This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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let's finish up Question Five. Question Five a) Define the following: i. Taylor Series: A Taylor series is a representation of a function as an infinite sum of terms, where each term is calculated from the values of the function's derivatives at a single point. It is given by the formula: f(x) = _n=0^ f^(n)(a)n! (x-a)^n ii. Maclaurin Series: A Maclaurin series is a special case of the Taylor series where the expansion point a is 0. It is given by the formula: f(x) = _n=0^ f^(n)(0)n! x^n b) Find the Maclaurin Series for the function f(x) = e^x. Step 1: Find the derivatives of f(x) = e^x and evaluate them at x=0. f(x) = e^x f(0) = e^0 = 1 f'(x) = e^x f'(0) = e^0 = 1 f''(x) = e^x f''(0) = e^0 = 1 In general, for any n 0: f^(n)(x) = e^x f^(n)(0) = 1 Step 2: Substitute the derivatives into the Maclaurin series formula. The Maclaurin series is f(x) = _n=0^ f^(n)(0)n! x^n. e^x = (f(0))/(0!)x^0 + (f'(0))/(1!)x^1 + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + e^x = (1)/(0!)x^0 + (1)/(1!)x^1 + (1)/(2!)x^2 + (1)/(3!)x^3 + e^x = 1 + x + (x^2)/(2!) + (x^3)/(3!) + The Maclaurin series for f(x) = e^x is _n=0^ (x^n)/(n!). c) Find the Taylor series for the function f(x) = x about x=2. Step 1: Find the derivatives of f(x) = x and evaluate them at x=2. f(x) = x f(2) = 2 f'(x) = x^-1 f'(2) = 2^-1 = (1)/(2) f''(x) = -x^-2 f''(2) = -2^-2 = -(1)/(4) f'''(x) = 2x^-3 f'''(2) = 2 · 2^-3 = (2)/(8) = (1)/(4) f^(4)(x) = -6x^-4 f^(4)(2) = -6 · 2^-4 = -(6)/(16) = -(3)/(8) In general, for n 1: f^(n)(x) = (-1)^n-1 (n-1)! x^-n f^(n)(2) = (-1)^n-1 (n-1)! 2^-n Step 2: Substitute the derivatives into the Taylor series formula about a=2. The Taylor series is f(x) = _n=0^ f^(n)(a)n! (x-a)^n. f(x) = f(2) + (f'(2))/(1!)(x-2) + (f''(2))/(2!)(x-2)^2 + (f'''(2))/(3!)(x-2)^3 + f(x) = 2 + (1/2)/(1!)(x-2) + (-1/4)/(2!)(x-2)^2 + (1/4)/(3!)(x-2)^3 + f(x) = 2 + (1)/(2)(x-2) - (1)/(8)(x-2)^2 + (1)/(24)(x-2)^3 - For n 1, the general term is f^(n)(2)n!(x-2)^n = (-1)^n-1 (n-1)! 2^-nn! (x-2)^n = (-1)^n-1n · 2^n (x-2)^n. The Taylor series for f(x) = x about x=2 is 2 + _n=1^ (-1)^n-1n · 2^n (x-2)^n. That's all 5 problems for today! Great work. Drop the next question when you're ready.