Euler's TheoremDate: 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/ |
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