Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Euler's Theorem

Date: 05/09/2003 at 08:35:40
From: Ciaran Mc Keown
Subject: It is about Gauss' fundamental theorem of algebra.

In such a theorem Gauss suggests that for any polynomial equation of 
degree n there are n roots, although such may be imaginary. I would 
like to know the solution of the equation sine of a variable x equals 
2. 

I don't understand how there could be a solution real or imaginary to 
such an equation.


Date: 05/09/2003 at 09:19:57
From: Doctor Luis
Subject: Re: It is about Gauss' fundamental theorem of algebra.

Hi Ciaran,

The fundamental theorem of algebra, first proved by Gauss, only 
applies to polynomials. The function sin(x) isn't a polynomial; 
rather, it is a transcendental function.

You are absolutely right in that sin(x)=2 doesn't have any real 
solutions, nor purely imaginary solutions. However, solutions do exist 
and they are fully complex numbers.

To understand why solutions do exist, consider Euler's theorem, which 
states:

  e^(it) = cos(t) + i sin(t)

Therefore,

  e^(-it) = cos(-t) + i sin(-t)
          = cos(t) - i sin(t)

Subtracting these two equations, we get

  e^(it) - e^(-it) = 2 i sin(t)

or
            e^(it) - e^(-it)
  sin(t) = ------------------
                 2 i

Now, let's find what value of t gives us sin(t) = 2
       e^(it) - e^(-it)
  2 = -----------------
           2 i

  4i = e^(it) - e^(-it)

  e^(it) - 4i - e^(-it) = 0

Now, let's multiply both sides by e^(it)

  e^(2it) - 4i e^(it) - 1 = 0

Interesting. If you substitute y=e^(it), then y^2=e^(2it), this means 
that now we have the quadratic equation:

  y^2 - 4i y - 1 = 0

Using the quadratic formula, we can solve this quadratic:

       +4i +/- sqrt(-16 - 4(1)(-1))
  y = -----------------------------
                 2(1)

    = (4i +/- sqrt(-12))/2

    = 2i +/- i sqrt(3)

  y = (2 +/- sqrt(3)) i

But y = e^(it), therefore 

 it = ln(y) + (2pi*i)*n,   where n is any integer

the 2pi*i*n is there because e^z is periodic with period 2pi*i,
therefore when you take the inverse of e^z (z complex), you must add 
an integer multiple to the answer (giving us infinitely many possible 
complex number solutions).

Divide by i and you get

  t = 2pi*n - i ln(y)

To take the logarithm of y, remember that i = e^(i*pi/2). As for the 
coefficient(s) 2 +/- sqrt(3) they are both real.

Therefore,

  ln(y) = ln( (2 +/- sqrt(3)) e^(i*pi/2) )
        = ln(2 +/- sqrt(3)) + i*pi/2  +  2pi*i*m,   m any integer

Therefore, the full answer for sin(t) = 2 is

  t = 2pi * n - i (ln(2 +/- sqrt(3)) + i*pi/2  +  2pi*i*m)

    = (2pi*n + pi/2 + 2pi*m) - 

    = (2pi*(m+n) + pi/2) - i ln(2 +/- sqrt(3))

  t = (2pi*N + pi/2) - i ln(2 +/- sqrt(3))

where N = m+n can be any integer. (In this case the m,n since they
both contributed the same thing to the real part of t, but for
other problems, they might not.)

I'll write it again, just because it's such a nice answer:

   The solutions to the equation

         sin(t) = 2
   are
         t = (2pi*N + pi/2) - i ln(2 +/- sqrt(3))

   where N can be any integer. Note you can choose either + or -
   in the answer. For either choice, there are infinitely many
   such solutions.

Infinitely many solutions is quite a change from the fundamental 
theorem of algebra, which says polynomials have only as many solutions 
as their degree. The sine function is definitely not a polynomial!

As a final note, sin(x)=0 also has infinitely many solutions. They are 
x=2pi*n, where n is any integer. Does this look familiar? :)

I hope this answered your question. Let us know if you have any 
others.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 05/09/2003 at 09:33:05
From: Doctor Jerry
Subject: Re: It is about Gauss' fundamental theorem of algebra.

Hi Ciaran,

You want to solve sin(x)=2?  What's the connection with Gauss' 
Fundamental Theorem of Algebra? I see none.

If you enter 2 in your calculator and press the arcsin key, your 
calculator (or, at least, many calculators) will return (1.57,-1.32), 
which is a complex number.

The sine of a complex number z=x+i*y is often defined by

  sin(z) = [e^{i*z} - e^{-i*z}]/(2i).

If we then set 2=sin(z) we can solve without great difficulty 
(quadratic formula) for e^{i*z}, finding

  e^{i*z}=(2+sqrt(3))*i

Taking ln of both sides gives

  z=(1.57,-1.32), approximately.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Trigonometry

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/