Chapter 12: Q16P (page 598)

Question:Use the Section 15 recursion relations and (17.4) to obtain the following recursion relations for spherical Bessel functions. We have written them for $j_{n}$ , but they are valid for $y_{n}$ and for the $h_{n} {(\frac{d}{dx})}^{"} [x^{- n} j_{n} {(x)}^{n}] " = x^{- n} j_{n + 1} (x)$

Short Answer

Expert verified

The resultant answer is $\frac{d}{dx} [x^{n + 1} j_{n} {(x)}^{n}] = (x^{n + 1} j_{n - 1} (x))$

Step by step solution

Concept of Spherical Bessel functions:

Spherical Bessel function is given as follows:

$h_{n}^{(1)} (x) = j_{n} (x) + {iy}_{n} (x)$

With:

$j_{n} (x) = x^{n} {(- \frac{1}{x} \frac{d}{dx})}^{n} (\frac{\sin (x)}{x}) y_{n} (x) = - x^{n} {(- \frac{1}{x} \frac{d}{dx})}^{n} (\frac{\cos (x)}{x})$

Formula used:

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

Consider jn(x)=∑r=0∞(-1)rΓ(r+1)Γ(r+1+n)(x2)2r+n and simplify it :

Consider and simplify the equation to obtain:

$(x^{- n} j_{n} (x)) = (x^{- n}) \sum_{r = 0}^{\infty} \frac{{(- 1)}^{r}}{Γ (r + 1) Γ (r + 1 + n)} {(\frac{x}{2})}^{2 r + n} = \sum_{r = 0}^{\infty} \frac{{(- 1)}^{r}}{Γ (r + 1) Γ (r + 1 + n)} {(\frac{x}{2})}^{2 r + n} {(x)}^{2 r} \frac{d}{dx} (x^{- n} j_{n} (x)) = \sum_{r = 1}^{\infty} \frac{{(- 1)}^{r}}{Γ (r + 1) Γ (r + 1 + n)} {(\frac{1}{2})}^{2 r + n} (2 r) {(x)}^{2 r - 1} = \frac{1}{x^{n}} \sum_{r = 1}^{\infty} \frac{{(- 1)}^{r} r}{Γ (r + 1) Γ (r + 1 + n)} {(\frac{1}{2})}^{2 r + n - 1 - 1 + 1} {(x)}^{2 r + n - 1 - 1 + 1} \frac{d}{dx} (x^{- n} j_{n} (x)) = \frac{1}{x^{n}} \sum_{r - 1 = 0}^{\infty} \frac{{(- 1)}^{r} r}{Γ (r + 1) Γ (r + 1 + n)} {(\frac{1}{2})}^{2 r - 2 + n + 1} {(x)}^{2 r - 2 + n + 1}$

Put value of r - 1 = t in the equation:

Substitute for (r- 1) in equation to obtain:

$\frac{d}{dx} (x^{- n} j_{n} (x)) = \frac{1}{x_{n}} \sum_{t = 0}^{\infty} \frac{{(- 1)}^{t + 1} (t + 1)}{Γ (t + 1 + 1) Γ (t + 1 + 1 + n)} {(\frac{1}{2})}^{2 t + n + 1} {(x)}^{2 t + n + 1} = \frac{1}{x_{n}} \sum_{t = 0}^{\infty} \frac{{(- 1)}^{t + 1} (t + 1)}{(t + 1) Γ (t + 1) Γ (t + 1 + 1 + n)} {(\frac{1}{2})}^{2 t + n + 1} {(x)}^{2 t + n + 1} = - x^{- n} \sum_{t = 0}^{\infty} \frac{{(- 1)}^{t}}{Γ (t + 1) Γ (t + 1 + 1 + n)} {(\frac{1}{2})}^{2 t + n + 1} {(x)}^{2 t + n + 1} = - x^{- n} j_{n + 1} (x)$

Hence, it is proved.

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Question:Use the Section 15 recursion relations and (17.4) to obtain the following recursion relations for spherical Bessel functions. We have written them for $j_{n}$ , but they are valid for $y_{n}$ and for the $h_{n} {(\frac{d}{dx})}^{"} [x^{- n} j_{n} {(x)}^{n}] " = x^{- n} j_{n + 1} (x)$

Short Answer

Step by step solution

Concept of Spherical Bessel functions:

Consider jn(x)=∑r=0∞(-1)rΓ(r+1)Γ(r+1+n)(x2)2r+n and simplify it :

Put value of r - 1 = t in the equation:

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Question:Use the Section 15 recursion relations and (17.4) to obtain the following recursion relations for spherical Bessel functions. We have written them for jn, but they are valid forynand for thehn(ddx)"[x-njn(x)n]"=x-njn+1(x)

Short Answer

Step by step solution

Concept of Spherical Bessel functions:

Consider jn(x)=∑r=0∞(-1)rΓ(r+1)Γ(r+1+n)(x2)2r+n and simplify it :

Put value of r - 1 = t in the equation:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Fluids

Mechanics and Materials

Conservation of Energy and Momentum

Oscillations

Electromagnetism

Radiation

Study anywhere. Anytime. Across all devices.

Question:Use the Section 15 recursion relations and (17.4) to obtain the following recursion relations for spherical Bessel functions. We have written them for $j_{n}$ , but they are valid for $y_{n}$ and for the $h_{n} {(\frac{d}{dx})}^{"} [x^{- n} j_{n} {(x)}^{n}] " = x^{- n} j_{n + 1} (x)$