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If D,E,$\delta,\mu$ be the operators with usual meaning and if hD=U, where h is the interval of differencing,How to prove the following relations between operators:-

1]$\frac{U}{\delta}=\frac{2}{\delta} \sinh^{-1}\frac{\delta}{2} = 1-\frac{\delta^2}{24}+\frac{3\delta^4}{640}-\frac{5\delta^6}{7168}+o(\delta^8)$
2]$\frac{U}{\delta\mu}=\frac{U}{\delta}\big(1+\frac{\delta^2}{4}\big)^{-\frac12} =1-\frac{\delta^2}{6}+\frac{\delta^4}{30}-\frac{\delta^6}{140}+o(\delta^8)$

I am unable to prove the utmost R.H.S. in both 1] and 2].
i-e $1-\frac{\delta^2}{24}\cdots$ in 1]

and $1-\frac{\delta^2}{6}+\cdots$ in 2]

orion
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Win_odd Dhamnekar
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1 Answers1

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U=hD $\delta=E^{\frac12}-E^{\frac{-1}{2}}$

${D_y}_{x}$= 1st derivative of ${y}_{x}$

h is the interval of differencing.$E\equiv 1+\Delta$

$\frac{U}\delta=\frac{2}{\delta}sinh^{-1}\frac{\delta}{2}$

Standard Result:-$sinh^{-1}(x)=x-\frac{1*x^3}{2*3}+\frac{1*3*x^5}{2*4*3}-\frac{1*3*5*x^7}{2*4*6*7}+0(x^9)$

Simplify:-$sinh^{-1}(x)=x-\frac16 x^3+\frac3{40}x^5-\frac5{112}x^7+0(x^9)$

Substituting $\frac{\delta}{2}$

$sinh^{-1}(x)=x-\frac16[\frac{\delta}{2}]^3+\frac3{40}[\frac{\delta}{2}]^5-\frac5{112}[\frac{\delta}{2}]^7+0(\frac{\delta}{2})^9$

Simplify:-$sinh^{-1}(x)=\frac{\delta}{2}-\frac1{48}\delta^3+\frac3{1280}\delta^5-\frac5{14336}\delta^7+0(\delta)^9$

Multiplying by $\frac2{\delta}$

$\frac2{\delta}sinh^{-1}(x)=1-\frac1{24}\delta^2+\frac3{640}\delta^4-\frac5{7168}\delta^6+0(\delta)^8$

Likewise we can also solve second $\frac{U}{\delta\mu}$ But it requires more lengthy calculations.If anyone knows $sinh U$, it would be better.

Win_odd Dhamnekar
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