Derivative Identity
\[\frac{d}{dx}\left[x^{-n}J_n(x)\right]=-x^{-n}J_{n+1}(x).\]
\[\frac{d}{dx}\left[x^{-n}J_n(x)\right]\]\[=-x^{-n}J_{n+1}(x).\]
This identity increases the order of the Bessel function by one.
Bessel Functions
Half-Order Bessel Functions Using a Derivative Identity
Bessel functions of half-integral order are developed through two IOE past-paper proofs. Begin with the standard half-order result and apply the derivative identity twice.
UnitSeries Solutions and Special Functions
TopicBessel Functions
Question Type4 Mark Problems
Theory Required for These Problems
Standard Half-Order Result
\[J_{1/2}(x)=\sqrt{\frac{2}{\pi x}}\sin x.\]
Note: The complete fraction \(\frac{2}{\pi x}\) remains inside the square root.
First Increase in Order
Put \(n=\frac12\) to obtain \(J_{3/2}(x)\). Simplify it before increasing the order again.
Second Increase in Order
Put \(n=\frac32\) and use \(J_{3/2}(x)\). Differentiate carefully, collect the sine terms, and factor out \(\sqrt{\frac{2}{\pi x}}\).
Method used: Start with \(J_{1/2}(x)\), use the derivative identity to obtain \(J_{3/2}(x)\), apply the identity again, differentiate, and simplify.
Continue through the IOE Engineering Mathematics II Digital Workbook. This bessel functions page follows the Engineering Mathematics II course sequence in the Tribhuvan University curriculum.
66
4 MarksProof of the Formula for J5/2(x)
PAST-PAPER REFERENCE: TU IOE • 2083 Baisakh • Back (New Course) • ENSH 151 • Q. 15 OR • 4 Marks
Prove that
\[J_{5/2}(x)=\sqrt{\frac{2}{\pi x}}\left\{\frac{3-x^2}{x^2}\sin x-\frac3x\cos x\right\},\]
\[J_{5/2}(x)=\sqrt{\frac{2}{\pi x}}\]\[\left\{\frac{3-x^2}{x^2}\sin x-\frac3x\cos x\right\},\]
where the symbols have their usual meanings.
Solution
Bessel derivative identityUse the identity
\[\frac{d}{dx}\left[x^{-n}J_n(x)\right]=-x^{-n}J_{n+1}(x).\]Also,
\[J_{1/2}(x)=\sqrt{\frac{2}{\pi x}}\sin x.\]Put \(n=\frac12\) in the identity:
\[J_{3/2}(x)=-x^{1/2}\frac{d}{dx}\left[x^{-1/2}J_{1/2}(x)\right].\]Now,
\[x^{-1/2}J_{1/2}(x)=\sqrt{\frac2\pi}\frac{\sin x}{x}.\]Therefore,
\[J_{3/2}(x)=\sqrt{\frac{2}{\pi x}}\left(\frac{\sin x}{x}-\cos x\right).\]
\[J_{3/2}(x)=\sqrt{\frac{2}{\pi x}}\]\[\left(\frac{\sin x}{x}-\cos x\right).\]
Next, put \(n=\frac32\):
\[J_{5/2}(x)=-x^{3/2}\frac{d}{dx}\left[x^{-3/2}J_{3/2}(x)\right].\]Substituting \(J_{3/2}(x)\),
\[J_{5/2}(x)=-x^{3/2}\sqrt{\frac2\pi}\frac{d}{dx}\left(\frac{\sin x}{x^3}-\frac{\cos x}{x^2}\right).\]
\[J_{5/2}(x)=-x^{3/2}\sqrt{\frac2\pi}\]\[\frac{d}{dx}\left(\frac{\sin x}{x^3}-\frac{\cos x}{x^2}\right).\]
Differentiate the bracket:
\[\frac{d}{dx}\left(\frac{\sin x}{x^3}-\frac{\cos x}{x^2}\right)=\frac{3\cos x}{x^3}+\frac{\sin x}{x^2}-\frac{3\sin x}{x^4}.\]
\[\frac{d}{dx}\left(\frac{\sin x}{x^3}-\frac{\cos x}{x^2}\right)\]\[=\frac{3\cos x}{x^3}+\frac{\sin x}{x^2}-\frac{3\sin x}{x^4}.\]
Therefore,
\[J_{5/2}(x)=\sqrt{\frac{2}{\pi x}}\left(\frac{3\sin x}{x^2}-\sin x-\frac{3\cos x}{x}\right).\]
\[J_{5/2}(x)=\sqrt{\frac{2}{\pi x}}\]\[\left(\frac{3\sin x}{x^2}-\sin x-\frac{3\cos x}{x}\right).\]
Combining the sine terms,
\[J_{5/2}(x)=\sqrt{\frac{2}{\pi x}}\left[\frac{3-x^2}{x^2}\sin x-\frac3x\cos x\right].\]
\[J_{5/2}(x)=\sqrt{\frac{2}{\pi x}}\]\[\left[\frac{3-x^2}{x^2}\sin x-\frac3x\cos x\right].\]
\[J_{5/2}(x)=\sqrt{\frac{2}{\pi x}}\left[\frac{3-x^2}{x^2}\sin x-\frac3x\cos x\right].\]
\[J_{5/2}(x)=\sqrt{\frac{2}{\pi x}}\]\[\left[\frac{3-x^2}{x^2}\sin x-\frac3x\cos x\right].\]
67
4 MarksProof of the Formula for J5/2(x)
PAST-PAPER REFERENCE: TU IOE • 2082 Bhadra • Regular (New Course) • ENSH 151 • Q. 15 OR • 4 Marks
Prove that
\[J_{5/2}(x)=\sqrt{\frac{2}{\pi x}}\left\{\frac{3-x^2}{x^2}\sin x-\frac3x\cos x\right\},\]
\[J_{5/2}(x)=\sqrt{\frac{2}{\pi x}}\]\[\left\{\frac{3-x^2}{x^2}\sin x-\frac3x\cos x\right\},\]
where the symbols have their usual meanings.
Solution
Bessel derivative identityUse the identity
\[\frac{d}{dx}\left[x^{-n}J_n(x)\right]=-x^{-n}J_{n+1}(x).\]Also,
\[J_{1/2}(x)=\sqrt{\frac{2}{\pi x}}\sin x.\]Put \(n=\frac12\) in the identity:
\[J_{3/2}(x)=-x^{1/2}\frac{d}{dx}\left[x^{-1/2}J_{1/2}(x)\right].\]Now,
\[x^{-1/2}J_{1/2}(x)=\sqrt{\frac2\pi}\frac{\sin x}{x}.\]Therefore,
\[J_{3/2}(x)=\sqrt{\frac{2}{\pi x}}\left(\frac{\sin x}{x}-\cos x\right).\]
\[J_{3/2}(x)=\sqrt{\frac{2}{\pi x}}\]\[\left(\frac{\sin x}{x}-\cos x\right).\]
Next, put \(n=\frac32\):
\[J_{5/2}(x)=-x^{3/2}\frac{d}{dx}\left[x^{-3/2}J_{3/2}(x)\right].\]Substituting \(J_{3/2}(x)\),
\[J_{5/2}(x)=-x^{3/2}\sqrt{\frac2\pi}\frac{d}{dx}\left(\frac{\sin x}{x^3}-\frac{\cos x}{x^2}\right).\]
\[J_{5/2}(x)=-x^{3/2}\sqrt{\frac2\pi}\]\[\frac{d}{dx}\left(\frac{\sin x}{x^3}-\frac{\cos x}{x^2}\right).\]
Differentiate the bracket:
\[\frac{d}{dx}\left(\frac{\sin x}{x^3}-\frac{\cos x}{x^2}\right)=\frac{3\cos x}{x^3}+\frac{\sin x}{x^2}-\frac{3\sin x}{x^4}.\]
\[\frac{d}{dx}\left(\frac{\sin x}{x^3}-\frac{\cos x}{x^2}\right)\]\[=\frac{3\cos x}{x^3}+\frac{\sin x}{x^2}-\frac{3\sin x}{x^4}.\]
Therefore,
\[J_{5/2}(x)=\sqrt{\frac{2}{\pi x}}\left(\frac{3\sin x}{x^2}-\sin x-\frac{3\cos x}{x}\right).\]
\[J_{5/2}(x)=\sqrt{\frac{2}{\pi x}}\]\[\left(\frac{3\sin x}{x^2}-\sin x-\frac{3\cos x}{x}\right).\]
Combining the sine terms,
\[J_{5/2}(x)=\sqrt{\frac{2}{\pi x}}\left[\frac{3-x^2}{x^2}\sin x-\frac3x\cos x\right].\]
\[J_{5/2}(x)=\sqrt{\frac{2}{\pi x}}\]\[\left[\frac{3-x^2}{x^2}\sin x-\frac3x\cos x\right].\]
\[J_{5/2}(x)=\sqrt{\frac{2}{\pi x}}\left[\frac{3-x^2}{x^2}\sin x-\frac3x\cos x\right].\]
\[J_{5/2}(x)=\sqrt{\frac{2}{\pi x}}\]\[\left[\frac{3-x^2}{x^2}\sin x-\frac3x\cos x\right].\]