[SOLVED] Fractional Indices and Negative Numbers

[GingerRobot]

Appears to work fine for positive bases Barand, but once we move into negatives, i think there are some problems.

 

For instance, if you input (-1)^0.5 you get 1; though the root of -1 is i.

[Barand]

OK modified it output the "i" if neg and power < 1

[GingerRobot]

Unfortunately i dont think its quite that simple.

 

For example, you should get answers with the following inputs of:

 

(-1)^1/2 = i

(-1)^1/3 = -1

(-1)^1/4 = 0.70711+0.7011i - the numerical values are actually sin(45 degrees) if you're interested

 

At present, your code returns i for all of these inputs.

 

I might take a look later and see if i can come up with a solution. I dont think it's going to be easy though.

[Barand]

so it's neg base number and even roots that get the "i" ?

 

(It's 40 years since I studied  maths so I'm a bit rusty)

[GingerRobot]

It seems so. When raising a negative number to a fractional power, if the denominator is even, you get a complex answer. If it is odd, you get a real answer.

 

I can't seem to figure out how it all works to be honest though.

[akitchin]

as far as i know, the order of operations dictates that the numerator of a rational exponent is evaluated first (assuming one re-writes the expression as a power beneath a root sign).  by that logic, the only case where a^(m/n) will produce a complex number is when a < 0, m is odd, n is even.

 

regardless, whether a root is complex or not is rather a matter of definitions and semantics in the case of rational exponents, not actual programming.  might as well save yourself a headache and take the absolute value of 'a'.  as you yourself pointed out, one could simply multiply any given rational expression by 2/2 and obtain an identical exponent (based on group theory) which won't jar your script.  you could always add an 'i' to the final output if you want to, but it would be a largely arguable point.

[btherl]

However, im fairly sure there is nothing to say that you have to use the fractional power inside the bracket. As far as i know (x^(1/b))^a is identical to (x^b)^(1/a).

 

Definitely one for my maths teacher i think.

 

At your service

 

Take the rule here:

 

(x^m)^n = x^(mn)

 

From this it follows that

 

(x^(1/b))^a

= x^(a/b)

 

(x^a)^(1/b)

= x^(a/b)

 

I assume that's what you meant, not (x^b)^(1/a)

 

These rules certainly don't work for b=0, and I would have doubts once things start getting negative..

[GingerRobot]

Sorry, yes you're right. I put the fraction with the wrong letter. However, you agreeing that

 

(x^(1/b))^a is the same as (x^a)^(a/b) ?

 

In which case, i think im back where i started.

 

Also, whilst you say the rules don't hold for b=0, that is simply due to the fact that x/0 is undefined - however you manipulate the expression, at some point you must divide by b, making the calculation impossible.

[btherl]

(x^(1/b))^a is the same as (x^a)^(a/b)

 

(x^(1/b))^a is the same as (x^a)^(1/b), rather

 

Also, whilst you say the rules don't hold for b=0, that is simply due to the fact that x/0 is undefined - however you manipulate the expression, at some point you must divide by b, making the calculation impossible.

 

That's what I meant, sorry for not being clearer.

 

Anyway, what is your current goal, when you say you are back where you started?  Are you still thinking about the apparent paradox in your earlier post?  I think the solution there is that you simply can't take the 3 and 6 inside, because that rule doesn't apply for x < 0.  So you have to use other methods to solve, and those other methods will give you a correct result.

[GingerRobot]

(x^(1/b))^a is the same as (x^a)^(1/b), rather

 

Whoops. Though it was 7 in the morning -thats my excuse anyway

 

As for current goal, i think i've satisfied my original motivation for posting, which was that i'd noticed that php couldn't deal with negative bases and fractional powers.

 

I've largely just been trying to understand, from a mathematical point of view, how/why it works/doesn't work. I think ill mark this as solved now, although im sure im going to continue my searches on the internet.

 

And thanks everyone, for all your help.

[AndyB]

I think ill mark this as solved now, although im sure im going to continue my searches on the internet.

 

Solved: I assume this is a real solution, not an imaginary one

[GingerRobot]

Haha. I think it probably is an imaginary solution! Im imagining that i understand it all perfectly

[cooldude832]

The PEAR extension has an add on that can carry variables through a math function and return an answer and do complex numbers such as doing 4(x-2)^2 it would return 4x^2-8X+8  but it requires installation

[Barand]

OK, I won't say anything about that returned result.

[AndyB]

OK, I won't say anything about that returned result.

 

New math rulez

[cooldude832]

New Math?  Math can't be "new" because its not like it didn't existed.  Better term would be newly discovered math.  Imaginary/complex numbers aren't actually new at all they have been around forever (in the sense of being known and used)

[btherl]

I think Barand is talking about this:

 

4(x-2)^2

= 4(x-2)(x-2)

= 4(x^2 - 2x -2x + 4)

= 4(x^2 - 4x + 4)

= 4x^2 - 16x + 16

 

No complex numbers involved there..

 

The lesson is, don't do maths in your head and drive!  The same rule applies when not driving.