Expand f(x)=cases1, x in (frac32,2), 3-x, x in(2,3) cases
Expand $f(x)=\begin{cases}1, & x\in\left(\frac32,2\right)\\3-x, & x\in(2,3)\end{cases}$ in a sine series on $\left(\dfrac32,3\right)$.
Explanation of Method
Shifting to $t=x-\frac32\in(0,\frac32)$ turns this into a standard half-range sine problem with half-period $l=\frac32$. The two pieces (constant, then linear) each contribute an elementary integral; combining and simplifying — using $\sin\frac{n\pi}2$ evaluated via $\cos(n\pi/3)$-type reasoning — yields a coefficient with both a $1/n$ piece and an $\sin(n\pi/3)/n^2$ piece.
- Shift to $t=x-\frac32$ and set up the half-range sine series with $l=\frac32$.
- Integrate both pieces (constant on $(0,\frac12)$, linear on $(\frac12,\frac32)$) and combine.
- Translate back to $x$ and simplify using $\sin(\theta-n\pi)=(-1)^n\sin\theta$.
Step 1 · Setup
Let $t=x-\frac32$, $l=\frac32$; then $g(t)=f(t+\tfrac32)=1$ on $(0,\frac12)$ and $g(t)=\frac32-t$ on $(\frac12,\frac32)$. The sine series is $g(t)=\sum b_n\sin\frac{n\pi t}l$, $b_n=\frac2l\int_0^lg(t)\sin\frac{n\pi t}l\,dt$.
Step 2 · Computing $b_n$
Direct integration by parts on both pieces and combining gives the closed form $$ b_n = \frac{2\pi n+3\sin\frac{n\pi}3}{\pi^2n^2} = \frac2{\pi n}+\frac{3\sin\frac{n\pi}3}{\pi^2n^2}. $$ (Verified via independent symbolic integration and numerically: $b_1\approx0.89986$, $b_2\approx0.38412$, $b_3\approx0.21221=\frac2{3\pi}$, matching a direct numerical integration of $g(t)$ exactly. An earlier attempt at this integral produced a substantially different — and incorrect — closed form; the formula above has been independently re-derived and verified.)
Step 3 · Translating back to $x$
Since $\sin\frac{n\pi t}l=\sin\left(\frac{2n\pi x}3-n\pi\right)=(-1)^n\sin\frac{2n\pi x}3$: $$ \boxed{f(x) = \sum_{n=1}^\infty(-1)^n\left[\frac2{\pi n}+\frac{3\sin\frac{n\pi}3}{\pi^2n^2}\right]\sin\frac{2n\pi x}3} $$ (Verified numerically at $x=1.6,1.9,2.3,2.7$: partial sums to $3000$ terms match $f(x)$ to 3+ decimal places.)
Summary Table
| Property | Value |
|---|---|
| Half-period | $l=3/2$ |
| Coefficient (in $t$) | $b_n=\dfrac2{\pi n}+\dfrac{3\sin(n\pi/3)}{\pi^2n^2}$ |
| Series (in $x$) | $f(x)=\sum(-1)^nb_n\sin\dfrac{2n\pi x}3$ |
Additional Notes
- The coefficient splits naturally into a "step function" part ($2/(\pi n)$, decaying like $1/n$, reflecting the jump discontinuity where the constant piece meets the linear piece) plus a "smooth correction" part ($3\sin(n\pi/3)/(\pi^2n^2)$, decaying like $1/n^2$) — this two-scale structure is typical whenever a piecewise function mixes a discontinuous jump with a continuous linear taper.
- Since $\sin(n\pi/3)$ cycles through $0,\pm\frac{\sqrt3}2,\pm\frac{\sqrt3}2,0,\ldots$ with period $6$ in $n$, the correction term's sign and magnitude follow a six-term repeating pattern, superimposed on the steady $1/n$ decay of the dominant term.
- This is the most algebraically involved half-range problem in this collection, combining the non-standard-interval technique of Problem 327 with the piecewise-integration technique used throughout Problems 328–330 and 349.