f(x)=xlnx，
f( 1)=0
f'=lnx+ 1
f''= 1/x
f'''=- 1/x？
f^(4)=2！ /x？
f^(5)=-3！ /x^4
f^(k)=(- 1)^k.(k-2)！ /x^(k- 1)
(x- 1) 5 coefficient f (5) (1)/5! =-3! /5! =- 1/20
Choose a.
36: Series σ (n = 1, ∞) x n/2 n n n/n! Convergence radius of.
If x=2, it becomes σ (n = 1, ∞) n n/n! , divergent;
Sterling formula uses factorial n! ≈√(2nπ)(n/e)^n
σ(n= 1,∞)x^n/2^n.n^n/n！
≈σ(n= 1,∞)x^n/2^n.n^n/√(2nπ)(n/e)^n
=σ(n= 1,∞)(x/2)^n)./√(2nπ)/e^n
=σ(n= 1,∞)(ex/2)^n)./√(2nπ)
=( 1/√(2π))σ(n= 1,∞)(ex/2)^n)/n
(ex/2)^n<； 1，ex/2 & lt； 1，x & lt2/e，
Convergence radius 2/e
Choose c;
37: Use approximate formula:
sinx=x-x？ /3! , | x |< 1 (radian), what is the numerical error?
Use the remainder formula.
rn=f^(n+ 1)(θ)(x-x0)^(n+ 1)/(n+ 1)！
The fourth term of the series is 0, there is no error, and the remainder of R4 is considered.
R4=f^(5)(θ)x^5/5！
f=sinx,f'=cosx,f''=-sinx,f'''=-cosx,f^(4)=sinx,f^(5)=cosx
R4=cos(θ)x^5/5！ & ltcos( 1)/ 120=0.0045
Choose C.