Connection between linear/quadratic/cubic/logarithmic convergence and function? 
*

*From the way linear/quadratic/cubic convergence of a sequence are
defined, I wonder why they are called linear/quadratic/cubic, in the
sense of some connections to linear/quadratic/cubic functions. 
Here are the definitions of linear/quadratic/cubic convergence of a
sequence in my words based on  Wikipedia

Suppose that the sequence $\{x_k\}$ converges to the number $L$.
  Suppose $q > 1$.
When $\lim_{k\to \infty} \frac{|x_{k+1}-L|}{|x_k-L|^q} = μ$ and $μ ∈  (0, 1)$,  we say that the sequence (Q-)converges linearly if
  $q=1$,
  quadratically if $q=2$, and cubically if $q=3$.


*Similarly, how is logarithmic convergence connected to a logarithm
function? The definition of logarithmic convergence is from the same
link to Wikipedia:

If the sequences converges sublinearly and additionally $
     \lim_{k\to \infty} \frac{|x_{k+2} - x_{k+1}|}{|x_{k+1} - x_k|} = 1, $ then it is said the sequence $\{x_k\}$ converges logarithmically
  to $L$.

I found a plot of linear, linear, quadratic and logarithmic rates of convergence for an example in Wikipedia, which seems to
    suggest some connection, although it is not clear to me how they are
    connected:

Thanks for clarification!
 A: It's simply that $x^q$ is a linear function of $x$ if $q=1$, a quadratic function of $x$ if $q=2$ and a cubic function of $x$ if $q=3$.  Convergence "logarithmically" is different, and that definition really doesn't have any necessary connection to a logarithm function.
A: In my opinion, the best way to motivate this terminology is to use series and decimal-base (or other base) expansions. (Let me assume that base-expansions don't end in a string of infinitely many $9$s). 
Suppose that $S$ is some positive real number that lies strictly between $0$ and $1$ (for simplicity). Thus we can write
$$S=.d_1d_2d_3d_4\ldots=\sum_{k=1}^\infty d_k(10)^{-k}$$
for some unique sequence $\{d_k\}$ from $\{0, 1,2, 3, 4, 5, 6, 7,8, 9\}$.
Let us define
$$S_n=\sum_{k=1}^n d_k(10)^{-k}=.d_1d_2\ldots d_n.$$
Then we can show that $\{S_n\}$ converges to $S$ linearly as long as $\{d_k\}$ is sufficiently nice. Namely, let's assume that $\{d_k\}$ does not have infinitely many zeroes. Then for sufficiently large $n$ (at least to pass all zeroes) we have
$$\frac{|S-S_{n+1}|}{|S-S_n|}<\frac{1\cdot 10^{-(n+1)}}{d_{n+1}\cdot 10^{-(n+1)}}\leq 1\,.$$
This is just a special case of something more general. Namely, if we set
$$S_n=\sum_{k=1}^{mn+b}d_k(10)^{-k}$$
for a pair of worth-while natural $m$ and $b$, we can show that $S_n$ converges to $S$ linearly as well.

Thus linear convergence is supposed to encapsulate the idea that the decimal expansions of the terms of the sequence tend to the decimal expansion of the limit in a linear fashion.

Going further with this, if we set
$$S_n=\sum_{k=1}^{n^2}d_k(10)^{-k}$$
we could show that $S_n$ converges to $S$ quadratically (for many different $S$ without oddities in their expansion). Since
$n^2-(n-1)^2=2n-1$
this roughly explains why quadratically converging sequences typically display terms whose decimal expansions are "twice" as good (i.e. twice as many correct decimal digits) as the prior term in the sequence.
I'm inclined to suspect that if we set
$$S_n=\sum_{k=1}^{\lfloor\log_2(n)\rfloor}d_k(10)^{-k}$$
then we'd see logarithmic convergence.
A: Alright, this thread is like 5 years old, but here goes:
Suppose you have several methods to find the root of the function $f(x) = x(x-4)^2$.  The first method you try is the bisection method.  Set up bounds of -0.5 and 1 (don't use symmetric bounds for this example) and here's the results of the first few iterations:
$1.0, -0.5, 0.25, -0.125, 0.0625, -0.03125, 0.015625, -0.0078125, 0.00390625$
Now, we obviously know that this is converging to $0$, but consider the number of iterations it takes to find a "correct" digit.  The first $0$ is found in iteration 1, the second $0$ is found in iteration 3, and the third is found in iteration 6.  We could go on like this, but given the particular properties of base10, we'd know that it takes about 3-4 iterations to find a new correct digit.
That's linear convergence.
Let's try quadratic, Newton's Method:
Our guess will be at x = 1.
$1, -2, -0.8, -0.2, -0.0173, -0.000149, -1.11*10^{-8}$
Now, look at how it took 2 iterations to find the first zero.  Then 2 more to find the second one.  Then 1 iteration to find the next 2 zeros.  Then 1 iteration to find 4 more. The precision doubles each term. So why is that called quadratic convergence and not exponential convergence, or something like that?
Think of the derivative of the linear function and the quadratic function:
$f(x) = c, f(x) = 2x$
If "x" here represents the order of magnitude of the error in the method for a given iteration, then you can see here that these derivatives correlate pretty well to the type of convergence it has.
Why the originators of the term decided to name it this way is beyond me, but that's the logic behind it.  I'd assume that logarithmic convergence is just an extension of that idea:
Consider the sequence $\frac{1}{n}$.  If you do some work with that, you'd see that it is logarithmically convergent. The first correct digit comes at iteration 2: $\frac{1}{2} = 0.5$, then iteration 11: $\frac{1}{11} = 0.09...$, then iteration 101, and so on.
The derivative of the logarithm is $f(x) = \frac{1}{x}$.  If you think of "x" represents the same thing as before, then you can see why it somewhat makes sense.
