02.42calculatinge

Calculating e

{Not part of the course}

Euler's Number, e, is an irrational number.

… … $e \approx 2.7182818 \dots$

e can be calculated in a number of ways. Here are two of them.

Method 1

… … $e = \begin{matrix} \lim \\ k \rightarrow \infty \\ \end{matrix} \left( 1 + \dfrac{1}{k} \right)^k \qquad \text{ for } k \in Z$

… … ie

… … $k = 2 \qquad \qquad e \approx \left( 1 + \dfrac{1}{2} \right)^2 = \left( \dfrac{3}{2} \right)^2 = 2.25$

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… … $k = 3 \qquad \qquad e \approx \left( 1 + \dfrac{1}{3} \right)^3 = \left( \dfrac{4}{3} \right)^3 = 2.37037037$

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… … $k = 4 \qquad \qquad e \approx \left( 1 + \dfrac{1}{4} \right)^4 = \left( \dfrac{5}{4} \right)^4 = 2.44140625$

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… … etc

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This is best done in a spreadsheet:

02.42table1.GIF

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Note: In a similar way, we can find the value of ex for any value of x

.

… … $e^x = \begin{matrix} \lim \\ k \rightarrow \infty \\ \end{matrix} \left( 1 + \dfrac{x}{k} \right)^k$

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Method 2

… … $e = 1 + \begin{matrix} \lim \\ k \rightarrow \infty \\ \end{matrix} \sum\limits_{n=1}^{k} { \dfrac{1}{n!} } = 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} + \dots$

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… … recall that $n! = n(n-1)(n-2) \dots (3)(2)(1)$

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… … ie

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… … $k = 2 \qquad \qquad e \approx 1 + \dfrac{1}{1!} + \dfrac{1}{2!} = 1 + 1 + \dfrac{1}{2} = 2.5$

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… … $k = 3 \qquad \qquad e \approx 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} = 1 + 1 + \dfrac{1}{2} + \dfrac{1}{6} = 2.6666667$

.

… … $k = 4 \qquad \qquad e \approx 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} = 1 + 1 + \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{24} = 2.7083333$

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… … etc

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Again, using a spreadsheet:

02.42table2.GIF

This formula is clearly much more efficient.

  • It approaches the exact value of e much more quickly.
    • After 12 iterations this method has calculated e correct to 9 decimal places.
    • The previous method took 10,000,000 iterations to get e accurate to 6 decimal places.

Note In a similar way, we can find the value of ex for any value of x

.

… … $e^x = 1 + \begin{matrix} \lim \\ k \rightarrow \infty \\ \end{matrix} \sum\limits_{n=1}^{k} { \dfrac{x}{n!} } = 1 + \dfrac{x}{1!} + \dfrac{x}{2!} + \dfrac{x}{3!} + \dfrac{x}{4!} + \dots$

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Note Of course, the quickest way to calculate e is to put $e^1$ on your calculator!!!

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