Find the coefficient of x^3 in the Taylor series about x = 0 for f(x) = sin(2x) (А) 23 (в) -43 4/3 D) -83 (E) -23
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- Determine the coefficients of the x3 term of the Taylor series about the point 0 of the function f(x) = 1/(3 -x)Find the first 7 terms for the Taylor Series centered at 0 f(x)=(1)/sqrt(1-x2) Then integrate to get the 7th degree polynomial for sin-1xDetermine the taylor series f(x)= ln(3+4x) about x=0; and n=5
- What is the Taylor Series of f(x)=ex about x=0?expand the function f(x) = sin(x) / (coshx +2) in a taylor series around the origin going up to x^3. calculate (f.1) from this series and compare to the exact answer obtained form the calculator.find the first four nonzero terms of the Taylor series generated by ƒ at x = a.70. ƒ(x) = 1/(1 - x) at x = 2