I Derivation for the indicial exponent in the Frobenius method

Kloo
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I'm reading a book called Asymptotic Methods and Perturbation Theory, and I came across a derivation that I just couldn't follow. Maybe its simple and I am missing something. Equation 3.3.3b below. y(x) takes the form A(x)*(x-x0)^α and A(x) is expanded in a taylor series.

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Use \begin{split}<br /> \left(\sum_{n=0}^\infty a_nx^n \right)\left(\sum_{n=0}^\infty b_nx^n \right) &amp;= <br /> \sum_{n=0}^\infty \sum_{m=0}^\infty a_n b_m x^{n+m} \\<br /> &amp;=\sum_{k=0}^\infty \left(\sum_{n=0}^n a_n b_{k-n}\right) x^k. \end{split}
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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