If you are comparing this to previous editions (like the 4th), the significant updates include:
The workhorse of adaptive filtering. The 5th edition updates the classical LMS analysis with:
Published in 2013, the 5th edition isn’t just a reprint. Haykin updated the text to bridge classical theory with modern machine learning concepts.
Key updates include:
A masterstroke of exposition. Haykin demonstrates that the RLS algorithm is a special case of the Kalman filter. This unified view helps engineers transition from adaptive filtering to state-space estimation.
(a) Taking expectations on both sides of the weight update equation, we have
$$E[\mathbfw(n+1)] = E[\mathbfw(n)] + \mu E[e(n) \mathbfx(n)]$$ simon haykin adaptive filter theory 5th edition pdf
Using the definition of the error signal, we can rewrite $e(n)$ as
$$e(n) = d(n) - \mathbfw^T(n)\mathbfx(n)$$
Substituting this into the expression for $E[\mathbfw(n+1)]$, we get
$$E[\mathbfw(n+1)] = E[\mathbfw(n)] + \mu E[(d(n) - \mathbfw^T(n)\mathbfx(n)) \mathbfx(n)]$$
Expanding the product and taking expectations, we obtain
$$E[\mathbfw(n+1)] = E[\mathbfw(n)] + \mu (E[d(n)\mathbfx(n)] - E[\mathbfx(n)\mathbfx^T(n)]E[\mathbfw(n)])$$ If you are comparing this to previous editions
$$= E[\mathbfw(n)] + \mu (E[d(n)\mathbfx(n)] - \mathbfRE[\mathbfw(n)])$$
(b) For a white noise input signal with variance $\sigma_x^2$, the autocorrelation matrix is
$$\mathbfR = \sigma_x^2 \mathbfI = \beginbmatrix \sigma_x^2 & 0 \ 0 & \sigma_x^2 \endbmatrix$$
The cross-correlation vector between the input signal and the desired response is
$$E[d(n)\mathbfx(n)] = E[(\alpha x(n) + v(n)) \beginbmatrix x(n) \ x(n-1) \endbmatrix] = \beginbmatrix \alpha \sigma_x^2 \ 0 \endbmatrix$$
Substituting these expressions into the result from part (a), we get Key updates include: A masterstroke of exposition
$$E[\mathbfw(n+1)] = E[\mathbfw(n)] + \mu (\beginbmatrix \alpha \sigma_x^2 \ 0 \endbmatrix - \sigma_x^2 \beginbmatrix 1 & 0 \ 0 & 1 \endbmatrix E[\mathbfw(n)])$$
Let $\mathbfw(n) = [w_1(n), w_2(n)]^T$. Then
$$E[w_1(n+1)] = E[w_1(n)] + \mu (\alpha \sigma_x^2 - \sigma_x^2 E[w_1(n)])$$
$$E[w_2(n+1)] = E[w_2(n)] - \mu \sigma_x^2 E[w_2(n)]$$
These equations describe the mean behavior of the adaptive filter.