How to find the approximate value by using the first five terms of the Maclaurin series?
#sum_(n=0)^oo e^2#
I know that the formula is #sum_(n=0)^oo (e^2/(n!))*x^n = e^3.2 + (e^3.2/1)x + (e^3.2/2)x^2+...#
...but I have to check my answer using a calculator. How do I do that exactly? I know that #a=0# , but what do I do about x to approximate it?
I know that the formula is
...but I have to check my answer using a calculator. How do I do that exactly? I know that
1 Answer
Using
# e^2 ~~ 7 #
Explanation:
The Maclaurin Series for
# e^x = 1 + x + (x^2)/(2!) + (x^3)/(3!) + (x^4)/(4!) + (x^5)/(5!) + ... #
So the power series sought for
# e^2 = 1 + 2 + (2^2)/(2!) + (2^3)/(3!) + (2^4)/(4!) + (2^5)/(5!) + ... #
If, as instructed, we restrict ourself to the first five terms, we get:
# e^2 ~~ 1 + 2 + (2^2)/(2!) + (2^3)/(3!) + (2^4)/(4!)#
# \ \ \ = 1 + 2 + (4)/(2) + (8)/(6) + (16)/(24)#
# \ \ \ = 3 + 2 + 4/3 + 2/3#
# \ \ \ = 5 + 6/3 #
# \ \ \ = 5 + 2 #
# \ \ \ = 7 #
We can compare this to a calculator answer:
# e^2 = 7.389056 ... #
And for further comparison, if we take further terms we get:
# 6 # terms:# 109/15 = 7.266666 ... #
# 7 # terms:# 331/45 = 7.355555 ...#
# 8 # terms:# 155/21 = 7.380952 ...#