Friday, July 19, 2013

Do A^x = 0 and A^x + B^x = 0 have roots?

In this blog entry we will look at two equations and see if there are solutions to them.

Things to Know:
In the realm of complex numbers, i = √(-1).

For the function f(x) = ln x:

If x > 0 and x is real, then ln x is also real and ln x = ln x.

If x < 0 and x is real, then ln x is complex and ln x = ln |x| + i * π.

If x is a complex number, such that x = a + bi, then:
ln x = ln r + i * θ where:
r = √(a^2 + b^2)
θ = atan(b/a) where -π ≤ θ ≤ π

Laws of Logarithm Needed:

Where a and b are any numbers:
ln (a^b) = b * ln a
ln(a * b) = ln a + ln b
ln(a / b) = ln a - ln b

And (a + b*i)^n = r^n * e^(i * π * θ) = r^n * (cos(n*θ) + i * sin(n*θ))

A^x = 0

(I) A^x = 0

Start by taking the logarithm of both sides:

(II) ln (A^x) = ln 0
(III) x * ln A = ln 0

And theoretically,

(IV) x = ln 0/ln A

Immediately we see a problem. The expression ln 0 is undefined. The function ln x approaches negative infinity when x approaches zero from the right side (that is x > 0 and calculating ln x as x is decreasing).

However, when x approaches from the left side, ln x approaches -infinity + i * π. There is no complex number makes the statement A^x = 0 true.

Therefore, A^x has no solutions. There is no x that makes statement (IV) in this section true.

A^x + B^x = 0

Let's see if we can get a closed form solution for x.

(I) A^x + B^x = 0

Assume that B can be rewritten as B = A^n., where n is any power. Then:

(II) A^x + (A^n)^x = 0
(III) A^x + (A^x)^n = 0

Factoring out A^x:

(IV) A^x * (1 + (A^x)^(n-1) ) = 0

From (IV) either:
A^x = 0, which can not happen sense from the previous section, A^x has no roots, real or complex.

So we have to look at the other possibility: (1 + (A^x)^(n-1) ) = 0.

Then:
(V) (A^x)^(n-1) = -1
(VI) A^x = -1^(1/(n-1))

The polar representation for -1 + 0i is 1*e^(i*π), a general solution for (VI) is:

(VII) A^x = cos((n-1) * π)+ i * sin((n-1)*π)

If n is an integer, then the roots of -1 can be found by the formula:

(-1)^(1/n) = cos((2*k+1)*π/n) + i*sin((2*k+1)*π/n) for k
for k=0,1,2,...,n-1

(Source: Zwillinger, Daniel. CRC Standard Mathematical Tables and Formulae, 32nd Edition. CRC Press, 2012)

From (VII) of this section, let α = cos((n-1) * π) and β = sin((n-1)*π) and

(VIII) A^x = α + i * β
(IX) ln A^x = ln (α + i * β)
(x) x * ln A = ln(α + i * β)

Hence, a general solution to A^x + B^x = 0 where B = A^n is:

(XI) x = ln(α + i * β)/ln A
Where α = cos((n-1) * π) and β = sin((n-1)*π)

There are no real solutions to A^x + B^x = 0, only complex ones.

An Example
2^x + 8^x = 0

Since 8 = 2^3,

2^x + (2^3)^x = 0
2^x + (2^x)^3 = 0
2^x * (1 + (2^x)^2) = 0

There is no solution for 2^x = 0, so we look to (1 + (2^x)^2) = 0.

(1 + (2^x)^2) = 0
(2^x)^2 = -1

The square roots of -1 are i and -i.

One solution is:
2^x = i
x * ln 2 = ln i
x = ln i / ln 2 ≈ 2.266180071i

The other is:
2^x = -i
x * ln 2 = ln -i
x = ln -i / ln 2 ≈ -2.266180071i

And that is a way to find the roots of A^x + B^x = 0.

Until next time, Eddie

This blog is property of Edward Shore. 2013