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

HP 32S II Statistical Formulas

HP 32S II Statistical Formulas Statistics Formulas The 32S II can store formulas for evaluations, including formulas involving stat variable...