Determine the Taylor series for the function f(x)=ex at x=−1 .Select the correct answer below:ex=ex+ex(x+1)+ex2!(x+1)2+ex3!(x+1)3+⋯+exn!(x+1)n+⋯ ex=e−1+e−1(x+1)+e−12(x+1)2+e−13(x+1)3+⋯+e−1n(x+1)n+⋯ ex=e−1+e−1(x−1)+e−12!(x−1)2+e−13!(x−1)3+⋯+e−1n!(x−1)n+⋯ ex=e−1+e−1(x+1)+e−12!(x+1)2+e−13!(x+1)3+⋯+e−1n!(x+1)n+⋯ FEEDBACK
Question
Determine the Taylor series for the function f(x)=ex at x=−1 .Select the correct answer below:ex=ex+ex(x+1)+ex2!(x+1)2+ex3!(x+1)3+⋯+exn!(x+1)n+⋯ ex=e−1+e−1(x+1)+e−12(x+1)2+e−13(x+1)3+⋯+e−1n(x+1)n+⋯ ex=e−1+e−1(x−1)+e−12!(x−1)2+e−13!(x−1)3+⋯+e−1n!(x−1)n+⋯ ex=e−1+e−1(x+1)+e−12!(x+1)2+e−13!(x+1)3+⋯+e−1n!(x+1)n+⋯ FEEDBACK
Solution
The correct answer is: ex=e−1+e−1(x+1)+e−12!(x+1)2+e−13!(x+1)3+⋯+e−1n!(x+1)n+⋯
Here's the step-by-step solution:
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The Taylor series for a function f(x) about x=a is given by: f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ... + f^n(a)(x-a)^n/n! + ...
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For the function f(x) = e^x, its derivative of any order is e^x. So, f'(x) = e^x, f''(x) = e^x, f'''(x) = e^x, ..., f^n(x) = e^x.
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We want the Taylor series at x = -1, so a = -1. Therefore, f(a) = e^-1, f'(a) = e^-1, f''(a) = e^-1, f'''(a) = e^-1, ..., f^n(a) = e^-1.
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Substituting these values into the Taylor series formula, we get: e^x = e^-1 + e^-1(x+1) + e^-1(x+1)^2/2! + e^-1(x+1)^3/3! + ... + e^-1(x+1)^n/n! + ...
So, the correct answer is ex=e−1+e−1(x+1)+e−12!(x+1)2+e−13!(x+1)3+⋯
Similar Questions
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