Question

How would you use the Maclaurin series for `e^-x` to calculate `e^0.1`?

Answer 1

`e^(0.1) ~~ 1.10517 (5dp)`

Explanation:

We know that

`e^(-x) = 1 - x + x^2/(2!)-x^3/(3!)+cdots`

is an alternating series and for `absx < 1` this series is absolutely convergent, so the error handled by cutting off after the `n` term is smaller than the higher order term left.

If we need to calculate `e^(-0.1)` within a precision `delta` then we solve first

`delta le x^n/(n!)` or `delta le (0.1)^n/(n!)` for instance, if `delta = 0.00001` then `n=4` would suffice. After that, `e^(0.1) = 1/e^(-0.1)`

`:. e^(0.1) = e^(-(-0.1))`
`:. e^(0.1) = 1-(-0.1)+(-0.1)^2/(2!)-(-0.1)^3/(3!)+(-0.1)^4/(4!) +`(higher terms)
`:. e^(0.1) ~~ 1+0.1+0.01/2+0.001/6+0.0001/24`
`:. e^(0.1) ~~ 1.1+0.005+0.00016667+0.000041667`
`:. e^(0.1) ~~ 1.10517083 ...`
`:. e^(0.1) ~~ 1.10517 (5dp)`

Compare with the calculator answer `e^0.1=1.10517091 ...`