91Ó°ÊÓ

Question: Find the coefficients of the Taylor series for the following function centered at point a: \(f(x) = e^x\), with \(a = 0\). Step 1: Determine the derivatives of the function at point a. The first few derivatives of the function \(f(x) = e^x\) are: \(f'(x) = e^x\) \(f''(x) = e^x\) \(f'''(x) = e^x\) And so on. Now, evaluate these derivatives at \(a = 0\): \(f(0) = e^0 = 1\) \(f'(0) = e^0 = 1\) \(f''(0) = e^0 = 1\) \(f'''(0) = e^0 = 1\) The pattern continues for higher derivatives as well. Step 2: Write down the general definition of the coefficients. The general coefficient \(c_n\) of the Taylor series is given by: $$c_n = \frac{f^{(n)}(0)}{n!}$$ Step 3: Substitute the derivatives evaluated at a into the coefficient definition formula. Since all derivatives of the function \(f(x) = e^x\) evaluated at a (= 0) are equal to 1, we can substitute this value into the formula: $$c_n = \frac{1}{n!}$$ Step 4: Find the coefficients of the Taylor series expansion. Using the equation obtained in Step 3, we can find the coefficients of the Taylor series for the function \(f(x) = e^x\) centered at \(a = 0\): \(c_0 = \frac{1}{0!} = 1\) \(c_1 = \frac{1}{1!} = 1\) \(c_2 = \frac{1}{2!} = \frac{1}{2}\) \(c_3 = \frac{1}{3!} = \frac{1}{6}\) And so on. The coefficients of the Taylor series follow a pattern: \(c_n = \frac{1}{n!}\) for \(n = 0, 1, 2, 3, ...\).