Created on 2024-06-20Asked by James Martin (Solvelet student)
Use Taylor's theorem to approximate f(x)=sin(x) near x=0 up to the third-order term.
Solution
To use Taylor's theorem to approximate f(x)=sin(x) near x=0 up to the third-order term: 1. **Compute the derivatives of f(x)=sin(x):** f(0)=sin(0)=0.f′(x)=cos(x),f′(0)=cos(0)=1.f′′(x)=−sin(x),f′′(0)=−sin(0)=0.f′′′(x)=−cos(x),f′′′(0)=−cos(0)=−1. 2. **Form the Taylor series up to the third-order term:** f(x)≈f(0)+f′(0)x+2!f′′(0)x2+3!f′′′(0)x3. 3. **Substitute the values:** sin(x)≈0+1⋅x+2!0x2+3!−1x3=x−6x3. 4. **Result:** The approximation of sin(x) near x=0 up to the third-order term is: sin(x)≈x−6x3.Solved on Solvelet with Basic AI Model
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Definition Taylor's Theorem states is that a Function can be approximated such that the approximation is taken as a sum of its polynomial terms using its derivatives at a single point and an error term is given to approximate the amount of error. Here is an Example: If the function is f(x), then Taylor's Theorem would be f(x)=f(a)+f′(a)(x−a)+2! f′′(a)(x−a)2+⋯+Rn(x); where Rn(x) is the remainder
