question
Let f(x) be a polar coordinate function defined from 0 to pi
Is there a translation that pulls f outward (like in the sense f(x) + c does) without altering the polar coordinate derivative?
context (good for understanding my reasons for asking but not actually relevant for answering)
The reason I ask is becuase I am attempting to comstruct a polar coordinate version of a cartesian construct known as a jump series which is a function that when subtracted from a piecewise continuous function, creates a new composed function that is continuous. Part of its requirements is that the derivative changes knowhere.
Therefore, in order to recreate it, I need to know the basic manner in which I can shift a polar coordinate function inward/outward without changing the deivative. Using this, I will be able to develop a formalized sense of precisely what "jump series" would mean in polar coordinates.
Let's see why this is not possible in general. Let $f(\theta) = 1$. The plot of $f$ is the unit circle. It is not possible to transform $f$ so that its plot moves one unit to the right and it is parameterized in the same way by $\theta$. Where does the line $\theta = \pi/4$ intersect the circle of radius 1 centered at $x=1,y=0$? At the point $x=1,y=1$, which in the original parameterization is given by $\theta = \pi/2$. So there's no hope of getting the same parameterization, and therefore no hope of preserving the slope of a tangent line at a given value of $\theta$ after translation.
Now suppose we give up on that particular condition and simply try to find some $g$ whose plot is a translation of the plot of $f$. (Note that as a result, the tangent slopes will be equal at corresponding points.) Well, this is possible if we move the plot one unit to the right like before, but if we try to move it two units to the right, we run into problems again. For instance, there is no intersection of our desired curve with the line $\theta = \pi/2$. So $g$ cannot have $\pi/2$ in its domain. In fact, we must define $g$ on two disjoint sets in $\mathbb R$, one for the upper half of the circle and one for the lower half. But this is getting further and further from what you probably think of as a function, and is not likely to be useful for defining any sort of jump function.