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Is the taylor expansion of the difference of functions (more specifically the difference of the same function at different points) simply the difference of the taylor expansions?

Since that may be vague, what I am looking at is $$F(\overline{x} +d\overline{x}) -F(\overline{x}) \tag{1}$$ Perhaps I could view this as a composition of functions or something?

The line after this (in the book I am reading) say "The first order approximation can be calculated by successive use of first order Taylor expansions for $F$ and $G$ based on the derivatives at $\overline{x}$

Now, I haven't mentioned $G$ in the question, but if more clarification is needed I can provide it. Basically this is referring to an optimization problem where $G$ is the left hand of the constraint, so we know that $F_1 = \lambda G_1(\overline{x}) $

Full disclosure (and this may help), I have not worked through to the end result the book discusses yet, and I suspect I may simply need to use the taylor expansion of $F(\overline{x} +d\overline{x})$ and substitute it into $(1)$, which I am about to try. However, I am wondering whether there is a formula for if I wanted to do the tayl$F(\overline{x} +d\overline{x}) - F(\overline{x})$

majmun
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  • "Is the taylor expansion of the difference of functions simply the difference of the taylor expansions?" Short answer: yes. You might be overthinking this. Then: "I suspect I may simply need to use the Taylor expansion of $F(\overline{x} +d\overline{x})$ and substitute it into $(1)$" That seems like the simplest method. – Antonio Vargas Oct 23 '15 at 18:49
  • I just did the math and indeed all I had to do was substitute into $(1)$. What I was really interested was whether "the taylor expansion of the difference of functions simply the difference of the taylor expansions?", so thank you for telling me that it is. – majmun Oct 23 '15 at 18:50

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If you just write out

$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$

$g(x)=\sum_{n=0}^{\infty}\frac{g^{(n)}(a)}{n!}(x-a)^n$,

and add and combine terms, you get

$f(x)+g(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)+g^{(n)}(a)}{n!}(x-a)^n$.

Differentiation is of course linear so $f^{(n)}(a)+g^{(n)}(a)=(f(a)+g(a))^{(n)}$, which gives

$f(x)+g(x)=\sum_{n=0}^{\infty}\frac{(f(a)+g(a))^{(n)}}{n!}(x-a)^n$.

charlestoncrabb
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