I am not sure I understand what you are suggesting.
Are you suggesting to think of the spline-like curve as a function that with domain [0,1]. For simplicity, let us simply say that original curve is defined between 0 and 1 on the x axis, so t(x) gives you the curve without additional work.
If you simply change the domain of t to be [0,a], then you have (trivially) produced an otherwise identical function (call it t') with a different endpoint; but I am not sure where the scaling t' by 1/a comes in. If you literally mean take t'(x)/a, then (unless a is 1) that is clearly a different function. If you mean take t'(x/a) then you have arrived back at the original function, but compressed along the x axis.
As I mentioned above, simply changing the domain without scaling does "work". The problem is that it is not obvious that the resulting curve is "spline-like". Admittedly, to my knowledge, "spline-like" has never been formally defined. However, we defined a spline by the fixed points through which it passes.
For example, suppose t(x) is the spline-like curve under discussion. We should have t = SplineLike((x1,y1),(x2,y2),...,(xn,yn)). Where SplineLike is a function that takes a set of points, and produces a spline-like curve that passes through said points.
Now consider the curve t' = SplineLike((x1,y1),(x2,y2),...,(a,t(a))). That is to say, take the previous set of points, and move the last point to somewhere else on the resulting curve. For simplicities sake, make the additional assumption that (a,t(a)) is still the rightmost point.
We are looking for a higher order function SplineLike such that t'(x) = t(x) where t' is defined. Additionally, SplineLike should produce functions that "feel" like a spline. Since the splines we are talking about are quadratic, a reasonable criteria for this is that the resulting functions are continuous, and have a continuous derivative.
Providing a way to go straight from t to t' does not provide any insight to the higher order SplineLike function; or even if such a function exists.
Are you suggesting to think of the spline-like curve as a function that with domain [0,1]. For simplicity, let us simply say that original curve is defined between 0 and 1 on the x axis, so t(x) gives you the curve without additional work.
If you simply change the domain of t to be [0,a], then you have (trivially) produced an otherwise identical function (call it t') with a different endpoint; but I am not sure where the scaling t' by 1/a comes in. If you literally mean take t'(x)/a, then (unless a is 1) that is clearly a different function. If you mean take t'(x/a) then you have arrived back at the original function, but compressed along the x axis.
As I mentioned above, simply changing the domain without scaling does "work". The problem is that it is not obvious that the resulting curve is "spline-like". Admittedly, to my knowledge, "spline-like" has never been formally defined. However, we defined a spline by the fixed points through which it passes.
For example, suppose t(x) is the spline-like curve under discussion. We should have t = SplineLike((x1,y1),(x2,y2),...,(xn,yn)). Where SplineLike is a function that takes a set of points, and produces a spline-like curve that passes through said points.
Now consider the curve t' = SplineLike((x1,y1),(x2,y2),...,(a,t(a))). That is to say, take the previous set of points, and move the last point to somewhere else on the resulting curve. For simplicities sake, make the additional assumption that (a,t(a)) is still the rightmost point.
We are looking for a higher order function SplineLike such that t'(x) = t(x) where t' is defined. Additionally, SplineLike should produce functions that "feel" like a spline. Since the splines we are talking about are quadratic, a reasonable criteria for this is that the resulting functions are continuous, and have a continuous derivative.
Providing a way to go straight from t to t' does not provide any insight to the higher order SplineLike function; or even if such a function exists.