- #1
DivGradCurl
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I'm supposed to show that
[tex] J_0 (x) = \sum _{n=0} ^{\infty} \frac{\left( -1 \right)^{n} x^{2n}}{2^{2n} \left( n! \right) ^2 } [/tex]
satisfies the differential equation
[tex] x^2 J_0 ^{\prime \prime} (x) + x J_0 ^{\prime} (x) + x^2 J_0 (x) = 0 [/tex]
Here's what I've got:
[tex] x^2 J_0 ^{\prime \prime} (x) = x^2 \sum _{n=2} ^{\infty} \frac{\left( -1 \right)^{n} \left( 2n \right) \left( 2n-1 \right) x^{2n-2}}{2^{2n} \left( n! \right) ^2 } = \sum _{n=0} ^{\infty} \frac{\left( -1 \right)^{n+2} \left( 2n+4 \right) \left( 2n+3 \right) x^{2n+4}}{2^{2n+4}\left[ \left( n+2 \right) ! \right] ^2 } = \frac{x^2}{2} - x^2 J_0 (x) + x J_1 (x) [/tex]
[tex] x J_0 ^{\prime} (x) = x \sum _{n=1} ^{\infty} \frac{\left( -1 \right)^{n} \left( 2n \right) x^{2n-1}}{2^{2n}\left( n! \right) ^2} = \sum _{n=0} ^{\infty} \frac{\left( -1 \right)^{n+1} \left( 2n+2 \right) x^{2n+2}}{2^{2n+4}\left[ \left( n+1 \right) ! \right] ^2 } = - x J_1 (x) [/tex]
[tex] x^2 J_0 (x) = x^2 \sum _{n=0} ^{\infty} \frac{\left( -1 \right)^{n} x^{2n}}{2^{2n} \left( n! \right) ^2 } = \sum _{n=0} ^{\infty} \frac{\left( -1 \right)^{n} x^{2n+2}}{2^{2n} \left( n! \right) ^2 } = x^2 J_0 (x) [/tex]
Then, I get
[tex] x^2 J_0 ^{\prime \prime} (x) + x J_0 ^{\prime} (x) + x^2 J_0 (x) = \frac{x^2}{2} - x^2 J_0 (x) + x J_1 (x) - x J_1 (x) + x^2 J_0 (x) = \frac{x^2}{2} [/tex]
which is not correct, except for when [tex] x=0 [/tex]. Can anyone help me find where I made a mistake?
Thank you very much.
[tex] J_0 (x) = \sum _{n=0} ^{\infty} \frac{\left( -1 \right)^{n} x^{2n}}{2^{2n} \left( n! \right) ^2 } [/tex]
satisfies the differential equation
[tex] x^2 J_0 ^{\prime \prime} (x) + x J_0 ^{\prime} (x) + x^2 J_0 (x) = 0 [/tex]
Here's what I've got:
[tex] x^2 J_0 ^{\prime \prime} (x) = x^2 \sum _{n=2} ^{\infty} \frac{\left( -1 \right)^{n} \left( 2n \right) \left( 2n-1 \right) x^{2n-2}}{2^{2n} \left( n! \right) ^2 } = \sum _{n=0} ^{\infty} \frac{\left( -1 \right)^{n+2} \left( 2n+4 \right) \left( 2n+3 \right) x^{2n+4}}{2^{2n+4}\left[ \left( n+2 \right) ! \right] ^2 } = \frac{x^2}{2} - x^2 J_0 (x) + x J_1 (x) [/tex]
[tex] x J_0 ^{\prime} (x) = x \sum _{n=1} ^{\infty} \frac{\left( -1 \right)^{n} \left( 2n \right) x^{2n-1}}{2^{2n}\left( n! \right) ^2} = \sum _{n=0} ^{\infty} \frac{\left( -1 \right)^{n+1} \left( 2n+2 \right) x^{2n+2}}{2^{2n+4}\left[ \left( n+1 \right) ! \right] ^2 } = - x J_1 (x) [/tex]
[tex] x^2 J_0 (x) = x^2 \sum _{n=0} ^{\infty} \frac{\left( -1 \right)^{n} x^{2n}}{2^{2n} \left( n! \right) ^2 } = \sum _{n=0} ^{\infty} \frac{\left( -1 \right)^{n} x^{2n+2}}{2^{2n} \left( n! \right) ^2 } = x^2 J_0 (x) [/tex]
Then, I get
[tex] x^2 J_0 ^{\prime \prime} (x) + x J_0 ^{\prime} (x) + x^2 J_0 (x) = \frac{x^2}{2} - x^2 J_0 (x) + x J_1 (x) - x J_1 (x) + x^2 J_0 (x) = \frac{x^2}{2} [/tex]
which is not correct, except for when [tex] x=0 [/tex]. Can anyone help me find where I made a mistake?
Thank you very much.