How to expand in powers of delta?

Question

I am working through Probability Theory: The Logic of Science, by E.T. Jaynes, and am stuck at equation 16.1, where he performs an expansion in powers of $\delta$. For full context, the preceding paragraph and equation is:

Thus, if the sampling distribution has the functional form $f(\frac{x}{\sigma})$, and $x$ and $\sigma$ are already known to be about equal to $x_0$ and $\sigma_0$, we are really making inferences about the small corrections $q \equiv x - x_0$ and $\delta \equiv \sigma - \sigma_0$. Expanding in powers of $\delta$ and keeping only the linear term, we have:
$$\frac{x}{\sigma} = \frac{x_0 + q}{\sigma_0 + \delta} = \frac{1}{\sigma_0}(x - \theta + \dots)$$ where $\theta \equiv \frac{x_0 \delta}{\sigma_0}$.

I am confused as to what he means by "expanding in powers of $\delta$", and not sure how he arrives at the expression on the right. I have been trying different Taylor series expansions, specifically:

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

Letting $a = \delta$, but still I am not able to derive the correct result.

Is a Taylor series expansion what Jaynes is referring to in this case? Should I be working with a multivariate Taylor series, since $f$ is parameterized by both $x$ and $\sigma$? If not, could anyone point me in the right direction to what he is referring to?

For reference, I have checked the usual sources at this point (wikipedia, google, math stack exchange), but to no avail. I think the main issue is not quite understanding what Jaynes means by "expanding in powers of..." . As far as I know, generally expanding in powers of something refers to the expression being raised to ascending powers; i.e. in the standard Taylor series expansion:

$$f(a) + f'(a)(x-a) + f''(a)\frac{(x-a)^2}{2} + \dots$$

we could say "expanding in powers of $(x-a)$".

Any clarity/insight here would be greatly appreciated, and I can provide any additional context that is needed. Thank you!

Answer 1

Let

$$f(\sigma)=\frac {x}{\sigma}, \>\>\>\>\>\>f'(\sigma)=-\frac {x}{\sigma^2}$$

Apply the first-order Tayler expansion,

$$f(\sigma)=f(\sigma_0)+f'(\sigma_0)(\sigma-\sigma_0)=\frac {x}{\sigma_0}-\frac {x}{\sigma_0^2}\delta=\frac {1}{\sigma_0}\left(x-\frac{x\delta}{\sigma_0} \right)=\frac {1}{\sigma_0}\left(x-\theta\right)$$

where,

$$\theta = \frac{x\delta}{\sigma_0}= \frac{x_0\delta}{\sigma_0}+ \frac{q\delta}{\sigma_0}= \frac{x_0\delta}{\sigma_0}$$

because $q$ is a small correction and the joint quadratic correction term $ \frac{q\delta}{\sigma_0}$ is neglected.

Answer 2

This expansion in powers of $\delta$ is the Taylor expansion, with $x=\sigma$, $a=\sigma_0$ and $\delta=x-a$.

You have $$ \frac{x}{\sigma} = \frac{x}{\sigma_0 + \delta} = \frac{x}{\sigma_0}\frac{1}{1-(-\frac{\delta}{\sigma_0})} = \frac{x}{\sigma_0}\sum_{n=0}^\infty (-\frac{\delta}{\sigma_0})^n = \frac{x}{\sigma_0}\sum_{n=0}^\infty \frac{(-1)^n}{\sigma_0^n} \delta^n = \\ = \frac{x}{\sigma_0} - \frac{x\delta}{\sigma_0^2} + \frac{x\delta^2}{\sigma_0^3} + \dots$$

It is called an expansion in powers of $\delta$ because what you get is a power series in $\delta$.

Answer 3

We write

$$\frac x{\sigma_0+\delta}=\frac x{\sigma_0}\frac1{1+\dfrac{\delta}{\sigma_0}}$$

and by the Taylor development $$\frac x{\sigma_0}\frac1{1+\dfrac{\delta}{\sigma_0}}=\frac x{\sigma_0}\left(1-\frac\delta{\sigma_0}+\left(\frac{\delta}{\sigma_0}\right)^2-\cdots\right)\approx\frac1{\sigma_0}\left(x-\frac{x\delta}{\sigma_0}\right).$$

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It seems that there is a confusion between $x$ and $x_0$.