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 i = non-i and an exponent question
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Cervantes




PostPosted: Thu Nov 03, 2005 2:18 pm   Post subject: i = non-i and an exponent question

A few weeks ago I thought this up:

Posted Image, might have been reduced in size. Click Image to view fullscreen.

What's the problem?

One of my friends tried to explain it using BEDMAS. Brackets before exponents. I forget now just where he said the problem was, but I retort to him by saying that brackets are not needed. Expressions can also be written in this format, which does not contain any brackets:
Posted Image, might have been reduced in size. Click Image to view fullscreen.
What is the value of the above expression? Is it 3^(3^3) = 3^27? Or is it (3^3)^3 = 27^3 = 3^9. (Multiplying exponents.)

In other words, do you work from left to right & bottom to top or right to left & top to bottom?
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Boo-chan




PostPosted: Thu Nov 03, 2005 2:45 pm   Post subject: (No subject)

ok, (-x)^.5=(-1)^.5 *x^.5=x^.5(i) for x>=0

Your statement is true for x<=0.
The problem is that in squaring the expression then taking the fourth root of the answer your simply dropping part of the final solution. Its the equivalent of stating:

(-x)^.5=(|-x|)^.5

The correct way to do it is:

(-x)^.5= (-x)^(2^.25)=...=x^.5 for x<=0
& =(-1)^.5 *x^.5=x^.5(i) for x>=0

The simplest way to explain why the logic is wrong is that when manipulationg a mathematical equation you can't change its answer, which you have done by neglecting the solution for all x>0.

Edit:I forgot that the two solutions overlap at 0. Assuming that 0^.5=0.
Cervantes




PostPosted: Thu Nov 03, 2005 5:27 pm   Post subject: (No subject)

First the little one:
Boo-chan wrote:

The correct way to do it is:

(-x)^.5= (-x)^(2^.25)=...=x^.5 for x<=0

I don't think (-x)^.5 = (-x)^(2^.25), since the right side of that equation ~= (-x)^1.189. I assume you meant to put a ' * ' in there.

Now the more important one:
Boo-chan wrote:
ok, (-x)^.5=(-1)^.5 *x^.5=x^.5(i) for x>=0

Your statement is true for x<=0.

Is it really?
If x < 0, we have a real number on the left-most side of the equation, but we have an imaginary number on the right-most side of the equation. The equation has the same problem, regardless of the value for x.
Naveg




PostPosted: Thu Nov 03, 2005 6:06 pm   Post subject: (No subject)

In the absence of brackets, like operations are evaluated in the order they appear. Therefore, 3^3^3 is evaluated as (3^3)^3=27^3=3^9. Brackets, in such as case, are merely to make the expression easier to read.

Now, let's take a look at the original dilemma:

Let @ be read as "square root of"

(-x)^.5 is equal to the square root of -x, or i@x, an imaginary number.

((-x)^2)^.25, performing the operations according to order of operations, resolves to the fourth root of (-x)^2, a positive number. Since the term under the radical is not negative, the resulting solution is a real number.

So (-x)^.5 != ((-x)^2)^.25
Cervantes




PostPosted: Thu Nov 03, 2005 6:28 pm   Post subject: (No subject)

Naveg wrote:
((-x)^2)^.25, performing the operations according to order of operations, resolves to the fourth root of (-x)^2, a positive number. Since the term under the radical is not negative, the resulting solution is a real number.

So (-x)^.5 != ((-x)^2)^.25


Congratulations, you've discovered the problem. Wink

That's the whole point of this. I've got an expression on the left which is an imaginary number when x > 0, and an expression on the right which is a real number when x > 0. (When x < 0, the left side is real and the right side is imaginary.)

What you've got in words up there is pretty much the next two equations in my little line. I know the left side doesn't equal the right side, but I've used the exponent laws to, seemingly, make them equal.

Back to the drawing board!

Here's the reason I was wondering about the order of operations of the exponents. It's essentially just an expansion of the first step.
(-x)^.5 = (-x)^(2*.25) = (-x)^2^.25

By the exponent laws, this should be valid, since an exponent raised to another exponent multiplies the exponents together. And no brackets are needed in the right side of this equation. Or are they?
Naveg




PostPosted: Thu Nov 03, 2005 6:32 pm   Post subject: (No subject)

The laws of exponents do not apply to imaginary numbers. (-x)^2^.25 cannot be solved correctly using exponent laws since, if it is, the resulting value is imaginary.

In fact, I'm pretty sure the laws of exponents do not always work for negative real bases, or fractional/irrational exponents at all. I may be wrong though.
Boo-chan




PostPosted: Thu Nov 03, 2005 7:56 pm   Post subject: (No subject)

Yes, I did mean to put a * in there.... I'm only good at math when I write it down on paper.

But I don't see the other problem. If x>0 then (-x)<0 and vice versa. Therefore, if x<0 then (-x)^.5 is the same as (|x|)^.5 so there isn't any problem with your solution. Its only when x>0 that you need to worry about the imaginary numbers.
Paul




PostPosted: Thu Nov 03, 2005 9:07 pm   Post subject: (No subject)

woot, testing out my new software, ignore me
Posted Image, might have been reduced in size. Click Image to view fullscreen.
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Brightguy




PostPosted: Fri Nov 04, 2005 1:07 pm   Post subject: Re: i = non-i and an exponent question

Naveg wrote:
In the absence of brackets, like operations are evaluated in the order they appear. Therefore, 3^3^3 is evaluated as (3^3)^3=27^3=3^9.

Exponentiation is actually handled from right-to-left. So 3^3^3 = 3^27 = 7.6*10^12
Cervantes




PostPosted: Fri Nov 04, 2005 2:22 pm   Post subject: (No subject)

Boo-chan wrote:

But I don't see the other problem. If x>0 then (-x)<0 and vice versa. Therefore, if x<0 then (-x)^.5 is the same as (|x|)^.5 so there isn't any problem with your solution. Its only when x>0 that you need to worry about the imaginary numbers.


If x > 0:
(-x)^.5 =/= (x)^.5, as that is an imaginary number on the left and a real number on the right. Negative under the root on the left, positive under the root on the right.

If x < 0:
(-x)^.5 =/= (x)^.5, as the number on the left is a real number and the number on the right is imaginary. On the left, it is a positive under the root (negative * negative = positive). On the right, it is a negative number under the root.

I don't know why you're using absolute bars.

Brightguy: Thanks for the response! Not that I'm doubting you (really, I have no idea either way), but did you get that from a reliable source, or is that your personal opinion?
Brightguy




PostPosted: Fri Nov 04, 2005 4:06 pm   Post subject: Re: i = non-i and an exponent question

Well, defining a^b^c^d as ((a^b)^c)^d ends up being a lot less useful, since if we wanted that we would normally just write a^(bcd) (for real numbers).

So right-to-left is the standard... a^b^c^d means a^(b^(c^d)), but there's no need to write the parenthesis. e.g. for power towers... the page doesn't even bother to point out how the powers are evaluated because it's clear.

I googled for several pages on "order of operations", and most just say "simplify brackets, then powers...". A few even do say "work left to right" for all operations! But Wikipedia comes to the rescue again.
Cervantes




PostPosted: Fri Nov 04, 2005 7:30 pm   Post subject: (No subject)

Awesome. Thanks for that, brightguy!
Brightguy




PostPosted: Sat Nov 05, 2005 8:00 am   Post subject: Re: i = non-i and an exponent question

Basically your problem is that you overlook the fact that taking the 4th root of a number has more than one solution. The only perfectly valid step is the second one. In your first step, you raise the number to the fourth and then take the fourth root. However, by doing that, you created two more possible values for the expression. Likewise, in your third step, you remove two possible values.

Notice that, for any x, (real or complex) the value before the first step will equal the value before the third step. (One of the possible values, that is - remember when you take the fourth root, there are 4 possibilities).

If you haven't worked with complex roots before... Well, any complex number can be written in the form r*e^(iθ), for some nonnegative real r and real θ.

Now, when written in that form, the nth roots of r*e^(iθ) will be
r^(1/n)*e^(i(θ+2kpi)/n), for k=0,1,2,...,(n-1)
where r^(1/n) is the positive nth root of r (since r is a nonnegative real).
Andy




PostPosted: Mon Nov 07, 2005 6:14 pm   Post subject: (No subject)

you do realize that you cant take both (x) and (-x) to the 1/2 because thats taking the square root, and unless x = 0, then one of x and -x will be negative.
Cervantes




PostPosted: Mon Nov 07, 2005 6:46 pm   Post subject: (No subject)

Andy wrote:
you do realize that you cant take both (x) and (-x) to the 1/2 because thats taking the square root, and unless x = 0, then one of x and -x will be negative.

Very Happy
Immediately after showing the series of 'equations', Cervantes wrote:
What's the problem?


Brightguy wrote:
In your first step, you raise the number to the fourth and then take the fourth root. However, by doing that, you created two more possible values for the expression. Likewise, in your third step, you remove two possible values.

I see that there should be plus-minus signs in front of things, but I don't see how they would alter the innards of those equations.
I'll look at it some more, later. Smile

And no, I have not dealt with complex numbers in that... complexity... before.
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