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
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
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.
To solve these questions, we will use the power series expansion of the exponential function. Let's go step by step:
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 necessary.
The power series expansion for e^x is given by: e^x = 1 + x/1! + x^2/2! + x^3/3! + ...
To calculate e, we substitute x = 1 into the power series expansion and add up the terms until we reach the desired precision (4 significant figures).
e = 1 + 1/1! + 1^2/2! + 1^3/3! + ...
= 1 + 1 + 1/2 + 1/6 + ...
To get 4 significant figures, we need to add up terms until the magnitude of the next term becomes smaller than the desired precision. In this case, we stop adding terms when the magnitude of the next term is smaller than 0.0001 (because 0.0001 is the fourth decimal place).
Continuing the calculation:
e ≈ 1 + 1 + 1/2 + 1/6
≈ 2.500
So, e ≈ 2.500 to 4 significant figures.
2. Calculate e^3 to 2 significant figures by using as many terms of the power series expansion as necessary.
Again, we use the power series expansion for e^x:
e^x = 1 + x/1! + x^2/2! + x^3/3! + ...
To calculate e^3, we substitute x = 3 into the power series expansion and add up the terms until we reach the desired precision (2 significant figures).
e^3 = 1 + 3/1! + 3^2/2! + 3^3/3! + ...
= 1 + 3 + 9/2 + 27/6 + ...
To get 2 significant figures, we add up terms until the magnitude of the next term becomes smaller than the desired precision. In this case, we stop adding terms when the magnitude of the next term is smaller than 0.01 (because 0.01 is the second decimal place).
Continuing the calculation:
e^3 ≈ 1 + 3 + 9/2 + 27/6
≈ 13.500
So, e^3 ≈ 13.500 to 2 significant figures.
Remember to show all calculations by carrying them out step by step and considering the desired precision. Good luck with your assessment!