Upper Bound on codes (McEliece) I try to understand the proof of McEliece, Rodemich, Rumsey, Welch for a new upper bound of a Code.
They define
$P(x)=\frac{ 2 }{ t+1 } \begin{pmatrix} n \\ t \end{pmatrix} \left[K_{ t+1 }(x)K_{t}(a)-K_{t}(x)K_{t+1}(a) \right] \sum_{k=0}^t{\frac{ K_{k}(x)K_{k}(a) }{\begin{pmatrix} n \\ k \end{pmatrix}  }  } $
with Krawtchouk Polynomial
$K_{k}(x) = \sum_{j=0}^k{ (-1)^{j} \begin{pmatrix} x \\ j \end{pmatrix} \begin{pmatrix} n-x \\ k-j \end{pmatrix} (q-1)^{k-j}}$.
They proof the upper bound
$M_{LP  }(n,d)\leq H_{2}(\frac{ 1 }{ 2 }-\sqrt{ \delta(1-\delta) })$,
where H is the entropy function.
I do understand the proof itself, but I'm having issues with the following intermediate step:
$M_{LP  }(n,d)\leq \frac{ P(0) }{ \lambda_{0} }$
$\lambda_{0}$ is given by
$\lambda_{0}=-\frac{2  }{t+1  }\begin{pmatrix} n \\ t \end{pmatrix}K_{t+1}(a)K_{t}(a)$.
I tried to compute $\lambda_{0}$ on my own but I got no result.
It is known that:
$\lambda_{k}=q^{ -n }\sum_{i=0}^n{ P(i) }K_{i}(k)$
and so it follows for q=2
$\lambda_{0}=\frac{ 1 }{ 2^{n} } \sum_{i=0}^n P(i) \begin{pmatrix} n \\ i \end{pmatrix} $
$=\frac{ 1 }{ 2^{n} } \sum_{i=0}^n \begin{pmatrix} n \\ i \end{pmatrix} \frac{ 2 }{t+1  } \begin{pmatrix} n \\ t \end{pmatrix}  \left\{ K_{t+1}(i)K_{t}(a)-K_{t}(i)K_{t+1}(a)    \right\} \sum_{k=0}^t{ \frac{ K_{k}(i)K_{k}(a) }{   \begin{pmatrix} n \\ k \end{pmatrix}  } }$
Can somebody help me compute $\lambda_{0}$?
 A: So, you have that $$\lambda _0 = \binom{n}{t}\frac{1}{2^{n-1}(t+1)}\sum _{k=0}^t\sum _{i=0}^n\binom{n}{i}\left (K_{t+1}(i)K_{t}(a)-K_{t+1}(a)K_{t}(i)\right )\frac{K_{k}(i)K_{k}(a)}{\binom{n}{k}}.$$
Notice that I changed the order of the sums. Take out the dependence on $k$ from the sum of $i$ and distribute to get
$$\lambda _0 = \binom{n}{t}\frac{1}{2^{n-1}(t+1)}\sum _{k=0}^t\frac{1}{\binom{n}{k}}\sum _{i=0}^n\binom{n}{i}\left (K_{t+1}(i)K_{t}(a)K_{k}(i)K_{k}(a)-K_{t+1}(a)K_{t}(i)K_{k}(i)K_{k}(a)\right ).$$
Now distribute the sum to get two sums
$$\sum _{i=0}^n\binom{n}{i}K_{t+1}(i)K_{t}(a)K_{k}(i)K_{k}(a)=K_{t}(a)K_{k}(a)\sum _{i=0}^n\binom{n}{i}K_{t+1}(i)K_{k}(i)$$ and
$$-\sum _{i=0}^n\binom{n}{i}K_{t+1}(a)K_{t}(i)K_{k}(i)K_{k}(a)=-K_{t+1}(a)K_{k}(a)\sum _{i=0}^n\binom{n}{i}K_{t}(i)K_{k}(i).$$
These are the Orthogonal identities*, so for the first one you get $0$ because there is no $k=t+1$(notice that $k\leq t$) and the second sum is just $K_{t+1}(a)K_{k}(a)2^n\binom{n}{t}$ just when $k=t$. This means that the whole sum evaluates to
$$\lambda _0=\binom{n}{t}\frac{1}{2^{n-1}(t+1)}\sum _{k=0}^t\frac{-K_{t+1}(a)K_{k}(a)2^n\binom{n}{t}\delta _{k,t}}{\binom{n}{t}}=-\frac{2}{t+1}\binom{n}{t}K_{t+1}(a)K_{k}(a)=-\frac{2}{t+1}\binom{n}{t}K_{t+1}(a)K_{t}(a).$$
That is it.
Orthogonal identity* $$\sum _{i = 0}^n\binom{n}{i}(q-1)^iK_r(i)K_{s}(i)=q^n(q-1)^r\binom{n}{r}\delta _{r,s}.$$
