04.52calculatinge

Calculating e

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{Not part of the course}

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Euler's Number, e, is an irrational number. $e \approx 2.7182818 \dots$

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

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Method 1

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

ie

… k = 2 … … $e \approx \left( 1 + \dfrac{1}{2} \right)^2 = \left( \dfrac{3}{2} \right)^2 = 2.25$
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… k = 3 … … $e \approx \left( 1 + \dfrac{1}{3} \right)^3 = \left( \dfrac{4}{3} \right)^3 = 2.37037037$

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

etc

This is best done in a spreadsheet:

  • The Difference column shows the difference between the calculated e and the correct value of e
  • This gives an idea of the accuracy of the calculated value of e
04.52table1.GIF

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

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

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ie

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

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

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

etc

Again, using a spreadsheet:

04.52table2.GIF

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

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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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