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?


1 Answer 1


If you know the Taylor expansion of the elementary function and the rules to compose one Taylor expansion into another, you can write the Taylor expansion of every function which is expressed by the composition of elementary functions, without computing any derivative.


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