✔ 最佳答案
e^x=1+x+x^2/2!+x^3/3!+...+x^N/N!+x^(N+1)e^z/(N+1)!
where z between (0,x)
The error is x^(N+1)e^z/(N+1)!
Now x=1/2, e^z will get the maximum value when z=1/2
The upper bound of the error is
1/2^(N+1)e^(1/2)/(N+1)!
correct to 4 decimal places.
then
1/2^(N+1)e^(1/2)/(N+1)!<=1/10^4
1/2^(N+1)16487<=(N+1)!
the least value of N is 6
e^x=1+x+x^2/2!+x^3/3!+x^4/4!+x^5/5!+x^6/6!
So the least number of terms required to estime e^(1/2) correct to 4 decimal places is equal to 7
NOTE
when x=1/2
e^1/2
=1.648721271
1+x+x^2/2!+x^3/3!+x^4/4!+x^5/5!+x^6/6!
=1.648719618
the approximation is equal to 4 decimal places!!
