[8] [Numerical differentiation and round-off error] In this problem we'll show that numerical differentiation can be unstable in the presence of round off error. Consider the 2nd order centered difference approximation to the first derivative; in exact arithmetic we know that |ƒ'(xo) — ƒ (xo + h) — ƒ(xo — h) 2h M < -h² (1) where M = maxä[a,b] |ƒ'''(x)|. Assume now that h can be represented exactly (no floating point error), but, in contrast, that f(x₁+h) and f(xo-h) have some error in their floating point representation. Denote these floating point approximations f(xo + h) := Fl(f(x + h)), ƒ(xo – h) := Fl(f(x − h)) and suppose that €1, €2 are the floating point errors in each value; that is: ƒ (xo – h) = f(xo — h) + €2. ƒ (xo + h) = f(xo + h) + €₁, - (a) Use the triangle inequality and (1) to show that the error in the centered difference approximation in the presence of round error satisfies: M f(xo + h) — f(xo – h) | P (16) - - | ≤ 11 1² + 1 ƒ'(xo) -h² 2h h where = €2- €₁/2. (b) As h gets smaller, notice that part of the error from (a) will vanish, while the other part will increase. Use differential calculus to find h such that the error derived in part (a) will be minimized.

Find the equation of the tangent line to the curve at the given point using implicit differentiation. Cardioid: (x2 + y2+ y)² = x² + y² at (1, 0) y The xy-coordinate plane is given. The curve starts at the point (1, 0), goes up and left becoming less steep, changes direction at the approximate point (0.4, 0.2), goes down and left becoming more steep, passes through the origin, sharply changes direction at the origin, goes up and left becoming less, steep, changes direction at the goes down and left becoming more steep, crosses the x-axis at the point (-1, 0), changes direction at the approximate point (-1.3, -0.7), goes down and right becoming less steep, changes direction at the approximate point (0, -2), goes up and right becoming more steep, changes direction at the approximate point (1.3, -0.7), goes up and left becoming less steep, and stops at the point (1, 0). oproximate point (-0.4, 0.2),

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