# How do I obtain this Taylor series?

For context, this was obtained from the top of page 40 of these data assimilation notes.

Given that $$x^T,x^b$$ are $$n$$-vectors where $$x^T=(x_1,...x_n)^T$$ is the model state vector, $$x^b$$ is the background information vector, $$x^t$$ is the "truth" vector and $$H$$ can be a linear or non-linear operator.

How do I get the following Taylor series (truncated at the linear term)?

$$H(x^b+x^t-x^b) \approx H(x^b)+ \mathrm{H}(x^t-x^b)$$ where $$\mathrm{H}=\frac{dH}{dx}|_{x^b }$$.

It is my understanding that the notation $$\frac{dH}{dx}|_{x^b }=\frac{dH}{dx}(x=x^b )$$.

To be more specific, how can I use Taylor's theorem to get that approximation? What do I substitute into the following?

$$f(x)=f(x=a)+(x-a)f'(x=a)+...$$

Does this mean $$x=x^b-x^b+x^t$$ and therefore $$x=x^t$$ ?

$$H(x^b+x^t-x^b) \approx H(x^b-x^b+x^t=x^b) + (x-x^b)H'(x=x^b)$$

So this gives:

$$H(x^b+x^t-x^b) \approx H(x^t=x^b) + (x^t-x^b)H'(x^t=x^b)$$

Obviously I am doing something fundamentally wrong and doesn't make sense, but I can't seem to understand how to apply Taylor's theorem in more "general" settings like this.

• Using straight and italics $H$ is not the best idea. It took me a while to realize.
– user65203
Jul 27 '20 at 19:09

The development is around $$x^b$$, for the function value at $$x^t$$, and is absolutely standard.
$$H(x_t)\approx \left.H(x)\right|_{x=x_b}+\left.\nabla H(x)\right|_{x=x_b}(x_t-x_b)=H(x_b)+\mathrm H(x_t-x_b).$$
• So following $f(x)=f(x=a)+(x-a)f(x=a)$, did it make sense to say $x=x^b+x^t-x^b=x^t$ and for $f(x=a)$, $H'(x^t=x^b)=H'(x^t-x^b)$? What about the $(x-a)$ coefficient? I can understand your answer, but I find myself at a lost when fiting it back to that formulation. Jul 27 '20 at 19:28