Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)+ (x−1)2+ (x−1)3+ (x−1)4+ (x−1)5. NOTE: This exercise gives a method for writing a polynomial in powers of x−1 instead of powers of x. This may be useful, for example, if you are interested in the values of the polynomial near x=1 .

Accepted Solution

Compute the necessary values/derivatives of [tex]f(x)[/tex] at [tex]x=1[/tex]:[tex]f(1)=3[/tex][tex]f'(1)=2[/tex][tex]f''(1)=-12[/tex][tex]f'''(1)=-12[/tex][tex]f^{(4)}(1)=48[/tex][tex]f^{(5)}(1)=120[/tex]Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) [tex]f(x)[/tex] at [tex]x=1[/tex] by[tex]T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5[/tex][tex]T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5[/tex]###Another way of doing this would be to solve for the coefficients [tex]a,b,c,d,e,g[/tex] in[tex]f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5[/tex]by expanding the right hand side and matching up terms with the same power of [tex]x[/tex].