Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've purchased "A Source Book in Mathematics" some time ago and I'm still baffled by De Moivre's paper on his formula. We all know the famous

$$\{\cos(x) + i \sin(x)\}^n = \cos(nx)+i \sin(nx)$$

but rarely we know where did De Moivre derive this equation. I'll put a part of "The analytic solution of certain equations of the third, fifth, seventh, ninth and other higher uneven powers, by rule similir ot those called Cardan's" which the first part of the chapter on De Moivre's formula.

Let $n$ denote any number whatever, $y$ an unknown quantity or root of this equation, and $a$ the absolute known qunatity, or what is called the homogeneum comparationis; let also the relation between these be expressed by the equation:

$$ny + \frac{{{n^2} - 1}}{{2 \times 3}}n{y^3} + \frac{{{n^2} - 1}}{{2 \times 3}}\frac{{{n^2} - 9}}{{4 \times 5}}n{y^5} + \frac{{{n^2} - 1}}{{2 \times 3}}\frac{{{n^2} - 9}}{{4 \times 5}}\frac{{{n^2} - 25}}{{6 \times 7}}n{y^7} + \cdots = a$$

From the nature of this series it is manifest, that $n$ be taken as any odd integer, either positive or negative, then the series will term nate and the equation become one of those above mentioned, the root of which is:

\begin{equation} y = \frac{1}{2}\sqrt[n]{\sqrt {1 + {a^2}} + a} - \frac{1}{2}\frac{1}{\sqrt[n]{\sqrt {1 + {a^2}} + a} } \end{equation} or \begin{equation} y = \frac{1}{2}\root n \of {\sqrt {1 + {a^2}} + a} - \frac{1}{2}\root n \of {\sqrt {1 + {a^2}} - a} \end{equation} or \begin{equation} y = \frac{1}{2}\frac{1}{{\root n \of {\sqrt {1 + {a^2}} - a} }} - \frac{1}{2}\root n \of {\sqrt {1 + {a^2}} - a} > \end{equation} or \begin{equation} y = \frac{1}{2}\frac{1}{{\root n \of {\sqrt {1 + {a^2}} - a} }} - \frac{1}{2}\frac{1}{{\root n \of {\sqrt {1 + {a^2}} + a} }} \end{equation}

(...)

Again, if the terms of the preceeding equation be alternately positive and negative, which is the same thing, if the series be as follows:

$$ny + \frac{{1-{n^2}}}{{2 \times 3}}n{y^3} + \frac{{1-{n^2} }}{{2 \times 3}}\frac{{9-{n^2}}}{{4 \times 5}}n{y^5} + \frac{{1-{n^2}}}{{2 \times 3}}\frac{{9-{n^2}}}{{4 \times 5}}\frac{{25-{n^2} }}{{6 \times 7}}n{y^7} + \cdots = a$$

its roots will be

\begin{equation} y = \frac{1}{2}\sqrt[n]{a+\sqrt {{a^2}-1} } + \frac{1}{2}\frac{1}{\sqrt[n]{\sqrt {{a^2}-1} + a} } \end{equation} or \begin{equation} y = \frac{1}{2}\root n \of {\sqrt {{a^2}-1} + a} + \frac{1}{2}\root n \of {a-\sqrt {{a^2}-1} } \end{equation} or \begin{equation} y = \frac{1}{2}\frac{1}{{\root n \of {a-\sqrt {{a^2}-1} }}} + \frac{1}{2}\root n \of {a-\sqrt {{a^2}-1} } \end{equation} or \begin{equation} y = \frac{1}{2}\frac{1}{{\root n \of {a-\sqrt {{a^2}-1} } }} + \frac{1}{2}\frac{1}{{\root n \of {\sqrt {{a^2}-1} + a} }} \end{equation}

It should be noted that if $\frac{n-1}{2}$ is an odd number, the sign of the root when found must be changed to the contrary sign.

He then explains:

Now by what artifices these formulae were discovered will clearly appear from the demonstrarion of the following theorm. In an unit circle let $x$ denote the versed sine of any arc, and $t$ that of another; and let the former arc be to the latter as 1 to $n$. Then assuming two equations which may be regarded as known, namely

$$1-2z^n+z^{2n} = -2z^nt$$

$$1-2z+z^2 = -2zx$$ on eliminating $z$, there will arise an equation by which the realtion between $x$ and $t$ will be determined.

Why do those last equations are regarded as known?

My last quote will be on his Lemma 1 of Miscellanea Analytica.

If $l$ and $x$ are the cosines of two arcs $A$ and $B$ of a circle of radius unity, and if the first arc is to the second as the number $n$ is to unity then:

$$x = \frac{1}{2}\root n \of {l + \sqrt {{l^2} - 1} } + \frac{1}{2}\frac{1}{{\root n \of {l + \sqrt {{l^2} - 1} } }}$$

This (and the first formulas for roots he provides) remind me of

$$\eqalign{ & {\cosh ^{ - 1}}x = \log \left( {x + \sqrt {{x^2} - 1} } \right) \cr & {\sinh ^{ - 1}}x = \log \left( {x + \sqrt {{x^2} + 1} } \right) \cr} $$

which can be put as

$$\eqalign{ & \frac{{{{\cosh }^{ - 1}}x}}{n} = \frac{1}{n}\log \left( {x + \sqrt {{x^2} - 1} } \right) = \log \left( {\root n \of {x + \sqrt {{x^2} - 1} } } \right) \cr & \frac{{{{\sinh }^{ - 1}}x}}{n} = \frac{1}{n}\log \left( {x + \sqrt {{x^2} + 1} } \right) = \log \left( {\root n \of {x + \sqrt {{x^2} + 1} } } \right) \cr} $$

and thus yield:

$$\eqalign{ & y = \sinh nx \cr & y = \cosh nx \cr} $$

I'm most interested in the infinite series De Moivre gives and the expressions of the roots. I'm also interested in connection between the whole exposition and the fact that $ \sinh ix = i\sin x $ and $ \cosh ix = \cos x $ which seem to be the key to the matter. Finally, I'd like the trigonometrical claims to be addressed and explained. Thanks in advance.

share|improve this question
    
I presume you already know what a versed sine is... –  J. M. Feb 7 '12 at 1:03
    
I do know. $ver\sin \theta = 1- \cos\theta$ Although I might as well just clicked your link. –  Pedro Tamaroff Feb 7 '12 at 1:07
1  
de moivre knew (i) vieta's formula expressing $\cos kt$ and $\frac{\sin(kt)}{\sin t}$ as polynomials in $\cos t$, (ii) binomial formula. i am wondering if the combination of the two does not produce the formulae $\cos t + \sin t)^k = \cos(kt) + i\sin(kt)$ for positive integers $k.$ –  abel Feb 26 '12 at 19:31
add comment

2 Answers

Well when I was doing A-level further maths, I thought this result was obvious when I first saw it, but noone else seemed to think so. You can certainly prove it by induction if you wish but there is an intuitive way to see it (which is the same as induction really).

Firstly, when we have two complex numbers of modului $r_1,r_2$ and arguments $\theta_1, \theta_2$...the product will have modulus $r_1r_2$ and argument $\theta_1 + \theta_2$. Proving this boils down to the addition formulae for sin and cos.

Ask yourself, what happens when the two complex numbers are the same?

Well then the modulus will be $r^2$ and the argument will be $2\theta$.

From this it is intuitive that if you multiply a complex number by itself $n$ times then the modulus will be $r^n$ and the argument will be $n\theta$. This is exactly what De-Moivre is saying (but for complex numbers of modulus $1$).

Whether or not this is historically accurate I do not know but as I learned about this stuff it seemed a natural progression to me.

share|improve this answer
7  
De Moivre's formula was 40 years before Euler's formula for the exponential of a complex angle, which was 50 years before the geometric interpretation of complex number multiplication. These later developments make the formula obvious but discovering it with the tools available in 1700 is more difficult. –  zyx Apr 4 '12 at 2:15
add comment

The equation $\cos(nx) + i\sin(nx) = (\cos(x) + i\sin(x))^n$ holds if $n = 0$ or $n = 1$. Assume it holds for $n$. Then

$$(\cos(x) + i\sin(x))^{n+1} = (\cos(x) + i\sin(x))^n(\cos(x) + i\sin(x))$$ so $$(\cos(x) + i\sin(x))^{n+1}= (\cos(nx) + i\sin(nx))(\cos(x) + i\sin(x)). $$ Now multiply and assemble real and imaginary parts to obtain $$(\cos(x) + i\sin(x))^{n+1} = ( \cos(nx)\cos(x) - \sin(nx)\sin(x)) + i(\sin(nx)\cos(x) + \cos(nx)\sin(x)$$ Now apply the addition formulae for the result.

Help! Editors! I am not sure how to use the align environment here. Your help would be wonderfully appreciated.

share|improve this answer
    
This is not what I'm asking for although I appreciate the contribution. I want to know how we can derive that identity under De Moivre's ideas and our modern ideas (but thinking we don't know the formula even exists). –  Pedro Tamaroff Feb 7 '12 at 1:17
    
If one can use Euler's formula $e^{i\theta} = cos(\theta)+isin(\theta)$ the proof is straightforward. Assuming that you are seeking a proof without using Euler's formula. –  Kirthi Raman Mar 3 '12 at 17:05
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.