# Kronecker Delta in Legendre Series

1. Nov 21, 2015

### PatsyTy

Hello everyone,

I'm working through some homework for a second year mathematical physics course. For the most part I am understanding everything however there is one step I do not understand regarding the steps taken to solve for the coefficients of the Legendre series. Starting with setting the function I want to expand to

\begin{equation*}
f(x)=\sum_{l=0}^\infty c_l P_l(x)
\end{equation*}

multiplying both sides by $P_m(x)$ and integrating from $-1$ to $1$ with respects to $x$ gives me

\begin{equation*}
\int_{-1}^1 f(x) P_m(x) dx = \sum_{l=0}^\infty c_l \int_{-1}^1{P_l(x) P_m (x)dx} = \sum_{l=0}^\infty c_l \frac{2}{2l+1}\delta_{lm}
\end{equation*}

The issue I am having is the next step where we go from

\begin{equation*}
\end{equation*}

I looked up Kronecker Delta on wikipedia and found the property

\begin{equation*}
\sum_j \delta_{ij} a_j = a_i
\end{equation*}

however I do not understand it. I know I could just memorize this and apply it however I would like to understand where this comes from. I've tried looking up proofs of the properties of Kronecker Delta however I haven't had any luck finding them and my textbook only has a paragraph on the function. Does anyone here know of a good resource for reading up on the Kronecker Delta function?

2. Nov 21, 2015

### Staff: Mentor

There's really not much to it. $\delta{ij} = 1$ if i = j, and $\delta{ij} = 0$ otherwise.
So $\sum_{j = 0}^n \delta_{ij} a_j = 0 a_0 + 0 a_1 + \dots + 1a_i + 0a_{i + 1} + \dots + 0a_n$

3. Nov 21, 2015

### BvU

Hi Pat,

Would be surprised if there were any fat volumes on this subject. It's just a convenient way to pick one value out ouf a series. Allows elegant notation in formulas. (but now I'm short-changing the poor delta a little bit).

There's also the Levi-Civita symbol for the sign of permutations, or the DIrac delta function, a kind of analog equivalent to the Kronecker delta -- with its own peculiar properties. Mathematically in the realm of distributions, but for physicists almost a tangible good friend .

--

(Mark beat me to answering this one - but we agree).

4. Nov 21, 2015

### mathman

Definition: $\delta_{ij}=1, i=j; \delta_{ij}=0,i\ne j$

5. Nov 21, 2015

### nrqed

The key point is this: the Kronecker delta ensures that in the sum over l, the only term that will survive is when l=m (that is forced by the delta). All other terms are killed by the delta. So you take the expression under the sum (without the delta) and everywhere you see an "l", you replace it by "m".