Modeling #4 - Questions and Answers

Question 1 Find the exponential function that models the data in the table below. x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | y | 4/27 | 4/9 | 4/3 | 4 | 12 | 36 | 108 | 324 | What is the exponential regression of the data? y = _____ Answer Introduction Given a dataset where an exponential function y = ab^x where a and b are constants, are to be obtained. Taking the logarithm on both sides of the above equation, log y = log a + x log b ⇒ v = A + Bx where v = log y, A = log a, B = log b Hence, the exponential function is reduced to a linear function between v and x and can be solved by the least square method. By the principle of least squares, the normal equations for estimating A and B is given by, ∑v = nA + B∑x and ∑vx = A∑x + B∑x² Calculations The following table is formed to calculate the value of A and B x | y | v = log y | x² | vx -3 | 4/27 | -0.8293 | 9 | 2.4879 -2 | 4/9 | -0.3522 | 4 | 0.7044 -1 | 4/3 | 0.1249 | 1 | -0.1249 0 | 4 | 0.6021 | 0 | 0 1 | 12 | 1.0792 | 1 | 1.0792 2 | 36 | 1.5563 | 4 | 3.1126 3 | 108 | 2.0334 | 9 | 6.1002 4 | 324 | 2.5105 | 16 | 10.0420 Total | - | 6.7249 | 44 | 23.4014 Hence, putting the values obtained from the table to the normal equations we get, 6.7249 = 8A+4B...... (i) 23.4014 = 4A + 44B...... (ii) On solving the simultaneous equations, by the method of comparison, it is obtained that, from (i) A = (6.7249 - 4B)/8 ...... (iii) from (ii) A = (23.4014 - 44B)/4 ....... (iv) Comparing (iii) and (iv) (6.7249 - 4B)/8 = (23.4014 - 44B)/4 Hence solving the above equation, it is obtained that, B = 0.4771 → b = Antilog B = 2.9999 Then putting the value of B in (iii), it is obtained that, A = 0.6021 ⇒ a = Antilog A = 4.0004 Therefore, The required exponential regression of the data is y = 4.0004(2.9999)^x Question 2 Isaac evaluates the function modeling Tony's grandparents house value, h(t) = 10,000 * (2^t), at t=2.5. The variable t represents the number of decades since 1960. What is h(2.5)? Answer Determine the value of h(2.5) using the given function. h(t) = 10000 * 2^t function for the value of the house h(2.5) = 10000 * 2^2.5 substitute for the number of decades h(2.5) ≈ 10000 * 5.66 evaluate the exponent h(2.5) ≈ 56568.54 calculated value of the house. Question 3 Write an equivalent expression for 5(2x+y-3z) by modeling and by using the distributive property. Answer To find the equivalent expression using modeling, make a table with the terms of 2x + y -3z along the top and 5 on the left. Then multiply each term of 2x + y -3z with 5. | 2x | y | -3z | 5 | 10x | 5y | -15z | The equivalent expression is the sum of the products so the equivalent expression is 10x + 5y - 15z. The Distributive Property states that a(b + c) = ab + ac. Using the Distributive Property gives: = 5(2x) + 5(y) + 5(-3z) Distributive Property. = 10x + 5y - 15z Multiply. Question 4 Make sure the provided power series is a solution to the given differential equation by directly substituting it. [Hint: For a power x^(2n+1) let k = n + 1.] ∞ y = ∑ ((-1)^n x^2n) / (2^(2n) (n!)²) , xy'' + y' + xy = 0 n=0 Answer Step 1 Direct substitution can be used to confirm that the specified power series: ∞ y = ∑ ((-1)^n x^(2n) / (2^(2n) (n!)²) is a solution of xy" + y' + xy = 0. n=0 Step 2 xy" + y' + xy = ∞ ∑ (-1)^n * (2n)(2n-1)x^(2n-2) / (2^(2n) (n!)²) + ∑ (-1)^n * (2n)x^(2n-1) / (2^(2n) (n!)²) + ∑ (-1)^n * x^(2n+1) / (2^(2n) (n!)²) = 0 n=0 n=0 n=0 ∞ ∑ (-1)^n * (x^2n) / (2^(2n) * (n!)²) * 2n(2n - 1)x⁻² + (2n)x⁻¹ + x = 0 Now let us substitute k = n+1 for the power x^(2n+1) xy" + y' + xy = ∑ (-1)^n 2n(2n-1)x^(2n) + ∑ (-1)^n (2n)x^(2n-1) + ∑ (-1)^(n-1) x^(2n) / (2^(2n) * (n!)²) = 0 xy" + y' + xy = ∑ (-1)^(k-1) 2(2k-1)x^(2k-2) / (2^(2k) (k!)²) + ∑ (-1)^k 2kx^(2k-1) / (2^(2k) (k!)²) + ∑ (-1)^(k-1)x^(2k-1) / (2^(2k-2)((k-1)!)²) = 0 THUS, xy" + y' + xy = 0. Hence, y = ∑ (-1)^n x^(2n) / (2^(2n) (n!)²) is a solution of xy" + y' + xy = 0. ∞ n=0