Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**SteveB****Member**- Registered: 2013-03-07
- Posts: 577

Following a recent thread in the "help me" section I became curious about the following alternative interpretation of one of the

questions. The person who started the thread was actually only wanting help with basic integer power functions, but it got

me thinking about what happens in more complicated questions like raising a negative number to a rational power:

[EDIT: My initial thoughts on this were a bit naive. I have tried to tidy up my many mistakes a bit, but there might be a few still.

A better question might have been to ask: It is all very well defining the result, but is it well defined, and are there good

continuous function properties to make raising a negative number to a non-integer power valid? Suppose for example that we

re-write the rational number as (4/6) or (6/9) etc. (as is not allowed conventionally) do we get the number of solutions using

complex numbers that the denominator dictates and is there a logical inconsistency here ?]

[EDIT: Actually the real solution is a valid answer. Complex number solutions do not in fact really resolve the problem that

I had identified. The problem is bound to occur with a negative base raised to a non integer power, given that there is an

alternating sign created by the fact that with an even number of multiplications of a negative number the result is a positive

number, but with an odd number of terms the result is negative. You can in fact get a postive real solution in this case as

Mrwhy has pointed out. The number 3 as the denominator allows 3 solutions with two requiring i terms, and with one real solution.

However you can understand that I was a little uncomfortable that you cannot of course literally split an alternating sign sequence

to include only 2 thirds of a negative term. Also I do still think that, although this is rather unconventional, you could argue that

if you do not specify that the fraction has to be cancelled into a pair of coprime integers then chaos is possible because an infinite

number of solutions is possible for the same fraction because the number of solutions depends upon the denominator.]

(EDIT: At this point I should have realised that I have divided both sides by y and not noted that I also have to deal with

the case where y=0 and find that x = cube root(16) my calculator confused me initially into thinking that the real solution

was not valid.)

In other words: (to the limits of the calculator display)

As a result of the comment below I will also add:

Also I did realise afterwards that the modulus and angle/argument method could be used:

*Last edited by SteveB (2013-04-28 04:05:56)*

Offline

**Mrwhy****Member**- Registered: 2012-07-02
- Posts: 52

A real number answer is cube root of 16

It was only the TOOL you used that could not handle a negatine thing to a fractional power

Offline

**SteveB****Member**- Registered: 2013-03-07
- Posts: 577

You have got a good point there.

It seems that the real solution does work after all.

If you cube the real solution you get 16, then take the negative square root.

One of my graphics calculators does not even accept the input, and the other gives the first of my complex number answers.

Offline

**Mrwhy****Member**- Registered: 2012-07-02
- Posts: 52

I think we now have ALL the solutions

For we have Z cubed=16

and every equation with ODD power (like we have power 3 here) has at least ONE real answer

The complex answers, like z=x+iy come ONLY IN PAIRS (because the square root of minus one is PLUS or MINUS i

Here is an interesting variant of your equation

Suppose iZ=(-4) power 2/3

Then squareroot(iZ) = minus cuberoot 16

Can you solve that one?

How many square roots does i have?

Offline

**SteveB****Member**- Registered: 2013-03-07
- Posts: 577

Here is an interesting variant of your equation

Suppose iZ=(-4) power 2/3

Using basically the same method with Z = x + yi

I have got these solutions:

(1) x = 0 , y = -2.51984209979

(2) x = 2.18224727194 , y = 1.25992104989

(3) x = -2.18224727194 , y = 1.25992104989

(Interestingly the division by i has had the effect of a multiplication by -i upon the solutions.)

Then squareroot(iZ) = minus cuberoot 16

Can you solve that one?

Fot this I have got:

(1) x = 0 , y = -6.34960420787

(2) x = 5.49891854799 , y = 3.17480210394

(3) x = -5.49891854799 , y = 3.17480210394

(This change has had the effect of a further multiplication by the positive real cube root of 16.)

How many square roots does i have?

I would say 2.

EDIT: I have looked at some Wikipedia entries on this topic and Wikipedia does explain this quite well. It appears, if I am

understanding things correctly, that there is a serious problem that occurs if you have a function like

where R is a real number that could be irrational. The problem is that because of alternating sign issues the function is not

well behaved in that very small changes in R can lead to wild changes in f(R). It is not therefore continuous, and you cannot

give a satisfactory definition of f(R) for an irrational number R.

I did try calculating (-1)^(pi) on my calculator that allows complex numbers. It did give an answer that was a complex

number, but I suspect that what it is doing is rounding pi off to a certain number of decimal places and then working it

out using the rational number definition. If the (+/-) sign (or real/non real status) of these answers depend upon whether the

numerator/denominator are odd/even in their lowest terms then we cannot converge to the correct answer with increasing accuracy.

You cannot really argue for instance that you can converge to a solution by taking increasingly large numbers of terms if when

doing so there are cliff edge gradients with very small changes in R.

When considered in modulus/argument form we can be certain of the modulus of a solution to a negative number raised to an

irrational power, but the argument (angle in a complex number diagram) can be argued to have an infinite number of possible

values with arbitary choice of 2(pi)n [with integer valued n] added to the angle before multiplying by the irrational number.

When we convert the angle back into the correct range there are going to be an endless (countably infinite) series of new solutions

for the angle. So really any angle is as good as any other on the circle that is created. Consistent inverses are probably not going to

trace the answer back to the original negative exponential base. I did originally have the naive idea that you might be able to put the

negative number through a function to a magic complex number and then back again with the same function using the reciprocal of

the power index, but if there are an infinite number of solutions to both functions then to say the result is ambiguous is putting it

rather mildly.

In fact strictly speaking the same logic applies even to a positive number raised to an irrational power, in that it has infinitely

many solutions derived by adding multiples of 2(pi)n and mutiplying by the irrational number then adjusting to the correct angle

range.

However it does of course give a clear unambiguous result provided we simply take the conventional positive real solution, with

of course no sign problem, and all of the required continuous convergence properties.

Looking on the bright side this has given me a quick refresher course in complex numbers, on the less bright side I made quite

a big mistake in suggesting that (-4)^(2/3) had no real solution, when the ordinary mathematical solution does actually work.

Had the problem been (-4)^(3/2) there would have been a necessity then for an imaginary term and so complex numbers would

have been much more relevant (no real solution to the square root of -64 is possible).

*Last edited by SteveB (2013-05-20 04:34:02)*

Offline