I have had to calculate the limits for the fraction 1-cos(x) over xsin(pi x) and found the answer to be 1 over 2pi. However, now i have to apply taylor series to the numerator and the denominator...
I have had to calculate the limits for the fraction 1-cos(x) over xsin(pi x) and found the answer to be 1 over 2pi. However, now i have to apply taylor series to the numerator and the denominator...
The answer requires the replacement of the numerator and denominator with a taylor series that will produce the same limit as when lhospitals is applied to the original fraction/equation
Consider the following function values http://oi61.tinypic.com/msoiuc.jpg also given http://oi57.tinypic.com/2cxux5h.jpg a ) Set up a system of equations...
I have already found out the Taylor polynomial of degree 3 f(x) = x³ + 2x² + x − 2 ----> f(1) = 2 f'(x) = 3x² + 4x + 1 -------> f'(1) = 8 f''(x) = 6x...
Note that (c) is independent component a) Find the 3rd Taylor polynomial for f(x)=x^3+2x^2+x-2 in vicinty of a=1 b) Talk about the error approach in (a) and how it could...
Find the Taylor series for f(x)=sin x expanded at about x=(pi/2) and prove that the series converges to sin x for all x
Find the Taylor Series of g(x) = cos(x2)/x, c=¶/4 ¶ = pi X