# [SOLVED]combining terms

#### dwsmith

##### Well-known member
Using the coefficients determined, combine the terms to arrive at the following approximation for the solution
$$u(r,\theta,\phi)\simeq 25 + \frac{75\sqrt{2}}{2}r\sin\theta\cos\phi + \frac{125}{\pi}r^2\sin^2\theta\cos 2\phi.$$

How do I combine them?

The coefficients are
For $\ell = 0$ and $m = 0$, we have
$$Y^0_0(\theta,\varphi) = \frac{1}{2\sqrt{\pi}}.$$
For $\ell = 1$ and $m = 1$, we have
\begin{alignat*}{3}
Y^{1}_1 & = & -\frac{1}{2}\sqrt{\frac{3}{2\pi}}(\cos^2\theta - 1)^{1/2}e^{i\varphi}\\
& = & -\frac{1}{2}\sqrt{\frac{3}{2\pi}}e^{i\varphi} \sin\theta
\end{alignat*}
From $\ell = 1$ and $m = 1$, we can obtain
$$Y^{-1}_1 = \frac{1}{2}\sqrt{\frac{3}{2\pi}}e^{-i\varphi}\sin \theta.$$
For $\ell = 1$ and $m = 0$, we have
$$Y^0_1 = \frac{\sqrt{3}}{2\sqrt{\pi}}\cos\theta.$$
For $\ell = 2$ and $m = 2$, we have
\begin{alignat*}{3}
Y^{2}_2 & = & \frac{3}{4}\sqrt{\frac{5}{6\pi}}(\cos^2\theta - 1)e^{2i\varphi}\\
& = & \frac{3}{4}\sqrt{\frac{5}{6\pi}}e^{2i\varphi}\sin^2\theta
\end{alignat*}
From $\ell = 2$ and $m = 2$, we can obtain
\begin{alignat*}{3}
Y^{-2}_2 & = & \frac{3}{4}\sqrt{\frac{5}{6\pi}}e^{-2i\varphi}\sin^2 \theta.
\end{alignat*}
For $\ell = 2$ and $m = 1$, we have
\begin{alignat*}{3}
Y^{1}_2 & = & -\frac{3}{2}\sqrt{\frac{5}{6\pi}}e^{i\varphi}(\cos^2\theta - 1)^{1/2}\cos\theta\\
& = & -\frac{3}{2}\sqrt{\frac{5}{6\pi}}e^{i\varphi}\sin \theta \cos\theta
\end{alignat*}
From $\ell = 2$ and $m = 1$, we can obtain
\begin{alignat*}{3}
Y^{-1}_2 & = & \frac{3}{2}\sqrt{\frac{5}{6\pi}}e^{-i\varphi}\sin \theta \cos\theta.
\end{alignat*}
For $\ell = 2$ and $m = 0$, we have
$$Y^0_{\ell} = \frac{1}{4}\sqrt{\frac{5}{2\pi}}(3\cos^2\theta - 1).$$

#### topsquark

##### Well-known member
MHB Math Helper
Using the coefficients determined, combine the terms to arrive at the following approximation for the solution
$$u(r,\theta,\phi)\simeq 25 + \frac{75\sqrt{2}}{2}r\sin\theta\cos\phi + \frac{125}{\pi}r^2\sin^2\theta\cos 2\phi.$$

How do I combine them?

The coefficients are
For $\ell = 0$ and $m = 0$, we have
$$Y^0_0(\theta,\varphi) = \frac{1}{2\sqrt{\pi}}.$$
For $\ell = 1$ and $m = 1$, we have
\begin{alignat*}{3}
Y^{1}_1 & = & -\frac{1}{2}\sqrt{\frac{3}{2\pi}}(\cos^2\theta - 1)^{1/2}e^{i\varphi}\\
& = & -\frac{1}{2}\sqrt{\frac{3}{2\pi}}e^{i\varphi} \sin\theta
\end{alignat*}
From $\ell = 1$ and $m = 1$, we can obtain
$$Y^{-1}_1 = \frac{1}{2}\sqrt{\frac{3}{2\pi}}e^{-i\varphi}\sin \theta.$$
For $\ell = 1$ and $m = 0$, we have
$$Y^0_1 = \frac{\sqrt{3}}{2\sqrt{\pi}}\cos\theta.$$
For $\ell = 2$ and $m = 2$, we have
\begin{alignat*}{3}
Y^{2}_2 & = & \frac{3}{4}\sqrt{\frac{5}{6\pi}}(\cos^2\theta - 1)e^{2i\varphi}\\
& = & \frac{3}{4}\sqrt{\frac{5}{6\pi}}e^{2i\varphi}\sin^2\theta
\end{alignat*}
From $\ell = 2$ and $m = 2$, we can obtain
\begin{alignat*}{3}
Y^{-2}_2 & = & \frac{3}{4}\sqrt{\frac{5}{6\pi}}e^{-2i\varphi}\sin^2 \theta.
\end{alignat*}
For $\ell = 2$ and $m = 1$, we have
\begin{alignat*}{3}
Y^{1}_2 & = & -\frac{3}{2}\sqrt{\frac{5}{6\pi}}e^{i\varphi}(\cos^2\theta - 1)^{1/2}\cos\theta\\
& = & -\frac{3}{2}\sqrt{\frac{5}{6\pi}}e^{i\varphi}\sin \theta \cos\theta
\end{alignat*}
From $\ell = 2$ and $m = 1$, we can obtain
\begin{alignat*}{3}
Y^{-1}_2 & = & \frac{3}{2}\sqrt{\frac{5}{6\pi}}e^{-i\varphi}\sin \theta \cos\theta.
\end{alignat*}
For $\ell = 2$ and $m = 0$, we have
$$Y^0_{\ell} = \frac{1}{4}\sqrt{\frac{5}{2\pi}}(3\cos^2\theta - 1).$$
I'm a bit confused about this one. The problem is very similar to using the dot product to separate terms. The usual
$$c_{nlm} = \int \psi^*_{nlm} u(r, \theta, \phi)~ d \tau$$

For example, the constant term is going deal with $$\psi _{000}$$ since there are no radial or angular terms. But even $$\psi _{000}$$ is going to contain a $$e^{-Zr/a_0}$$ factor (assuming a Hydrogen-like wavefunction for the radial part) and there is no exponential factor in your function.

-Dan

#### dwsmith

##### Well-known member
I'm a bit confused about this one. The problem is very similar to using the dot product to separate terms. The usual
$$c_{nlm} = \int \psi^*_{nlm} u(r, \theta, \phi)~ d \tau$$

For example, the constant term is going deal with $$\psi _{000}$$ since there are no radial or angular terms. But even $$\psi _{000}$$ is going to contain a $$e^{-Zr/a_0}$$ factor (assuming a Hydrogen-like wavefunction for the radial part) and there is no exponential factor in your function.

-Dan
I am confuse too. I am not sure what I am supposed to do but that is the question verbatim.

#### dwsmith

##### Well-known member
I am confuse too. I am not sure what I am supposed to do but that is the question verbatim.

I figure it out.

#### topsquark

##### Well-known member
MHB Math Helper
I figure it out.
I'm curious about your solution. Could you post the general idea? (I had a thought this morning about expanding the "missing" exponential. I'm wondering if that was the right approach.)

-Dan

#### dwsmith

##### Well-known member
I'm curious about your solution. Could you post the general idea? (I had a thought this morning about expanding the "missing" exponential. I'm wondering if that was the right approach.)

-Dan
I also had to use the spherical harmonics of the solution which wasn't apparent to me by the questions wording.

We ca expand our series out term by term
\begin{alignat*}{3}
u(r,\theta,\varphi) & = & \sum_{\ell = 0}^2\sum_{m = -\ell}^{\ell} A_{\ell,m}Y^m_{\ell}(\theta,\varphi).
\end{alignat*}
For $\ell = 0$, we have that $u(r,\theta,\varphi) = 25$.
For $\ell = 1$, we have that $u(r,\theta,\varphi) = 25 + r\frac{75}{\sqrt{2}}\sin\theta\frac{e^{i\varphi} + e^{-i\varphi}}{2}$.
Lastly, for $\ell = 2$, we have
$$u(r,\theta,\varphi) = 25 + \frac{75r}{\sqrt{2}}\sin\theta\cos\varphi + \frac{125r^2}{\pi}\sin^2\theta\cos 2\varphi = 25 + \frac{75\sqrt{2}}{2}r\sin\theta\cos\phi + \frac{125}{\pi}r^2\sin^2\theta\cos 2\phi.$$
That is, we can approximate $u(r,\theta,\varphi)$ when $\ell = 0,1,2$ to be
$$u(r,\theta,\phi)\simeq 25 + \frac{75\sqrt{2}}{2}r\sin\theta\cos\phi + \frac{125}{\pi}r^2\sin^2\theta\cos 2\phi.$$