Let $f:[0,1] \to \mathbb{R}$ be a contious function. Then we have the Fourier series of $f$:
$$f(x) \sim \sum_{n=0}^{\infty}a_n\sin(n\pi x)$$ with $\displaystyle a_n = 2 \int_0^1 f(z)\sin(n\pi z)dz.$
Can we get the expandation
$$f(x) \sim \sum_{n=0}^{\infty} b_n \sin \left(n+\dfrac{1}{2}\right)\pi x $$
If yes, how to get the coefficent $b_n$. Thanks for your help.
Yes, this can happen if you're trying to solve a boundary value problem $$\frac{\partial^2y}{\partial x^2}=\frac{\partial y}{\partial t}$$ $$y(0,t)=0$$ $$y^{\prime}(1,t)=0$$ $$y(x,0)=f(x)$$ So your eigenfunctions are now $$y_n(x,t)=\sin\left(n+\frac12\right)\pi xe^{-\left(n+\frac12\right)^2\pi^2t}$$ The eignefunctions are orthogonal because they are the solutions to a Sturm-Liouville eigenvalue problem, but we can check that if $n\ne m$ then $$\begin{align}\int_0^1\sin\left(n+\frac12\right)\pi x\sin\left(m+\frac12\right)\pi x\,dx&=\frac12\int_0^1\left[\cos(n-m)\pi x-\cos(n+m+1)\pi x\right]dx\\ &=\frac12\left[\frac{\sin(n-m)\pi x}{(n-m)\pi}-\frac{\sin(n+m+1)\pi x}{(n+m+1)\pi}\right]_0^1\\ &=0\end{align}$$ And $$\begin{align}\int_0^1\sin^2\left(n+\frac12\right)\pi x\,dx&=\frac12\int_0^1\left[1-\cos(2n+1)\pi x\right]dx\\ &=\frac12\left[x-\frac{\sin(2n+1)\pi x}{(2n+1)\pi}\right]_0^1=\frac12\end{align}$$ So we can see that $$b_n=2\int_0^1f(x)\sin\left(n+\frac12\right)\pi x\,dx$$