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(Answered): The flag of Nepal is the only national flag that is not a quadrilateral. What's more, its aspect ra ...



The flag of Nepal is the only national flag that is not a quadrilateral. Whats more, its aspect ratio (width : height) is apf = x -> c[1] + x*(C[2] + x*(C[3] + x*(C[4] + x*C[5]))) s = 6136891429688 t = 306253616715 u = 934861968 v = 20332617192 w =

The flag of Nepal is the only national flag that is not a quadrilateral. What's more, its aspect ratio (width : height) is approximately 1:1.219. The exact ratio is the smallest root of the quartic polynomial f(x) = 243356742235044x4 - 1325568548812608x? + 2700899847521244x2 - 2439951444086880x + 824634725389225, which happens to be X1 = 6136891429688 – 306253616715/2 - V118 - 48V2 (934861968 + 203326171922) 4506606337686 In Julia syntax, for your convenience, begin c = [824634725389225, -2439951444086880, 2700899847521244, -1325568548812608, 243356742235044] f = x -> c[1] + x*(C[2] + x*(C[3] + x*(C[4] + x*C[5] ) ) ) s = 6136891429688 t = 306253616715 u = 934861968 v = 20332617192 f = x -> c[1] + x*(C[2] + x*(C[3] + x*(C[4] + x*C[5]))) s = 6136891429688 t = 306253616715 u = 934861968 v = 20332617192 w = 4506606337686 xl - (s-t*sqrt(2) -sqrt(118-48*sqrt(2))*(u+v*sqrt(2)))/w end Your professor fiddled with the signs and found the fourth root (the largest root) to be X4 = 6136891429688 + 30625361671572 - V118 + 4872 (934861968 – 20332617192/2) 4506606337686 but he could not find the second and third roots. • Use the Intermediate Value Theorem and extended precision rational arithmetic Rational{BigInt} to find two intervals that are guaranteed to contain the two missing roots X2 and X3. • There are a few reasonable stopping criteria for iterative methods. Some are based on achieving a pre-determined absolute or relative error in the numerical approximation to a root. However, these stopping criteria are heuristic because we usually do not know a root before finding it. Instead, we shall use a stopping criterion based on the theory of absolute condition number condabs(f, e) (see Definition 1.3.2); in Julia, an iterative method can be enclosed in a while loop which will be exited upon satisfying the following inequality: abs(f(x)) > maximum( (abs(f(x+eps (x) )-f(x)), abs (f(x-eps (x) )-f(x)))) Explain how this stopping criterion is related to the absolute condition number. What does it guarantee? What does it not guarantee? • Write a Julia function that implements one of the following root-finders. . If your birth month 1 < MM <4, use the secant method, - If your birth month 5 < MM <8, use the Newton--Raphson method, . If your birth month 9 < MM < 12, use Halley's method. • Test your root-finder on the exact formulae for Xi and x4 in Float64. If x is a root given by the formulae above and x* an approximation, report |x - *"1/xl. • Use your root-finder to approximate X2 and xz for each of the subtypes. (AbstractFloat).


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