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 ...