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7m — 7m—1 Q~rT~, ï i 7m Q / n . \ ; \i r> . o . r7m+i it+l+m (it + 1 + m)(R + 2 + m)

It is easy to see that when Q = 0, ym =1, which means Bm = Am for all m. The perturbation theory approach expands ym for each m as a power series in Q.

From the solution when Q = 0 we immediately know the first term in the expansion: y^ = 1 • The remaining terms are determined order-by-order by substituting into (53) and collecting terms with the same power of Q:

The first-order (/ = 1) equations are easy to solve since the zeroth-order solutions are just unity:

An important limitation of the perturbation expansion is revealed by the first order solution. Consider the behavior of the sum:

m rm

For large w, the sum increases in magnitude logarithmically with m:

This means that no matter how small Q is, for large enough m the first order expansion will fail. This reflects a limitation of the perturbation expansion itself for this problem— stopping the expansion at any finite order will lead to a series valid only up to some maxi mum size m.

The only way to obtain a consistent expansion is to sum all orders of the series. Unfortunately, the equations (55) are difficult to solve exacdy, and even if they were possible to solve, it would be even more difficult to carry out the summation. However, it isn't difficult to figure out the dominant contribution at each order. It helps to first look at the equations for i = 2:

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