Proof for Cotes' Formulas

In today's blog, I will show how De Moivre's Formula can be used to establish Cotes' Formulas.

Lemma 1:

Let μ_n be the set of all nth roots of unity

Then:



Proof:

(1) We know that xⁿ - 1 has n root from the Fundamental Theorem of Algebra.

(2) We also note that for all ζ, we have (ζ)ⁿ = 1

(3) Based on #2, the Fundamental Theorem of Algebra gives us:

xⁿ - 1 = (x - ζ₁)*(x - ζ₂)*(x - ζ₃)*...*(x - ζ_n)

(4) Step #3 can then be restated as:



QED

Corollary 1.1:



Proof:

This follows from applying De Moivre's proof [see Corollary 1.1, here] for the existence of the roots of complex numbers with the result in Lemma 1 above.

QED

Corollary 1.2:



Proof:

This follows from applying De Moivre's proof [see Corollary 1.2, here] for the existence of the roots of complex numbers with the result in Lemma 1 above.

QED

Theorem 2:

If n is odd:


If n is even:

Proof:

(1) From Corollary 1.1 above, we have:



(2) Now, we can remove the k=0 case to get the following:



(3) Now, it is also possible to pair up roots. That is, let l = n - k.

(4) Then we have:

cos(2[(n-k)π]/n) =
=cos([2nπ - 2kπ]/n) =
=cos([2nπ/n] - [2kπ/n]) =
=cos(2π - 2kπ/n) =
cos(-2kπ/n) = cos(2kπ/n) [See Property 9, here]

sin(2[(n-k)π]/n) =
=sin([2nπ - 2kπ]/n) =
=sin([2nπ/n] - [2kπ/n]) =
=sin(2π - 2kπ/n) =
=sin(-2kπ/n) = -sin(2kπ/n) [See Property 4, here]

(5) So, pairing the roots of unity k with its n-k pair gives us:















(6) So, if n is odd, we have:



(7) If n is even, then at k=n/2 we have:

cos([2(n/2)π]/n) = cos(nπ/n) = cos(π) = -1 [See Property 6, here]

sin([2(n/2)π]/n) = sin(nπ/n) = sin(π) = 0 [See Property 1, here]

So:

[x - cos([2(n/2)π];/n) - isin([2(n/2)π]/n)] =

= x - (-1) - i*0 = x + 1

(8) So, if n is even, we have:



since we can see that n/2 is not included in any of the pairings:

k = 1 .. (n/2)-1

n-k = (n/2)+1 .. n-1

QED

Corollary 2.1:




Proof:

(1) Assume that n is odd.

(2) Then from Theorem 2 above, we have:



(3) Since n is odd, there exists m such that n = 2m+1.

(4) This then gives us:



(5) Setting x = (a/x) gives us:



(6) Multiplying both sides by x^2m+1 gives us:


(7) Assume that n is even.

(8) Then, from Theorem 2 above we have:



(9) Since n is even, there exists an integer m such that n = 2m.

(10) This then gives us:



(11) Setting x = (a/x) gives us:



(12) Multiplying both sides by x^2m gives us:


QED

Theorem 3:

If n is odd:



If n is even:



Proof:

(1) From Corollary 1.2 above, we have:



(2) Now, it is also possible to pair up roots. That is, let l = n - k -1.

(3) Then we have:

cos([2(n-k-1)+1]π/n) =
=cos([2n - 2k -1]π/n) =
=cos([2nπ/n] - [(2k+1)π/n] ) =
=cos(2π - (2k+1)π/n) =
cos(-[2k+1]π/n) = cos([2k+1]π/n) [See Property 9, here]

sin([2(n-k-1)+1]π/n) =
=sin([2n - 2k - 1]π/n) =
=sin([2nπ/n] - [(2k+1)π/n]) =
=sin(2π - (2k+1)π/n) =
=sin(-[2k+1]π/n) = -sin([2k+1]π/n) [See Property 4, here]

(4) So, pairing each roots of unity k with its n-k-1 pair gives us:















(5) If n is even, the we have:



(6) If n is odd, then at k=(n-1)/2:





and







(7) So:



= x - (-1) - i*0 = x + 1


(8) So, if n is odd, we have:



since we can see that (n-1)/2 is not included in any of the pairings:

k = 0 .. (n-1)/2 - 1

n-k-1 = (n-1)/2 + 1 .. n-1

since:

n-[(n-1)/2 - 1] - 1 =
= n - (n-1)/2 + 1 - 1 =
= (2n - [n-1])/2 =
= (n + 1)/2 =
= (n - 1 + 2)/2 =
(n-1)/2 + 1

QED

Corollary 3.1:




Proof:

(1) Assume that n is odd.

(2) Then from Theorem 3 above, we have:



(3) Since n is odd, there exists m such that n = 2m+1.

(4) This then gives us:



(5) Setting x = (a/x) gives us:



(6) Multiplying both sides by x^2m+1 gives us:



(7) Assume that n is even.

(8) Then, from Theorem 3 above we have:



(9) Since n is even, there exists an integer m such that n = 2m.

(10) This then gives us:



(11) Setting x = (a/x) gives us:



(12) Multiplying both sides by x^2m gives us:



QED

References

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001

Roots of Unity

For n ≥ 3, xⁿ = 1 has three or more distinct roots. While (1, -1) were well known during Abraham de Moivre's day. The other roots were not so well known. It took de Moivre's formula to establish these roots on a firm footing. Roger Cotes had independently discovered the properties of the roots of unity as demonstrated by his formulas. Unfortunately, he had died before he could explain many of his discoveries in detail. It was left to de Moivre to explain Cotes' formulas.

In a previous blog, I showed how de Moivre came up with his famous formula based on the work of Francois Viete. In today's blog, I will show how he provided a proof for the existence of the roots of unity.

In 1739, de Moivre used his formula to find the nth root for any complex number (a + bi).

Theorem 1: For any complex number, there are n distinct nth roots.

if a,b ≠ 0, then:



where:

0 ≤ k ≤ n-1

Proof:

(1) Let a,b be real numbers such that either a ≠ 0 or b ≠ 0

(2) We note first that:






(3) Since:



It is clear that:



And further that:



(4) Therefore, there exists φ such that [see here for review of cosine if needed]:



(5) Since cos²(φ) + sin²(φ) = 1 (see Corollary 2, here), we can see that:

sin²(φ) = 1 - cos²(φ) =



(6) So that we also have now:



(7) We can now conclude:



since:







(8) Using De Moivre's formula (see Theorem 1, here), we know that:





(9) From the basic properties of sin and cosine, we know that:

cos(φ + 2kπ) + isin(φ 2kπ) = cos(φ) + isin(φ)

(10) Multiplying √a² + b² to both sides gives us:







(11) Combining step #7 with step #10 gives us:



(12) Now, we know that each of the values are distinct for 0 ≤ k ≤ n-1 since none of the angles vary by a multiple of 2π.

QED

Corollary 1.1: Roots of Unity

There are n roots of unity such that:



where:

0 ≤ k ≤ n-1

Proof:

(1) Let a=1 and b=0

(2) Then a+bi = 1+0*i = 1

(3) Using Theorem 1 above, we have:



(4) Using step #4 and step #6 in Theorem 1 above, we have:



and



(5) So, φ = 0

(6) Since:



we are left with:



where:

0 ≤ k ≤ n-1

QED

Corollary 1.2: Roots of -1

There are n roots of unity such that:



where:

0 ≤ k ≤ n-1

Proof:

(1) Let a=-1 and b=0

(2) Then a+bi = -1+0*i = -1

(3) Using Theorem 1 above, we have:



(4) Using step #4 and step #6 in Theorem 1 above, we have:



and



(5) So, φ = π

(6) Since:



we are left with:





where:

0 ≤ k ≤ n-1

QED

References

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001

De Moivre's Famous Formula

The simple proof of DeMoivre's Formula by Leonhard Euler does not do justice to Abraham de Moivre's important breakthrough on the roots of unity. A person viewing Euler's proof might be surprised that Sir Isaac Newton often answered questions about Principia with: "Go to Mr. de Moivre, he knows these things better than I do."

In today's blog, I will show how de Moivre derived his formula based on Francois Viete's solution of Van Roomen's Problem.

Theorem 1: De Moivre's Formula



Proof:

(1) Let x = 2 cos(α)

(2) Let l = cos(nα)

(3) We can define z such that:



(4) So that:

zⁿ = l + √l² - 1

So that:

zⁿ - l = √l² - 1

and:

(zⁿ - l)² = l² - 1

(5) Working through the equation in step #4, we get:

(zⁿ - l)² = z²ⁿ - 2lzⁿ + l² = l² - 1

Subtracting l² - 1 from both sides gives us:

z²ⁿ - 2lzⁿ + 1 = 0

(6) Now, from a Viete's famous solution to Van Roomen's problem (see Corollary 1.1, here), we have:

2 cos(nα) = f_n(2 cos α)

where:

f_n(x) =



(7) Putting step #6 in step #5 gives us:

z²ⁿ - 2*(1/2)f_n(x)zⁿ + 1 = z²ⁿ - f_n(x)zⁿ + 1 =0

(8) For n=1, we have:

z²⁽¹⁾ - f₁(x)z¹ + 1 = 0

(9) Now using the definition for f_n(x) in step #6 above, we can see that:

f₁(x) = (-1)⁰(1/[1-0])([1-0]!/0![1 - 2*0]!)x^{1 - 2*0} =

= 1*(1/1)(1!/[0!1!])x¹ =

= x

(10) This then gives us:

z² - xz + 1 = 0

(11) Now using the quadratic equation to solve for z (see Theorem, here) gives us:

z = (x ± √x² - 4)/2 =

= x/2 ± √(x/2)² - 1

(12) Since x = 2cos(α), this gives us:

z = [2 cos(α)]/2 ± √[2 cos(α)/2]² - 1 =

= cos(α) ± √cos²(α) - 1

(13) Since cos²(x) + sin²(x) = 1 [See Corollary 2, here], we have:

z = cos(α) ± √cos²(α) - 1 =

= cos(α) ± √(-1)[1 - cos²(α)] =

= cos(α) ± √(-1)sin²(α) =

= cos(α) ± [√(-1)]sin(α)

(14) Since l = cos(nα), from step #2 above, we have:












(15) Putting step #13 and step #14 together gives us:



QED

You might notice that De Moivre's result is ambiguous. Since there are n roots, it is impossible that its result could only be two possible values. This problem was not evident until after there was deeper understanding of the fundamental theorem of algebra. Interestingly, this fundamental theorem was not well understood until after De Moivre's discovery of the roots of unity which I will talk about in my next blog.

The corrected version of de Moivre's formula is:

cos(nα) + isin(nα) = [cos(α) + isin(α)]ⁿ

References

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001

Euler's Proof of De Moivre Formula

De Moivre's Formula is easily proven using induction. The following proof was first given by Leonhard Euler in 1748. Abraham De Moivre never stated his formula exactly in this form.

Lemma 1: (cos α + isin α)(cos β + isin β) = cos(α + β) + isin(α + β)

Proof:

(1) cos(α + β) = cos(α)cos(β) - sin(α)sin(β) [See Theorem 2, here]

(2) sin(α + β) = cos(α)sin(β) + cos(β)sin(α) [See Theorem 1, here]

(3) isin(α + β) = cos(α)isin(β) + cos(β)isin(α)

(4) Since isin(α)*isin(β) = -sin(α)sin(β), we have:

cos(α + β) + isin(α + β) =

= cos(α)cos(β) - sin(α)sin(β) + cos(α)isin(β) + cos(β)isin(α) =

= cos(α)[cos(β) + isin(β)] + isin(α)[cos(β) + isin(β)] =

= [cos(α) + isin(α)][cos(β) + isin(β)]

QED

Theorem 1: De Moivre's Formula

(cos α + i sin α)ⁿ = cos(nα) + isin(nα) for all α ∈ R.

Proof:

(1) For n=1:

(cos α + i sin α)¹ = cos(1*α) + isin(1*α)

(2) Assume that the premise is true for all values up to n such that:

(cos α + i sin α)ⁿ = cos(nα) + isin(n&alpha) for all α ∈ R.

(3) (cos α + isin α)ⁿ⁺¹ =
(cos α + isin α)ⁿ(cos α + isin α)

(4) Using the assumption in step #2, we have:

(cos α + isin α)ⁿ(cos α + isin α) =

[cos(n α) + isin(n α)](cos α + isinα)

(5) Using Lemma 1 above, we have:

[cos(n α) + isin(n α)](cos α + isinα) =

cos(n α + α) + isin(nα + α) =

= cos ([n+1]α) + isin([n+1]α)

QED

References

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001