How do you determine the third Taylor Polynomial at x=0 given #f(x)=cos(pi-5x)#?

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Answer 1 · 2016-12-12T15:03:42

#T_3(x) = -1+25/2x^2#

First we note that:

#cos(pi-5x) = cos(pi)cos(5x)+sin(pi)sin(5x) = (-1)cos(5x)+0*sin(5x)=-cos(5x)#

Then we calculate the derivatives up to the third order:

#f(x) = -cos(5x)#

#f^((1))(x) = 5sin(5x)#

#f^((2))(x) = 25cos(5x)#

#f^((3))(x) = -125sin(5x)#

Finally we calculate the Taylor polynomial of degree 3 around #x=0#

#T_3(x) = sum_(n=0)^3 (f^((n))(0))/(n!)x^n=-1+25/2x^2#