Question

The exponential function can be represented as an infinity power series as follows.

e^x=1 + x/1! + x^2/2! + x^3/3!+⋯,-∞<x<∞

1. Calculate e (which is the same as e^1) to 4 significant figures by using as many terms of the power series expansion as is necessary. Show all calculations.
2. Calculate e^3 to 2 significant figures by using as many terms of the above power series expansion as is necessary. Show all calculation.

please help, this is sample question of assessment which I have tomorrow.

Thanks

Answer 1

e^1 --> substitute 1 wherever you see x:
e^1=1 + 1/1! + 1^2/2! + 1^3/3!+⋯,-∞<x<∞
Add as many terms as you need to, until the sum stays at the same 4 significant figures.

Similar idea for x=2

Answer 2

for e^1
sum(1) = 1
sum(2) = 1 + 1 = 2
sum(3) = 2 + 1/2! = 1/2 = 2.5
sum(4) = 2.5 + 1/3! = 2.5 + 1/6 = 8/3 = 2.666..
sum(5) = 8/3 + 1/4! = 8/3 + 1/24 = 65/24 = 2.70833..
sum(6) = 65/24+1/5! = 65/24+1/120 = 163/60=2.71666..
sum(7) = 163/60+1/6! =163/60+1/720 =1951/720=2.7180555...
sum(8) = 1951/720+1/5040 = 2.718253..
sum(9) = 2.718253.. + 1/40320 = 2.718278...

The first 4 significant figures have not changed from sum(7) to sum(8), so 8 terms would do it

( 60 years ago, when I studied this, we of course had no calculators and had to do those divisions using pencil and paper. So easy now)

For e^3 your terms would be
1 + 3/1 + 9/2 + 27/6 + 81/24 + 243/120 + ..

notice the terms are converging much slower, so you will need a lot more terms. e^3 = appr 20.0855
Good luck
I made up a simple "BASIC" computer program (from the 1970's) and got 20.0855 after 15 terms.