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The Euler and De Moivre identities are the fundamental identities for deriving trigonometric formulas. From the identity e j θ = cos θ + j sin θ and the conjugate identity e - j θ = ( e j θ ) * = cos θ - j sin θ , we have the Euler identities for cos θ and sin θ :

cos θ = e j θ + e - j θ 2 sin θ = e j θ - e - j θ 2 j

These identities are illustrated in [link] .

Figure one is a cartesian graph with multiple labeled points connected by line segments.  Three points along with the origin with lines connecting them create a square, sitting diagonally with diagonal resting on the horizontal axis. The point in the fourth quadrant is labeled e^(-jθ). The point opposite the origin, and sitting on the horizontal axis in the positive direction, is labeled e^(jθ) + e^(-jθ). The point in the first quadrant, completing the square, is labeled e^(jθ). This point's horizontal and vertical values are labeled, measuring cosθ in the horizontal direction and sinθ in the vertical direction. A fourth point located on the vertical axis is labeled e^(jθ) - e^(-jθ). This point is connected to the vertex of the square that is located in the first quadrant. Figure one is a cartesian graph with multiple labeled points connected by line segments.  Three points along with the origin with lines connecting them create a square, sitting diagonally with diagonal resting on the horizontal axis. The point in the fourth quadrant is labeled e^(-jθ). The point opposite the origin, and sitting on the horizontal axis in the positive direction, is labeled e^(jθ) + e^(-jθ). The point in the first quadrant, completing the square, is labeled e^(jθ). This point's horizontal and vertical values are labeled, measuring cosθ in the horizontal direction and sinθ in the vertical direction. A fourth point located on the vertical axis is labeled e^(jθ) - e^(-jθ). This point is connected to the vertex of the square that is located in the first quadrant.
Euler's Identities

The identity e j θ = cos θ + j sin θ also produces the De Moivre identity:

( cos θ + j s i n θ ) n = ( e j θ ) n = e j n θ = cos n θ + j sin n θ .

When the left-hand side of this equation is expanded with the binomial expansion, we obtain the identity

k = 0 n n k ( cos θ ) n - k ( j sin θ ) k = cos n θ + j sin n θ .

Binomial Coefficients and Pascal's Triangle. The binomial coefficients ( k n ) in [link] are shorthand for the number

n k = n ! ( n - k ) ! k ! ' k = 0 , 1 , ... , n .

This number gives the coefficient of x n - k y k in the expansion of ( x + y ) n . How do we know that there are ( k n ) terms of the form x n - k y k ? One way to answer this question is to use Pascal's triangle, illustrated in [link] . Each node on Pascal's triangle shows the number of routes that terminate at that node. This number is always the sum of the number of routes that terminate at the nodes just above the node in question. If we think of a left-hand path as an occurrence of an x and a right-hand path as an occurrence of a y , then we see that Pascal's triangle keeps track of the number of occurrences of x n - k y k .

Figure two is a series of connected points forming pascal's triangle. From a top vertex, a point extends to the left and right at 45 degree angles. These lines continue only for a short distance, where they stop at a point and split to form two new lines at 45 degree angles to the left and right. This continues through four steps, forming a series of triangles and squares. Every line drawn towards the right is labeled, y, and every line drawn towards the left is labeled x. The points on the outside of the figure are labeled 1, and the points on the inside of the figure are labeled with the addition of the numbers to their upper-right and upper-left, 2 in the center, 3, and 3 below the 2, and 4, 6, 4 on the inside of the bottom row. Figure two is a series of connected points forming pascal's triangle. From a top vertex, a point extends to the left and right at 45 degree angles. These lines continue only for a short distance, where they stop at a point and split to form two new lines at 45 degree angles to the left and right. This continues through four steps, forming a series of triangles and squares. Every line drawn towards the right is labeled, y, and every line drawn towards the left is labeled x. The points on the outside of the figure are labeled 1, and the points on the inside of the figure are labeled with the addition of the numbers to their upper-right and upper-left, 2 in the center, 3, and 3 below the 2, and 4, 6, 4 on the inside of the bottom row.
Pascal's Triangle and the Binomial Coefficients

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Source:  OpenStax, A first course in electrical and computer engineering. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col10685/1.2
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