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Math  /  Calculus

Question13) (Bessel Function) The domain of Bessel function J3(x)=∑k=0∞(−1)kx2k+322k+1k!(k+3)!J_{3}(x)=\sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k+3}}{2^{2 k+1} k!(k+3)!} is A) (−1,1)(-1,1) B) (−∞,∞)(-\infty, \infty) C) (0,1)(0,1) D) (−1,0)(-1,0)

Studdy Solution

STEP 1

1. The given function is a Bessel function of the first kind, specifically J3(x) J_3(x) , defined by the series.
2. The domain of the Bessel function is determined by the convergence of the infinite series.
3. The series given is J3(x)=∑k=0∞(−1)kx2k+322k+1k!(k+3)! J_{3}(x)=\sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k+3}}{2^{2 k+1} k!(k+3)!} .
4. The convergence of the series depends on the behavior of the general term as k k approaches infinity.

STEP 2

1. Express the general term of the series.
2. Analyze the convergence of the series using the ratio test.
3. Determine the domain based on the result of the ratio test.

STEP 3

Express the general term of the series J3(x) J_3(x) .
The general term is: ak=(−1)kx2k+322k+1k!(k+3)! a_k = (-1)^k \frac{x^{2k+3}}{2^{2k+1} k! (k+3)!}

STEP 4

Apply the ratio test to determine the convergence of the series.
The ratio test involves finding the limit of ∣ak+1ak∣ \left| \frac{a_{k+1}}{a_k} \right| as k k approaches infinity: lim⁡k→∞∣ak+1ak∣ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right|
Substitute ak a_k and ak+1 a_{k+1} into the ratio: ∣ak+1ak∣=∣(−1)k+1x2(k+1)+322(k+1)+1(k+1)!(k+4)!(−1)kx2k+322k+1k!(k+3)!∣ \left| \frac{a_{k+1}}{a_k} \right| = \left| \frac{(-1)^{k+1} \frac{x^{2(k+1)+3}}{2^{2(k+1)+1} (k+1)! (k+4)!}}{(-1)^k \frac{x^{2k+3}}{2^{2k+1} k! (k+3)!}} \right|

STEP 5

Simplify the ratio of consecutive terms.
∣ak+1ak∣=∣(−1)k+1x2(k+1)+322(k+1)+1(k+1)!(k+4)!⋅22k+1k!(k+3)!(−1)kx2k+3∣ \left| \frac{a_{k+1}}{a_k} \right| = \left| \frac{(-1)^{k+1} x^{2(k+1)+3}}{2^{2(k+1)+1} (k+1)! (k+4)!} \cdot \frac{2^{2k+1} k! (k+3)!}{(-1)^k x^{2k+3}} \right|
=∣x2k+522k+3(k+1)(k+4)!⋅22k+1(k+3)!x2k+3∣ = \left| \frac{x^{2k+5}}{2^{2k+3} (k+1) (k+4)!} \cdot \frac{2^{2k+1} (k+3)!}{x^{2k+3}} \right|
=∣x2k+5(k+3)!22k+3(k+1)(k+4)!⋅22k+1x2k+3∣ = \left| \frac{x^{2k+5} (k+3)!}{2^{2k+3} (k+1) (k+4)!} \cdot \frac{2^{2k+1}}{x^{2k+3}} \right|
=∣x2x2k+3(k+3)!2222k+1(k+1)(k+4)(k+3)!⋅22k+1x2k+3∣ = \left| \frac{x^2 x^{2k+3} (k+3)!}{2^2 2^{2k+1} (k+1) (k+4) (k+3)!} \cdot \frac{2^{2k+1}}{x^{2k+3}} \right|
=∣x24(k+1)(k+4)∣ = \left| \frac{x^2}{4 (k+1)(k+4)} \right|

STEP 6

Evaluate the limit of the simplified ratio as k k approaches infinity.
lim⁡k→∞∣x24(k+1)(k+4)∣=lim⁡k→∞∣x24k2∣=0 \lim_{k \to \infty} \left| \frac{x^2}{4 (k+1)(k+4)} \right| = \lim_{k \to \infty} \left| \frac{x^2}{4 k^2} \right| = 0

STEP 7

Conclude the domain based on the ratio test result.
Since the limit of the ratio is 0 for all x x , the series converges for all x∈(−∞,∞) x \in (-\infty, \infty) .
Therefore, the domain of the Bessel function J3(x) J_3(x) is (−∞,∞)(- \infty, \infty), which corresponds to option B.
Solution: The domain of J3(x) J_3(x) is B\boxed{B}.

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