Applied Mathematics

( 0.5625 ) ≈ 1.75505 − 1.6875 = 0.06755 > 0 After four iterations, the estimated positive real root of the equation is x $x \approx 0.5625$ within the interval [ 0.5625 , 0.625 ] [0.5625, 0.625] [ 0.5625 , 0.625 ] . Part (B): Fixed Point and Existence Theorem Definition of Fixed Point A point $p$ in the domain of a function $g$ is called a fixed point of $g$ if and only if: $g(p) = p$ Theorem: Existence of a Fixed Point Theorem: Let g : [ a , b ] $g: [a,b] \rightarrow [a,b]$ be a continuous function. Then $g$ has at least one fixed point in $[a,b]$ . Proof: If $g(a) = a$ or $g(b) = b$ , then the fixed point exists trivially at the boundary. Suppose g ( a ) ≠ a g(a) \neq a g ( a )  = a and g ( b ) ≠ b g(b) \neq b g ( b )  = b . Since the domain and codomain of $g$ are both $[a,b]$ , the values of $g(x)$ must lie within $[a,b]$ for all x $x \in [a,b]$ . Consequently, we must have: g ( a ) > a $g(a) > a \quad \text{and} \quad g(b) < b$ We define a auxiliary function $h(x)$ on $[a,b]$ as: $h(x) = g(x) - x$ Since $g(x)$ and $x$ are continuous on $[a,b]$ , their difference $h(x)$ is also continuous on $[a,b]$ . Evaluating $h(x)$ at the boundary points $a$ and $b$ : $h(a) = g(a) - a > 0$ $h(b) = g(b) - b < 0$ Since $h(a) > 0$ and $h(b) < 0$ , the Intermediate Value Theorem (IVT) guarantees that there exists at least one real number p $p \in (a,b)$ such that: $h(p) = 0$ Substituting the definition of $h(x)$ : g ( p ) − p = 0 $g(p) - p = 0 \implies g(p) = p$ Thus, $g$ has at least one fixed point p $p \in [a,b]$ . 05 (A) Three forces P, Q, R act along the sides BC, CA, AB respectively of a triangle ABC. Prove that if PcosA + QcosB + RcosC = 0, then the line of action of the resultant passes through the circum centre of the triangle. (B) Forces X, Y, Z, P+X, Q+Y, P+Z act at a point in the directions of sides of a regular hexagon, taken one way round. Show that their resultant is a force P + Q in the direction of the force Q + Y.