2 \cdot \sum_ {n = - \infty} ^ {\infty} t ^ {n} \cdot J _ {n} ^ {\prime} (x) n = − ∞ ∑ ∞ t n + 1 ⋅ J n ( x ) − n = − ∞ ∑ ∞ t n − 1 ⋅ J n ( x ) = 2 ⋅ n = − ∞ ∑ ∞ t n ⋅ J n ′ ( x )
Since this is an equality of power series, the coefficients with the same t n t^n t n
n + 1 = k → ∑ n = − ∞ ∞ t n + 1 ⋅ J n ( x ) = ∑ k = − ∞ ∞ t k ⋅ J k − 1 ( x ) ≡ n + 1 = k \rightarrow \sum_ {n = - \infty} ^ {\infty} t ^ {n + 1} \cdot J _ {n} (x) = \sum_ {k = - \infty} ^ {\infty} t ^ {k} \cdot J _ {k - 1} (x) \equiv n + 1 = k → n = − ∞ ∑ ∞ t n + 1 ⋅ J n ( x ) = k = − ∞ ∑ ∞ t k ⋅ J k − 1 ( x ) ≡ n − 1 = m → ∑ n = − ∞ ∞ t n − 1 ⋅ J n ( x ) = ∑ m = − ∞ ∞ t m ⋅ J m + 1 ( x ) ≡ ∑ n = − ∞ ∞ t n ⋅ J n + 1 ( x ) n - 1 = m \rightarrow \sum_ {n = - \infty} ^ {\infty} t ^ {n - 1} \cdot J _ {n} (x) = \sum_ {m = - \infty} ^ {\infty} t ^ {m} \cdot J _ {m + 1} (x) \equiv \sum_ {n = - \infty} ^ {\infty} t ^ {n} \cdot J _ {n + 1} (x) n − 1 = m → n = − ∞ ∑ ∞ t n − 1 ⋅ J n ( x ) = m = − ∞ ∑ ∞ t m ⋅ J m + 1 ( x ) ≡ n = − ∞ ∑ ∞ t n ⋅ J n + 1 ( x )
Then,
∑ n = − ∞ ∞ t n ⋅ J n − 1 ( x ) − ∑ n = − ∞ ∞ t n ⋅ J n + 1 ( x ) = 2 ⋅ ∑ n = − ∞ ∞ t n ⋅ J n ′ ( x ) → \sum_ {n = - \infty} ^ {\infty} t ^ {n} \cdot J _ {n - 1} (x) - \sum_ {n = - \infty} ^ {\infty} t ^ {n} \cdot J _ {n + 1} (x) = 2 \cdot \sum_ {n = - \infty} ^ {\infty} t ^ {n} \cdot J _ {n} ^ {\prime} (x) \rightarrow n = − ∞ ∑ ∞ t n ⋅ J n − 1 ( x ) − n = − ∞ ∑ ∞ t n ⋅ J n + 1 ( x ) = 2 ⋅ n = − ∞ ∑ ∞ t n ⋅ J n ′ ( x ) → J n − 1 ( x ) − J n + 1 ( x ) = 2 ⋅ J n ′ ( x ) \boxed {J _ {n - 1} (x) - J _ {n + 1} (x) = 2 \cdot J _ {n} ^ {\prime} (x)} J n − 1 ( x ) − J n + 1 ( x ) = 2 ⋅ J n ′ ( x )
1 FORMULA:
d d t ∣ e ( x 2 ) ( t − 1 t ) = ∑ n = − ∞ ∞ t n ⋅ J n ( x ) → \frac {d}{d t} \left| e ^ {\left(\frac {x}{2}\right) \left(t - \frac {1}{t}\right)} = \sum_ {n = - \infty} ^ {\infty} t ^ {n} \cdot J _ {n} (x) \rightarrow \right. d t d ∣ ∣ e ( 2 x ) ( t − t 1 ) = n = − ∞ ∑ ∞ t n ⋅ J n ( x ) → e ( x 2 ) ( t − 1 t ) ⋅ x 2 ( 1 + 1 t 2 ) = ∑ n = − ∞ ∞ n t n − 1 ⋅ J n ( x ) → ( ∑ n = − ∞ ∞ t n ⋅ J n ( x ) ) ⋅ x 2 ( 1 + 1 t 2 ) = ∑ n = − ∞ ∞ n t n − 1 ⋅ J n ( x ) → e ^ {\left(\frac {x}{2}\right) \left(t - \frac {1}{t}\right)} \cdot \frac {x}{2} \left(1 + \frac {1}{t ^ {2}}\right) = \sum_ {n = - \infty} ^ {\infty} n t ^ {n - 1} \cdot J _ {n} (x) \rightarrow \left(\sum_ {n = - \infty} ^ {\infty} t ^ {n} \cdot J _ {n} (x)\right) \cdot \frac {x}{2} \left(1 + \frac {1}{t ^ {2}}\right) = \sum_ {n = - \infty} ^ {\infty} n t ^ {n - 1} \cdot J _ {n} (x) \rightarrow e ( 2 x ) ( t − t 1 ) ⋅ 2 x ( 1 + t 2 1 ) = n = − ∞ ∑ ∞ n t n − 1 ⋅ J n ( x ) → ( n = − ∞ ∑ ∞ t n ⋅ J n ( x ) ) ⋅ 2 x ( 1 + t 2 1 ) = n = − ∞ ∑ ∞ n t n − 1 ⋅ J n ( x ) → x 2 ⋅ ( ∑ n = − ∞ ∞ t n ⋅ J n ( x ) + ∑ n = − ∞ ∞ t n − 2 ⋅ J n ( x ) ) = ∑ n = − ∞ ∞ n t n − 1 ⋅ J n ( x ) ∣ ⋅ ( 2 x ) \frac {x}{2} \cdot \left(\sum_ {n = - \infty} ^ {\infty} t ^ {n} \cdot J _ {n} (x) + \sum_ {n = - \infty} ^ {\infty} t ^ {n - 2} \cdot J _ {n} (x)\right) = \sum_ {n = - \infty} ^ {\infty} n t ^ {n - 1} \cdot J _ {n} (x) \Bigg | \cdot \left(\frac {2}{x}\right) 2 x ⋅ ( n = − ∞ ∑ ∞ t n ⋅ J n ( x ) + n = − ∞ ∑ ∞ t n − 2 ⋅ J n ( x ) ) = n = − ∞ ∑ ∞ n t n − 1 ⋅ J n ( x ) ∣ ∣ ⋅ ( x 2 ) ∑ n = − ∞ ∞ t n ⋅ J n ( x ) + ∑ n = − ∞ ∞ t n − 2 ⋅ J n ( x ) = ∑ n = − ∞ ∞ ( 2 n x ) t n − 1 ⋅ J n ( x ) \sum_ {n = - \infty} ^ {\infty} t ^ {n} \cdot J _ {n} (x) + \sum_ {n = - \infty} ^ {\infty} t ^ {n - 2} \cdot J _ {n} (x) = \sum_ {n = - \infty} ^ {\infty} \left(\frac {2 n}{x}\right) t ^ {n - 1} \cdot J _ {n} (x) n = − ∞ ∑ ∞ t n ⋅ J n ( x ) + n = − ∞ ∑ ∞ t n − 2 ⋅ J n ( x ) = n = − ∞ ∑ ∞ ( x 2 n ) t n − 1 ⋅ J n ( x )
Since this is an equality of power series, the coefficients with the same t n t^n t n
n − 1 = m → ∑ n = − ∞ ∞ t n ⋅ J n ( x ) = ∑ m = − ∞ ∞ t m + 1 ⋅ J m + 1 ( x ) ≡ ∑ n = − ∞ ∞ t n + 1 ⋅ J n + 1 ( x ) n - 1 = m \rightarrow \sum_ {n = - \infty} ^ {\infty} t ^ {n} \cdot J _ {n} (x) = \sum_ {m = - \infty} ^ {\infty} t ^ {m + 1} \cdot J _ {m + 1} (x) \equiv \sum_ {n = - \infty} ^ {\infty} t ^ {n + 1} \cdot J _ {n + 1} (x) n − 1 = m → n = − ∞ ∑ ∞ t n ⋅ J n ( x ) = m = − ∞ ∑ ∞ t m + 1 ⋅ J m + 1 ( x ) ≡ n = − ∞ ∑ ∞ t n + 1 ⋅ J n + 1 ( x ) n − 1 = m → ∑ n = − ∞ ∞ t n − 2 ⋅ J n ( x ) = ∑ m = − ∞ ∞ t m − 1 ⋅ J m − 1 ( x ) ≡ ∑ n = − ∞ ∞ t n − 1 ⋅ J n − 1 ( x ) n - 1 = m \rightarrow \sum_ {n = - \infty} ^ {\infty} t ^ {n - 2} \cdot J _ {n} (x) = \sum_ {m = - \infty} ^ {\infty} t ^ {m - 1} \cdot J _ {m - 1} (x) \equiv \sum_ {n = - \infty} ^ {\infty} t ^ {n - 1} \cdot J _ {n - 1} (x) n − 1 = m → n = − ∞ ∑ ∞ t n − 2 ⋅ J n ( x ) = m = − ∞ ∑ ∞ t m − 1 ⋅ J m − 1 ( x ) ≡ n = − ∞ ∑ ∞ t n − 1 ⋅ J n − 1 ( x )
Then,
∑ n = − ∞ ∞ t n + 1 ⋅ J n + 1 ( x ) + ∑ n = − ∞ ∞ t n − 1 ⋅ J n − 1 ( x ) = ∑ n = − ∞ ∞ ( 2 n x ) t n − 1 ⋅ J n ( x ) → \sum_ {n = - \infty} ^ {\infty} t ^ {n + 1} \cdot J _ {n + 1} (x) + \sum_ {n = - \infty} ^ {\infty} t ^ {n - 1} \cdot J _ {n - 1} (x) = \sum_ {n = - \infty} ^ {\infty} \left(\frac {2 n}{x}\right) t ^ {n - 1} \cdot J _ {n} (x) \rightarrow n = − ∞ ∑ ∞ t n + 1 ⋅ J n + 1 ( x ) + n = − ∞ ∑ ∞ t n − 1 ⋅ J n − 1 ( x ) = n = − ∞ ∑ ∞ ( x 2 n ) t n − 1 ⋅ J n ( x ) → J n + 1 ( x ) + J n − 1 ( x ) = 2 n x ⋅ J n ( x ) \boxed {J _ {n + 1} (x) + J _ {n - 1} (x) = \frac {2 n}{x} \cdot J _ {n} (x)} J n + 1 ( x ) + J n − 1 ( x ) = x 2 n ⋅ J n ( x )
Conclusion,
{ J n − 1 ( x ) − J n + 1 ( x ) = 2 ⋅ J n ′ ( x ) J n − 1 ( x ) + J n + 1 ( x ) = 2 n x ⋅ J n ( x ) \boxed {\left\{ \begin{array}{l} J _ {n - 1} (x) - J _ {n + 1} (x) = 2 \cdot J _ {n} ^ {\prime} (x) \\ J _ {n - 1} (x) + J _ {n + 1} (x) = \frac {2 n}{x} \cdot J _ {n} (x) \end{array} \right.} { J n − 1 ( x ) − J n + 1 ( x ) = 2 ⋅ J n ′ ( x ) J n − 1 ( x ) + J n + 1 ( x ) = x 2 n ⋅ J n ( x )
We add and subtract the equation indicating system
{ J n − 1 ( x ) − J n + 1 ( x ) = 2 ⋅ J n ′ ( x ) J n − 1 ( x ) + J n + 1 ( x ) = 2 n x ⋅ J n ( x ) → { 2 J n − 1 ( x ) = 2 n x ⋅ J n ( x ) + 2 ⋅ J n ′ ( x ) ∣ ⋅ ( x 2 ) 2 J n + 1 ( x ) = 2 n x ⋅ J n ( x ) − 2 ⋅ J n ′ ( x ) ∣ ⋅ ( x 2 ) → \left\{ \begin{array}{l} J _ {n - 1} (x) - J _ {n + 1} (x) = 2 \cdot J _ {n} ^ {\prime} (x) \\ J _ {n - 1} (x) + J _ {n + 1} (x) = \frac {2 n}{x} \cdot J _ {n} (x) \end{array} \right. \to \left\{ \begin{array}{l} 2 J _ {n - 1} (x) = \frac {2 n}{x} \cdot J _ {n} (x) + 2 \cdot J _ {n} ^ {\prime} (x) \Bigg | \cdot \left(\frac {x}{2}\right) \\ 2 J _ {n + 1} (x) = \frac {2 n}{x} \cdot J _ {n} (x) - 2 \cdot J _ {n} ^ {\prime} (x) \Bigg | \cdot \left(\frac {x}{2}\right) \end{array} \right. \to { J n − 1 ( x ) − J n + 1 ( x ) = 2 ⋅ J n ′ ( x ) J n − 1 ( x ) + J n + 1 ( x ) = x 2 n ⋅ J n ( x ) → ⎩ ⎨ ⎧ 2 J n − 1 ( x ) = x 2 n ⋅ J n ( x ) + 2 ⋅ J n ′ ( x ) ∣ ∣ ⋅ ( 2 x ) 2 J n + 1 ( x ) = x 2 n ⋅ J n ( x ) − 2 ⋅ J n ′ ( x ) ∣ ∣ ⋅ ( 2 x ) → { x ⋅ J n − 1 ( x ) = n ⋅ J n ( x ) + x ⋅ J n ′ ( x ) x ⋅ J n + 1 ( x ) = n ⋅ J n ( x ) − x ⋅ J n ′ ( x ) \left\{ \begin{array}{l} x \cdot J _ {n - 1} (x) = n \cdot J _ {n} (x) + x \cdot J _ {n} ^ {\prime} (x) \\ x \cdot J _ {n + 1} (x) = n \cdot J _ {n} (x) - x \cdot J _ {n} ^ {\prime} (x) \end{array} \right. { x ⋅ J n − 1 ( x ) = n ⋅ J n ( x ) + x ⋅ J n ′ ( x ) x ⋅ J n + 1 ( x ) = n ⋅ J n ( x ) − x ⋅ J n ′ ( x )
The first equation of the resulting system is important for us.
x ⋅ J n − 1 ( x ) = n ⋅ J n ( x ) + x ⋅ J n ′ ( x ) ∣ ⋅ ( x n − 1 ) → x \cdot J _ {n - 1} (x) = n \cdot J _ {n} (x) + x \cdot J _ {n} ^ {\prime} (x) | \cdot (x ^ {n - 1}) \rightarrow x ⋅ J n − 1 ( x ) = n ⋅ J n ( x ) + x ⋅ J n ′ ( x ) ∣ ⋅ ( x n − 1 ) → x n ⋅ J n − 1 ( x ) = n ⋅ x n − 1 ⋅ J n ( x ) + x n ⋅ J n ′ ( x ) → [ n ⋅ x n − 1 ≡ d d x ( x n ) J n ′ ( x ) ≡ d d x ( J n ( x ) ) ] → x ^ {n} \cdot J _ {n - 1} (x) = n \cdot x ^ {n - 1} \cdot J _ {n} (x) + x ^ {n} \cdot J _ {n} ^ {\prime} (x) \rightarrow \left[\begin{array}{l}n \cdot x ^ {n - 1} \equiv \frac {d}{d x} (x ^ {n})\\J _ {n} ^ {\prime} (x) \equiv \frac {d}{d x} \big (J _ {n} (x) \big)\end{array}\right]\rightarrow x n ⋅ J n − 1 ( x ) = n ⋅ x n − 1 ⋅ J n ( x ) + x n ⋅ J n ′ ( x ) → [ n ⋅ x n − 1 ≡ d x d ( x n ) J n ′ ( x ) ≡ d x d ( J n ( x ) ) ] → x n ⋅ J n − 1 ( x ) = d d x ( x n ) ⋅ J n ( x ) + x n ⋅ d d x ( J n ( x ) ) → x n ⋅ J n − 1 ( x ) = d d x ( x n ⋅ J n ( x ) ) x ^ {n} \cdot J _ {n - 1} (x) = \frac {d}{d x} (x ^ {n}) \cdot J _ {n} (x) + x ^ {n} \cdot \frac {d}{d x} \big (J _ {n} (x) \big) \rightarrow x ^ {n} \cdot J _ {n - 1} (x) = \frac {d}{d x} \big (x ^ {n} \cdot J _ {n} (x) \big) x n ⋅ J n − 1 ( x ) = d x d ( x n ) ⋅ J n ( x ) + x n ⋅ d x d ( J n ( x ) ) → x n ⋅ J n − 1 ( x ) = d x d ( x n ⋅ J n ( x ) )
Conclusion,
d d x ( x n ⋅ J n ( x ) ) = x n ⋅ J n − 1 ( x ) − general formula \boxed {\frac {d}{d x} \big (x ^ {n} \cdot J _ {n} (x) \big) = x ^ {n} \cdot J _ {n - 1} (x) - \text {general formula}} d x d ( x n ⋅ J n ( x ) ) = x n ⋅ J n − 1 ( x ) − general formula
For n = 1 n = 1 n = 1
d d x ( x 1 ⋅ J 1 ( x ) ) = x 1 ⋅ J 1 − 1 ( x ) → d d x ( x ⋅ J 1 ( x ) ) = x ⋅ J 0 ( x ) \frac {d}{d x} \big (x ^ {1} \cdot J _ {1} (x) \big) = x ^ {1} \cdot J _ {1 - 1} (x) \rightarrow \boxed {\frac {d}{d x} \big (x \cdot J _ {1} (x) \big) = x \cdot J _ {0} (x)} d x d ( x 1 ⋅ J 1 ( x ) ) = x 1 ⋅ J 1 − 1 ( x ) → d x d ( x ⋅ J 1 ( x ) ) = x ⋅ J 0 ( x )
1) To prove this formula, we use the representation of the Bessel function as a power series:
J α ( x ) = ∑ n = 0 ∞ ( − 1 ) n Γ ( n + 1 ) Γ ( n + α + 1 ) ( x 2 ) 2 n + α J _ {\alpha} (x) = \sum_ {n = 0} ^ {\infty} \frac {(- 1) ^ {n}}{\Gamma (n + 1) \Gamma (n + \alpha + 1)} \left(\frac {x}{2}\right) ^ {2 n + \alpha} J α ( x ) = n = 0 ∑ ∞ Γ ( n + 1 ) Γ ( n + α + 1 ) ( − 1 ) n ( 2 x ) 2 n + α
(More information: https://en.wikipedia.org/wiki/Bessel_function)
Then, for α = 1 / 2 \alpha = 1 / 2 α = 1/2
J 1 / 2 ( x ) = ∑ n = 0 ∞ ( − 1 ) n Γ ( n + 1 ) Γ ( n + 1 / 2 + 1 ) ( x 2 ) 2 n + 1 / 2 = ∑ n = 0 ∞ ( − 1 ) n Γ ( n + 1 ) Γ ( n + 3 / 2 ) ( x 2 ) 2 n + 1 / 2 → J _ {1 / 2} (x) = \sum_ {n = 0} ^ {\infty} \frac {(- 1) ^ {n}}{\Gamma (n + 1) \Gamma (n + 1 / 2 + 1)} \left(\frac {x}{2}\right) ^ {2 n + 1 / 2} = \sum_ {n = 0} ^ {\infty} \frac {(- 1) ^ {n}}{\Gamma (n + 1) \Gamma (n + 3 / 2)} \left(\frac {x}{2}\right) ^ {2 n + 1 / 2} \rightarrow J 1/2 ( x ) = n = 0 ∑ ∞ Γ ( n + 1 ) Γ ( n + 1/2 + 1 ) ( − 1 ) n ( 2 x ) 2 n + 1/2 = n = 0 ∑ ∞ Γ ( n + 1 ) Γ ( n + 3/2 ) ( − 1 ) n ( 2 x ) 2 n + 1/2 → ( ( x 2 ) 1 / 2 ) ⋅ ∣ J 1 / 2 ( x ) = ∑ n = 0 ∞ ( − 1 ) n Γ ( n + 1 ) Γ ( n + 3 / 2 ) ( x 2 ) 2 n + 1 / 2 → \left(\left(\frac {x}{2}\right) ^ {1 / 2}\right) \cdot \left| J _ {1 / 2} (x) = \sum_ {n = 0} ^ {\infty} \frac {(- 1) ^ {n}}{\Gamma (n + 1) \Gamma (n + 3 / 2)} \left(\frac {x}{2}\right) ^ {2 n + 1 / 2} \rightarrow \right. ( ( 2 x ) 1/2 ) ⋅ ∣ ∣ J 1/2 ( x ) = n = 0 ∑ ∞ Γ ( n + 1 ) Γ ( n + 3/2 ) ( − 1 ) n ( 2 x ) 2 n + 1/2 → ( x 2 ) 1 / 2 ⋅ J 1 / 2 ( x ) = ∑ n = 0 ∞ ( − 1 ) n Γ ( n + 1 ) Γ ( n + 3 / 2 ) ( x 2 ) 2 n + 1 → \left(\frac {x}{2}\right) ^ {1 / 2} \cdot J _ {1 / 2} (x) = \sum_ {n = 0} ^ {\infty} \frac {(- 1) ^ {n}}{\Gamma (n + 1) \Gamma (n + 3 / 2)} \left(\frac {x}{2}\right) ^ {2 n + 1} \rightarrow ( 2 x ) 1/2 ⋅ J 1/2 ( x ) = n = 0 ∑ ∞ Γ ( n + 1 ) Γ ( n + 3/2 ) ( − 1 ) n ( 2 x ) 2 n + 1 →
Now we need to use several properties of the gamma function:
{ Γ ( z + 1 ) = z ⋅ Γ ( z ) Γ ( 1 2 + n ) = ( 2 n ) ! 4 n ⋅ n ! π Γ ( n + 1 ) = n ! , n = 0 , 1 , 2 , 3 , 4 , … \left\{ \begin{array}{c} \Gamma (z + 1) = z \cdot \Gamma (z) \\ \Gamma \left(\frac {1}{2} + n\right) = \frac {(2 n) !}{4 ^ {n} \cdot n !} \sqrt {\pi} \\ \Gamma (n + 1) = n!, n = 0, 1, 2, 3, 4, \ldots \end{array} \right. ⎩ ⎨ ⎧ Γ ( z + 1 ) = z ⋅ Γ ( z ) Γ ( 2 1 + n ) = 4 n ⋅ n ! ( 2 n )! π Γ ( n + 1 ) = n ! , n = 0 , 1 , 2 , 3 , 4 , …
(More information: https://en.wikipedia.org/wiki/Gamma_function)
Γ ( n + 3 2 ) = Γ ( ( 1 2 + n ) + 1 ) = ( n + 1 2 ) Γ ( 1 2 + n ) = 2 n + 1 2 ⋅ ( 2 n ) ! 4 n ⋅ n ! π = 2 n + 1 2 ⋅ ( 2 n ) ! ( 2 2 ) n ⋅ n ! π = = ( 2 n + 1 ) ! 2 2 n + 1 ⋅ n ! π \begin{array}{l} \Gamma \left(n + \frac {3}{2}\right) = \Gamma \left(\left(\frac {1}{2} + n\right) + 1\right) = \left(n + \frac {1}{2}\right) \Gamma \left(\frac {1}{2} + n\right) = \frac {2 n + 1}{2} \cdot \frac {(2 n) !}{4 ^ {n} \cdot n !} \sqrt {\pi} = \frac {2 n + 1}{2} \cdot \frac {(2 n) !}{(2 ^ {2}) ^ {n} \cdot n !} \sqrt {\pi} = \\ = \frac {(2 n + 1) !}{2 ^ {2 n + 1} \cdot n !} \sqrt {\pi} \\ \end{array} Γ ( n + 2 3 ) = Γ ( ( 2 1 + n ) + 1 ) = ( n + 2 1 ) Γ ( 2 1 + n ) = 2 2 n + 1 ⋅ 4 n ⋅ n ! ( 2 n )! π = 2 2 n + 1 ⋅ ( 2 2 ) n ⋅ n ! ( 2 n )! π = = 2 2 n + 1 ⋅ n ! ( 2 n + 1 )! π
Then,
x 2 ⋅ J 1 / 2 ( x ) = ∑ n = 0 ∞ ( − 1 ) n n ! ⋅ ( ( 2 n + 1 ) ! 2 2 n + 1 ⋅ n ! π ) ( x 2 ) 2 n + 1 → \sqrt {\frac {x}{2}} \cdot J _ {1 / 2} (x) = \sum_ {n = 0} ^ {\infty} \frac {(- 1) ^ {n}}{n ! \cdot \left(\frac {(2 n + 1) !}{2 ^ {2 n + 1} \cdot n !} \sqrt {\pi}\right)} \left(\frac {x}{2}\right) ^ {2 n + 1} \rightarrow 2 x ⋅ J 1/2 ( x ) = n = 0 ∑ ∞ n ! ⋅ ( 2 2 n + 1 ⋅ n ! ( 2 n + 1 )! π ) ( − 1 ) n ( 2 x ) 2 n + 1 → x 2 ⋅ J 1 / 2 ( x ) = ∑ n = 0 ∞ ( − 1 ) n ⋅ n ! ⋅ 2 2 n + 1 n ! ⋅ ( 2 n + 1 ) ! ⋅ π ⋅ x 2 n + 1 2 2 n + 1 = 1 π ∑ n = 0 ∞ ( − 1 ) n ⋅ x 2 n + 1 ( 2 n + 1 ) ! \sqrt {\frac {x}{2}} \cdot J _ {1 / 2} (x) = \sum_ {n = 0} ^ {\infty} \frac {(- 1) ^ {n} \cdot n ! \cdot 2 ^ {2 n + 1}}{n ! \cdot (2 n + 1) ! \cdot \sqrt {\pi}} \cdot \frac {x ^ {2 n + 1}}{2 ^ {2 n + 1}} = \frac {1}{\sqrt {\pi}} \sum_ {n = 0} ^ {\infty} \frac {(- 1) ^ {n} \cdot x ^ {2 n + 1}}{(2 n + 1) !} 2 x ⋅ J 1/2 ( x ) = n = 0 ∑ ∞ n ! ⋅ ( 2 n + 1 )! ⋅ π ( − 1 ) n ⋅ n ! ⋅ 2 2 n + 1 ⋅ 2 2 n + 1 x 2 n + 1 = π 1 n = 0 ∑ ∞ ( 2 n + 1 )! ( − 1 ) n ⋅ x 2 n + 1
Conclusion,
x 2 ⋅ J 1 / 2 ( x ) = 1 π ∑ n = 0 ∞ ( − 1 ) n ⋅ x 2 n + 1 ( 2 n + 1 ) ! \sqrt {\frac {x}{2}} \cdot J _ {1 / 2} (x) = \frac {1}{\sqrt {\pi}} \sum_ {n = 0} ^ {\infty} \frac {(- 1) ^ {n} \cdot x ^ {2 n + 1}}{(2 n + 1) !} 2 x ⋅ J 1/2 ( x ) = π 1 n = 0 ∑ ∞ ( 2 n + 1 )! ( − 1 ) n ⋅ x 2 n + 1
As we know
sin ( x ) = ∑ n = 0 ∞ ( − 1 ) n ⋅ x 2 n + 1 ( 2 n + 1 ) ! \sin (x) = \sum_ {n = 0} ^ {\infty} \frac {(- 1) ^ {n} \cdot x ^ {2 n + 1}}{(2 n + 1) !} sin ( x ) = n = 0 ∑ ∞ ( 2 n + 1 )! ( − 1 ) n ⋅ x 2 n + 1
(More information: https://en.wikipedia.org/wiki/Taylor_series)
Then,
x 2 ⋅ J 1 / 2 ( x ) = 1 π ⋅ sin ( x ) ∣ ⋅ ( 2 x ) → \sqrt {\frac {x}{2}} \cdot J _ {1 / 2} (x) = \frac {1}{\sqrt {\pi}} \cdot \sin (x) \Bigg | \cdot \left(\sqrt {\frac {2}{x}}\right) \rightarrow 2 x ⋅ J 1/2 ( x ) = π 1 ⋅ sin ( x ) ∣ ∣ ⋅ ( x 2 ) → J 1 / 2 ( x ) = 2 π x ⋅ sin ( x ) J _ {1 / 2} (x) = \sqrt {\frac {2}{\pi x}} \cdot \sin (x) J 1/2 ( x ) = π x 2 ⋅ sin ( x )
Commentary: The condition of this task contained a mistake, as it indicates the cosine.
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#339914 on Dec 2023