Substituting series expansions into each other
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Consider the formal power series f(x) = x + x^2/2 + x^3/6 + x^4/24 and g(x) = x - x*/2 + x^3/3 - x^4/4. Compute by hand the first five coefficients (i.e., up to the coefficient of x^4) of
(a) h(x) = x^f(x)
(d) k(x) = log(1 + g(x))
(c) m(x) = (h o k)(x)
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Solution Summary
We show how to solve the series expansions efficiently, without having to do tedious computations.
Solution Preview
We can just substitute f(x) and g(x) in the formulae and compute the expansions, but there are ways to simplify the algebra. Note that
f(x) = exp(x) - 1 + O(x^5) (1)
g(x) = Log(1+x) + O(x^5) (2)
We can write:
exp[f(x)] = f(x) + f(x)^2/2 + f(x)^3/6 + f(x)^4/24 + O(x^5) (3)
To compute the powers of f(x), we can use (1):
f(x)^2 = exp(2x) - 2 exp(x) + 1 + O(x^6)
= 1 + 2 x + 2 x^2 + 8 x^3/6 + 16 x^4/24 - 2 - 2 x - x^2 - 2 x^3/6 - 2 x^4/24 + 1 + O(x^5)
= x^2 + x^3 + 7 x^4/12 + O(x^5)
This is a bit easier than squaring the expansion of f(x) directly. To compute f(x)^3, it is easier to do this directly:
f(x)^3 = [x + x^2/2 + O(x^3)]^3 = x^3 + ...
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