Weierstrass Approximation Theorem/Lemma 1

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Theorem

Let $\map {p_{n, k} } t : \N^2 \times \closedint 0 1 \to \R$ be a real valued function defined by:

$\map {p_{n, k} } t := \dbinom n k t^k \paren {1 - t}^{n - k}$

where:

$n, k \in \N$
$t \in \closedint 0 1$
$\dbinom n k$ denotes the binomial coefficient.


Then:

$\ds \sum_{k \mathop = 0}^n k \map {p_{n, k} } t = n t$


Proof

From the binomial theorem:

\(\ds \paren {x + y}^n\) \(=\) \(\ds \sum_{k \mathop = 0}^n \binom n k y^k x^{n - k}\)
\(\ds \leadsto \ \ \) \(\ds 1\) \(=\) \(\ds \sum_{k \mathop = 0}^n \binom n k t^k \paren {1 - t}^{n - k}\) $y = t, ~x = 1 - t$
\(\ds 0\) \(=\) \(\ds \sum_{k \mathop = 0}^n \binom n k \paren {k t^{k - 1} \paren {1 - t}^{n - k} - t^k \paren{n - k} \paren {1 - t}^{n - k - 1} }\) Derivative with respect to $t$
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n k p_{n,k} \paren {\frac 1 t + \frac 1 {1 - t} } - \frac n {1 - t}\)
\(\ds \leadsto \ \ \) \(\ds n t\) \(=\) \(\ds \sum_{k \mathop = 0}^n k p_{n,k}\)

$\blacksquare$


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