What is the value of 2.34188136e-8
Answers
Answer:
Step-by-step explanation:
Calculating the Value of e
There are several ways to calculate the value of e. Let's look at the historical development.
Using a Binomial Expansion
If n is very large (approaches infinity) the value of `(1+1/n)^n`approaches e.
This is not an efficient way to find `e`. Even if we go out to n = 100,000, our value is only correct to the 4th decimal place.
`e~~[(1+1/n)^n]_(n=100000)` `=2.718268237`
Here's the graph demonstrating this expansion:
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e = 2.71828...`y=(1+1/n)^n`
Graph of `y=(1+1/n)^n`, showing the limit as `n->+-oo` is e.
Another Expansion
As n becomes very small, `(1+n)^(1"/"n)` approaches the value of e.
We can obtain reasonable accuracy with a very small value of n.
`e~~[(1+n)^(1"/"n)]_(n=0.000000001)` `=2.718281827`
Let's see the graph of the situation.
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e = 2.71828...`y=(1+n)^(1//n)`
Graph of `y=(1+n)^(1"/"n)`, showing the y-intecept (the limit as `x->0`) is e.
(There is actually a "hole" at n = 0. Can you understand why?)
Newton's Series Expansion for e
The series expansion for e is
`e^x=1+x+1/2x^2+1/6x^3+` `1/24x^4+...`
Replacing x with 1, we have:
`e=1+1+1/2(1)^2+` `1/6(1)^3+` `1/24(1)^4+...`
We can write this as:
`e=sum_(n=0)^oo(1/(n!))`
This series converges to give us the answer correct to 9 decimal places using 12 steps:
`e~~sum_(n=0)^12(1/(n!))=2.718281828`
Brothers' Formulae
Recently, new formulae have been developed by Brothers (2004) which make the calculation of e very efficient.
`e=sum_(n=0)^oo(2n+2)/((2n+1)!`
We only need 6 steps for 9 decimal place accuracy:
`e=sum_(n=0)^6(2n+2)/((2n+1)!)=` `2.718281828`
Graphical Demonstration of e
The area under the curve `y=1/x` between 1 and e is equal to `1` unit2.
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Area under the curve `y=1/x` between `1` and `e`.
Reference
Brothers, H.J. 2004. Improving the convergence of Newton's series approximation for e. College Mathematics Journal 35(January):34-39..
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