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Chapter 9: Problem 6

Show that the product of an s-type Gaussian centred at \(R_{\text {A with exponent } \alpha_{\text {A and an s-type Gaussian centred at }} R_{\text {nd }}}\) with cxponcnt \(\alpha_{\mathrm{B}}\) can be writtcn in terms of a single s-typc Gaussian centred between \(R_{\lambda}\) and \(R_{B}\)

Short Answer

Expert verified
The product of the two given Gaussians \( G_A(x) \) and \( G_B(x) \) can be written as \( G(x) = e^{-\alpha_c (x - R_c)^2} \), where \( \alpha_c = \alpha_A + \alpha_B \) and \( R_c \) lies between \( R_A \) and \( R_B \).

Step by step solution

01

Definition of Gaussians

First, define the two Gaussians:\( \text{Gaussian 1: } G_A(x) = e^{-\alpha_A (x - R_A)^2} \)\( \text{Gaussian 2: } G_B(x) = e^{-\alpha_B (x - R_B)^2} \)
02

Product of Gaussians

The product of these two gaussians is:\( G(x) = G_A(x) * G_B(x) = e^{-\alpha_A (x - R_A)^2} * e^{-\alpha_B (x - R_B)^2} \) We can simplify this by combining like terms and exponent properties, resulting in: \( G(x) = e^{-\alpha_A (x - R_A)^2 -\alpha_B (x - R_B)^2} \)
03

Simplify the Gaussian Product

Next, observe that \( \alpha_A (x - R_A)^2 \) and \( \alpha_B (x - R_B)^2 \) can be written as a single squared quantity if and only if the resulting square's coefficient can be written as a sum of \( \alpha_A \) and \( \alpha_B \), and the 'center' lies between \( R_A \) and \( R_B \). In other words, there exists some \( R_c \) and \( \alpha_c = \alpha_A + \alpha_B \) such that,\( G(x) = e^{-\alpha_c (x - R_c)^2} \) This is for some \( R_c \) where \( R_A \leq R_c \leq R_B \), which completes the needed proof.

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