Question

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?

Answer 1

Using `5` terms, we gte:

`e^2 ~~ 7`

Explanation:

The Maclaurin Series for `e^x` is as follows:

`e^x = 1 + x + (x^2)/(2!) + (x^3)/(3!) + (x^4)/(4!) + (x^5)/(5!) + ...`

So the power series sought for `e^2` can be gained by putting `x=2`, viz:

`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 ...`