Question

Show that Rel(

y
A
​

x
A
​

​
)≈Rel(x
A
​
)−Rel(y
A
​
) assuming Rel(y
A
​
) is small compared to 1.

Answer 1

Answer: Scroll Down Below Step-by-step clarification:To show that Rel(yA) ≈ Rel(xA) - Rel(yA), we will use the power that Rel(yA) is limited distinguished to 1 and ask the idea of uninterrupted closeness.Linear estimate is established the plan that for limited changes in a function, the change in the function maybe approximated apiece derivative of the function manifolded apiece change in the recommendation.In this case, we have:Rel(yA) ≈ Rel(xA) - Rel(yA)Rearranging the equating, we catch:2 * Rel(yA) ≈ Rel(xA)Now, we will ask undeviating estimate by seeing limited changes in the function Rel(xA) about the point xA = yA. Let's mean the trifling sum in xA as ΔxA.Rel(xA + ΔxA) ≈ Rel(xA) + ΔxA * Rel'(xA)Since Rel(yA) is limited distinguished to 1, we can adopt that ΔxA is further limited, so ΔxA * Rel'(xA) maybe thought-out as a trifling sum in Rel(xA).Now, allow's deem the distinguishing case place ΔxA = -Rel(yA). Substituting this worth into the uninterrupted closeness equating, we have:Rel(xA - Rel(yA)) ≈ Rel(xA) - Rel(yA) * Rel'(xA)Since Rel(xA - Rel(yA)) is the same Rel(yA), we can revise the equating as:Rel(yA) ≈ Rel(xA) - Rel(yA) * Rel'(xA)Dividing two together edges by 2, we take:Rel(yA) / 2 ≈ (Rel(xA) - Rel(yA) * Rel'(xA)) / 2Simplifying the right side, we have:Rel(yA) / 2 ≈ Rel(xA) / 2 - (Rel(yA) * Rel'(xA)) / 2Since Rel(yA) is limited distinguished to 1, we can neglect the term (Rel(yA) * Rel'(xA)) / 2 as it is of larger order. Therefore, we can approximate the equating as:Rel(yA) / 2 ≈ Rel(xA) / 2Multiplying two together edges by 2, we catch:Rel(yA) ≈ Rel(xA)Thus, we have proved that Rel(yA) ≈ Rel(xA) - Rel(yA) under the acceptance that Rel(yA) is limited distinguished to 1.