Q9P To show that, 聽蟿蟿虫/2J1/2(x)=s... [FREE SOLUTION]

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Chapter 12: Q9P (page 590)

To show that, $\sqrt{蟿蟿虫 /} 2 J_{1 / 2} (x) = \sin x$ .

Short Answer

Expert verified

Hence, $\sqrt{\frac{蟺虫}{2}} J_{\frac{1}{2}} (x) = \sin x$ is proved.

Step by step solution

Concept of the Infinite series:

Infinite series:

$J_{p} (x) = {鈭�}_{n = 0}^{鈭�} \frac{{(- 1)}^{n}}{鈱� (n + 1) 鈱� (n + 1 + p)} {(\frac{x}{2})}^{2 n + p}$

Calculation of the function ττx/2J1/2(x) :

Considering the provided statement to show that the provided infinite series for $J_{p} (x)$ converges for all the values of x by the ratio test.

The infinite series is,

$\sqrt{\frac{蟺 x}{2}} J_{\frac{1}{2}} (x) = \sin x$

Substituting the series for $\frac{1}{2}$ for p in the equation (1) as follows:

$\sqrt{\frac{蟺虫}{2}} J_{\frac{1}{2}} (x) = \sqrt{\frac{蟺虫}{2}} {鈭�}_{n = 0}^{鈭�} \frac{{(- 1)}^{n}}{鈱� (n + 1) 鈱� (n + 1 + \frac{1}{2})} {(\frac{x}{2})}^{2 蟺 + \frac{1}{2}} = \sqrt{\frac{蟺虫}{2}} {鈭�}_{n = 0}^{鈭�} \frac{{(- 1)}^{n}}{n! (n + \frac{1}{2}) (n + \frac{1}{2} - 1) . . . . . . . . . . \frac{1}{2} 鈱� (\frac{1}{2}) 2^{蟺} \sqrt{2}} x^{2 n} = {鈭�}_{n = 0}^{鈭�} \frac{{(- 1)}^{n}}{(2 n + 1)!} x^{2 n + 1} = \sin x$

Hence, $\sqrt{\frac{蟺 x}{2}} J_{\frac{1}{2}} (x) = \sin x is proved$

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91影视

To show that, $\sqrt{蟿蟿虫 /} 2 J_{1 / 2} (x) = \sin x$ .

Short Answer

Step by step solution

Concept of the Infinite series:

Calculation of the function ττx/2J1/2(x) :

One App. One Place for Learning.

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91影视

To show that, 蟿蟿虫/2J1/2(x)=sinx.

Short Answer

Step by step solution

Concept of the Infinite series:

Calculation of the function ττx/2J1/2(x) :

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Translational Dynamics

Waves Physics

Magnetism and Electromagnetic Induction

Fluids

Wave Optics

Geometrical and Physical Optics

Study anywhere. Anytime. Across all devices.

To show that, $\sqrt{蟿蟿虫 /} 2 J_{1 / 2} (x) = \sin x$ .