Parallel to line on $f(x)=1+\sin(x)/x$ I want to draw a curve on the top of the function $f(x)=1+\dfrac{\sin(x)}{x}$, but the curve should be equidistant (perpendicular distance from any point of the function $f(x)=1+\dfrac{\sin(x)}{x}$) from every point of the curve. Could anyone help in this regard? 
 A: I've arbitrarily decided to be indulgent today, so I'll write out in full how one might generate the parallel curve of a function. Adapting the formulae from the MathWorld link to the case of a function curve, we have the following parametric equations for a parallel curve to $y=f(x)$ with offset $h$:
$$\begin{align*}
x&=t\pm\frac{h f^\prime(t)}{\sqrt{1+f^\prime(t)^2}}\\
y&=f(t)\mp\frac{h}{\sqrt{1+f^\prime(t)^2}}
\end{align*}$$
For the OP's function, we have, after some simplification,
$$\begin{align*}
x&=t\pm\frac{h (t\cos\,t-\sin\,t)}{\sqrt{t^4+(t\cos\,t-\sin\,t)^2}}\\
y&=1+\frac{\sin\,t}{t}\mp\frac{ht^2}{\sqrt{t^4+(t\cos\,t-\sin\,t)^2}}
\end{align*}$$
Here's a plot of a bunch of parallels, with the starting curve marked in red:

As I've previously noted, for certain values of $h$, the curves are no longer functions, and start displaying cusps. On the other hand, even for the parallels that are still functions, it is rather difficult to obtain an explicit $y=f(x)$ expression, so you may have to be content with the parametric representation.
