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Incomect Question 1 0/1pts (Lesson 3.1: Solving a Differential Equation.) Suppose that f(:z) = . We know that if p is small, then 3 ~ flath)-f(z) flle) =—=—-. Using this expression with h = 0.01, find an approximate value for f'(1) a1 b.2.72 c.7.38 d. 14.93 We have f(z) ~ f(2+hz—f(2) = 2la+h) g2z h So using h =0.01, we have €202_g2 (1)~ St =14.93 Thus, the answer is (d).

(Lesson 3.1: Solving a Differential Equation.) Suppose that f(z) = e, What is the actual value of f(1)? % | b.e~ 2.72 c.e? ~ 17.39 d.2¢2 ~ 14.78 f'(z) = 2€*,s0 that f'(1) = 2¢?, and thus the answer is (d). e 14.93 f'(z) = 2e*,s0 that f'(1) = 2¢2, and thus the answer is (d).

Question 3 1/1pts | (Lesson 3.1: Solving a Differential Equation.) Consider the differential equation f'(z) = (z + 1) f(x) with f(0) = 1. What is the exact formula for f(z)? a. f(z) = €* b f(z) = e This takes a little work. The good news is that you can actually get the true answer using the technique of separation of variables. We have f@ _ f@) so that 1@ gy — ff(z)dz—fa:+1dz Which implies z+1, In(f(e)) = Z + 2 +C. 2 so that f(a:) — Ke> +z, where C and K are arbitrary constants. Setting f((]) = 1 implies that K = 1, so that the exact answer is , 22 . - =-+z, i.e., choice (c). the answer is f(z) =1 d. f(z) = cxp{uv2 + 21}

Question 4 1/1pts (Lesson 3.1: Solving Differential Equations.) Consider the differential equation f'(z) = (z + 1) f(z) with £(0) = 1. Solve for £(0.20) using Euler's approximation method with increment & = 0.01 for z € [0,0.20]. a. £(0.20) ~ 0.0 b. £(0.20) &~ 1.0 c. £(0.20) ~ 1.24

By previous question, the true answer is the answer is L fl@)=er™"" But our job is to use Euler to come up with an iterative approximation, s0 here it goes. As usual, we start with f@+h) = f(z) + hf'(@) = f2) + bz + 1) f(z) = f@)[L+ h(z +1)], from which we obtain the following table. z | Euler approx true f(z) 0.00 1.0000 1.0000 0.01 1.0100 1.0101 0.02 1.0202 1.0204 0.03 1.0306 1.0309 0.04 1.0412 1.0416 0.05 1.0521 1.0526 0.06 1.0631 1.0637 0.07 1.0744 1.0751 0.08 1.0859 1.0868 0.09 1.0976 1.0986 0.10 1.1096 1.01107 0.19 1.2287 1.2313 0.20 1.2433 1.2461 ‘Wow, what a good match! In any case, the answer is (c). Question 5 1/1pts (Lesson 3.2: Monte Carlo Integration.) Suppose that we want to use Monte Carlo integration to approximate I = fls 1% dz. If Uy, U, ..., U, areiid. Unif(0,1)'s, what's a good approximation [, for r

1 n 1 a i 0, In the notation of the lesson, the general approximation we've been using is — + e~ + <) S s T 1 1+U; Ll n 2 n 1 n kS so that the answer has simplified very nicely to (a). 2 n 1 b. 5 i 1+U; —~n 1 n Z<i=1 1327, 2 o 1 4= Y o

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