# how do you determine what the coefficients are on a taylor series expansion if the derivative is too hard to compute?

In a past lecture we talked about how you need to expand The Taylor series of a composed function based on what its input is. For, example:

$e^u$ where ${\color{red} u} = \cos x=1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4- \cdots$

We must expand $e^u$ about $u =1 = \cos 0$

Substituting $u=\cos x$ into the expansion for $e^u$

$e^{\color{red} u} = e + e( {\color{red} u} -1) + \frac {e}{2!}({\color{red} u} -1)^2+ \cdots$ Taylor expansion of $e^u$ abou u=1

$e^{\cos x} = e\cdot e^{\cos x -1 }$

Using the definition of the Taylor series together with repeated applications of the derivative of the original function, I can find what the coefficients are for expanding $\sin x$ at $x=3$

I've also seen cases where you can get around having to compute the derivative by rewriting the original function as an integral of a series, such as in the case of ln(1+x). So this can sometimes work using a substitution of variables as long as you stay within the radius of convergence of the function.

Another case I've run into is when you expand a function at infinity (if it converges):

If $\lim_{x\to\infty}f(x)=L$ is finite, then the `zeroth order term' in the expansion is $L$ and you substitute $z=\frac{1}{x}$ and expand the function at $z=0$. Finding the coefficients for the remaining terms is straightforward.

However, what happens when the derivative of the function is too difficult to compute? Is there a general approach to determine what the coefficients of the Taylor series are when this happens?

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