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Given a function f(x), I can simply compute its Maclaurin series as: $ f(x) = \sum\limits_{k = 0}^{\infty} \frac{f^{(k)} (0)}{k!} *x^k $ I find the series represenation of some function as $\cos(x), \sin(x), \exp(x) $ are already so common, that many people have memorized them.

However now functions can be composited and turned into something ugly. For example: The series represenation for $\ln(x+1)$ is well known, however the one for $\ln(1+\ln(1+x))$ no longer does look so nice.

I know that substitution goes fine with taylor, but is it possible to compute from the Maclaurin series of $\ln(n+1)$ (memorized) the Taylor Series of an ugly function (possible composition of "nice" or "well known" functions) ? For example $\ln(1+\ln(x+1))$ at $a= 0.5$ ?

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This will be rather difficult.

Let's first of all suppose there are no problems, i.e. the functions you are interested in are infinitely differentiable and continuous in the range of values you are interested in.

Then, even with the simple example that you know the Maclaurin series of $f(x)$ and want the Taylor series of $f(x)$ about another point $a$, this is a cumbersome endeavour. Here is the procedure.

With the known Maclaurin series (you memorized the $m_k$)

$f(x) = \sum\limits_{k = 0}^{\infty} \frac{f^{(k)} (0)}{k!} x^k = \sum\limits_{k = 0}^{\infty} m_k x^k$

you require the Taylor series

$f_a(x) = \sum\limits_{k = 0}^{\infty} c_k (x-a)^k$

Now the coefficients are given by

$c_k = \frac{f^{(k)} (a)}{k!}$

which in turn you can get by differentiating your MacLaurin series and evaluate at $a$:

$c_k = \frac{f^{(k)} (a)}{k!} = \sum\limits_{n = k}^{\infty} \frac{n!}{k! (n-k)!} m_n a^{n-k}$

E.g. $c_0 = \sum\limits_{n = 0}^{\infty} m_n a^{n}$

So you need to sum an infinite series to arrive at your new coefficients - big labour.

Another method is to identify the coefficients of all powers of $(x-a)$. To do so, write the MacLaurin series

$f(x) = \sum\limits_{k = 0}^{\infty} m_k ((x-a)+a)^k = \sum\limits_{k = 0}^{\infty} m_k \sum_{j=0}^k {k \choose j} (x-a)^j (a)^{k-j}$

This leads to the same result.

Things get worse if you consider $g(f(x))$.

I guess this already answers your question: it's not a smooth technique to calculate the Taylor expansion from the known MacLaurin series.

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