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?


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.

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

The development is around $x^b$, for the function value at $x^t$, and is absolutely standard.

$$H(x_t)\approx \left.H(x)\right|_{x=x_b}+\left.\nabla H(x)\right|_{x=x_b}(x_t-x_b)=H(x_b)+\mathrm H(x_t-x_b).$$

  • $\begingroup$ 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. $\endgroup$
    – mathnoob
    Jul 27 '20 at 19:28
  • $\begingroup$ @mathnoob: I don't think that my answer was of any help. $\endgroup$
    – user65203
    Jul 27 '20 at 19:30

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