Implied Functions - Printable Version

Implied Functions - Printable Version

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Implied Functions - 2147483647 - 2011-03-26

Original question

Second question

Third question:

How do you solve:

sqrt(a) + sqrt(b) = -1

sqrt(a) + sqrt(b) + 1 = 0

(a^(1/4) + i*sqrt(sqrt(b)+1))*(a^(1/4) - i*sqrt(sqrt(b)+1)) = 0

a^(1/4) = +/- i*sqrt(sqrt(b)+1)

This is where it becomes very problematic. There's no way to get rid of that one fourth power without eliminating the +/-, which was exactly what I was trying to avoid to begin with.

In essence, my question is this: can the following ever exist, either as a real number or a complex number?

sqrt(a) = -1

What happens when something that doesn't exist arises? Imaginary numbers were made to handle sqrt(-1). Limits were made to handle indeterminate forms. So what about this?

Implied Functions - Lozmaster - 2011-03-26

If you ever get 0=1 or something else impossible, you've probably divided by 0 somewhere. Here your problem is at the very start.

sqrt(a) = -sqrt(b)

Implies a= b = 0, otherwise its not true
Then your next step
sqrt(a)/sqrt(b) = -1
assumes that 0/0 = -1, which causes all sorts of crazy problems (like leading to 0 =1)

Implied Functions - DarkQThunder - 2011-03-26

You're dividing by (sqrt(a) + sqrt(b))=0

edit: ninja'd due to me thinking it was the squaring step not being reversible (skimmed it at first)

edit2: the solution assuming sqrt function gives only N*cis [0,180) in complex plane is: a = b = 0 (N*cis theta in [180,360) = -1*N*cis (theta - 180) which has the same square).

Implied Functions - 2147483647 - 2011-03-26

DarkQThunder Wrote:edit2: the solution assuming sqrt function gives only N*cis [0,180) in complex plane is: a = b = 0 (N*cis theta in [180,360) = -1*N*cis (theta - 180) which has the same square).

Where'd the N come from?

cis(θ Wink

= e^(iθ

= cos(θ

+i*sin(θ

Why is θ only be defined on [0,π Wink

and [π,2π

, considering that arccos(θ Wink

is limited to [0,π Wink

and arcsin(θ Wink

is limited to [-π/2,π/2)? They're clearly different. I don't see what you're doing, other than rewriting cis(θ Wink

in a different way. If you're trying to square it:

cis(θ Wink

^2 = e^(iθ Wink

^2 = e^(-θ^2)

That's just the Gaussian function.

Implied Functions - Noah - 2011-03-27

2147483647 Wrote:a^2 + sqrt(b^4*i^4) = c^2

a^2 + sqrt(i^4)*sqrt(b^4) = c^2

For complex numbers, the formula

[Image: 62mlsb2.png]

is not valid, because of the discontinuous nature of the square root in the complex plane.

2147483647 Wrote:In essence, my question is this: can the following ever exist, either as a real number or a complex number?

sqrt(a) = -1

What happens when something that doesn't exist arises? Imaginary numbers were made to handle sqrt(-1). Limits were made to handle indeterminate forms. So what about this?

No. No solutions exist to that equation. And no, imaginary numbers were not made to handle sqrt(-1), they were made to handle equations on the form x^2 = -y. The definition of i is not i = sqrt(-1), but rather, i^2 = -1. You cannot really use the former, though it sometimes work.

Noah

Implied Functions - DarkQThunder - 2011-03-27

N is a non-negative real number equal to the square root of the norm of the complex number you're taking the sqrt of.
Here's a better way to explain: Unless I'm being stupid, the square root function only returns complex numbers with non-negative coefficients for the imaginary term, therefor if you wish for two square roots to add to a real number, both numbers would need square roots in non-negative reals (because the range of the square root function takes the form n*cis [0, pi) )
P. S. idk why I decided to use degrees in my last post

EDIT: @ Noah, sqrt(i) = sqrt(2)/2 + sqrt(2)/2 * i, sqrt(-1) = i, sqrt(a) = -1 has no solution because -1 is outside the range of the square root function.

Implied Functions - 2147483647 - 2011-03-27

Noah Wrote:For complex numbers, the formula

is not valid, because of the discontinuous nature of the square root in the complex plane.

I haven't really worked with complex numbers, but when I plotted it, I don't see how the discontinuities affect the function other than separate the real from the imaginary parts:

Spoiler

Noah Wrote:No. No solutions exist to that equation. And no, imaginary numbers were not made to handle sqrt(-1), they were made to handle equations on the form x^2 = -y. The definition of i is not sqrt(i) = -1, but rather, i^2 = -1. You cannot really use the former, though it sometimes work.

i = sqrt(-1)

How is it not used to handle sqrt(-1)?

sqrt(i) = a + bi
i = (a+bi)^2 = a^2 + 2abi - b^2

a^2 - b^2 = (a+b)*(a-b) = 0
a = +/- b

2ab i = i
2*b^2*i = i
2*b^2 = 1
b = sqrt(2)/2 = a

sqrt(i) = sqrt(2)/2 + sqrt(2)/2*i

How can sqrt(i) = -1 sometimes work, when sqrt(i) is already defined?

DarkQThunder Wrote:the square root function only returns complex numbers with non-negative coefficients for the imaginary term, therefore if you wish for two square roots to add to a real number, both numbers would need square roots in non-negative reals (because the range of the square root function takes the form n*cis [0, pi) )
P. S. idk why I decided to use degrees in my last post

I'm actually not sure what you mean by the bolded. The range of n*cis(θ Wink

in the interval [0,π Wink

is just [-n,n) in the real plane and [0,n] in the imaginary plane. Wouldn't the real part imply that sqrt(a) can equal -a, because it's in the range?

Implied Functions - DarkQThunder - 2011-03-27

2147483647 Wrote:I'm actually not sure what you mean by the bolded. The range of n*cis(θ in the interval [0,π is just [-n,n) in the real plane and [0,n] in the imaginary plane. Wouldn't the real part imply that sqrt(a) can equal -a, because it's in the range?

Look at my specific interval, it doesn't include the n*cis(pi) end.

Implied Functions - Noah - 2011-03-27

DarkQThunder Wrote:EDIT: @ Noah, sqrt(i) = sqrt(2)/2 + sqrt(2)/2 * i, sqrt(-1) = i, sqrt(a) = -1 has no solution because -1 is outside the range of the square root function.

I meant the opposite, thanks - sqrt(-1) = i. Thanks for the notification. Smile

2147483647 Wrote:I haven't really worked with complex numbers, but when I plotted it, I don't see how the discontinuities affect the function other than separate the real from the imaginary parts:

Spoiler

The imaginary values are not continuous.

[Image: bSCQn.gif]

2147483647 Wrote:i = sqrt(-1)

How is it not used to handle sqrt(-1)?

Because sqrt(-1) = -i too. You could use the property sqrt(-1) = i, but then you also have to solve the equation for sqrt(-1) = -i as well. It is thus easier to just use the i^2 = -1 property. This also usually removes possible square root problems you might run into, such as complex square roots.

Noah

Implied Functions - 2147483647 - 2011-03-27

DarkQThunder Wrote:Look at my specific interval, it doesn't include the n*cis(π end.

I didn't include it either:

2147483647 Wrote:I'm actually not sure what you mean by the bolded. The range of n*cis(θ in the interval [0,π is just [-n,n) in the real plane and [0,n] in the imaginary plane. Wouldn't the real part imply that sqrt(a) can equal -a, because it's in the range?

I'm assuming you're talking about my bolded part. cis(θ Wink

's imaginary part is i*sin(θ Wink

, which increases from 0 to i*n at π/2 and then back to 0 at π. Therefore, it includes i*n. I meant [0,i*n] by the way.

Noah Wrote:The imaginary values are not continuous.

It looks like the function is only discontinuous when y=0 and x<0. If y=0, then the imaginary part disappears entirely. The graph seems to be saying that:

For x<0, lim sqrt(x+i*y) as y approaches -0 = -sqrt|x|

and

For x<0, lim sqrt(x+i*y) when x<0 as y approaches +0 = +sqrt|x|

I really don't see how this is possible, but okay. Let's say it is. There is indeed a discontinuity, and it only occurs at y=0 for x<0. I don't see how this disproves the identity:

[Image: 62mlsb2.png]

Take -1 for example. It's possible to split this up into sqrt(x)*sqrt(y) as long as x*y = -1, which pretty much puts them on "opposite" sides of the discontinuity. The discontinuity didn't seem to change anything. The identity still seems to hold for any -C, where C is a constant.

Noah Wrote:Because sqrt(-1) = -i too. You could use the property sqrt(-1) = i, but then you also have to solve the equation for sqrt(-1) = -i as well. It is thus easier to just use the i^2 = -1 property. This also usually removes possible square root problems you might run into, such as complex square roots.

What? i is defined to be sqrt(-1). The only way I can possibly see this being possible is:

sqrt(-1) = i

sqrt(-i^4) = i

sqrt(i^4)*sqrt(-1) = i

i^2*sqrt(-1) = i

-sqrt(-1) = i

sqrt(-1) = -i

You already said that this is false, because the property:

[Image: 62mlsb2.png]

doesn't hold in the complex domain.

Implied Functions - Noah - 2011-03-27

2147483647 Wrote:What? i is defined to be sqrt(-1). The only way I can possibly see this being possible is:

Spoiler

sqrt(-1) = i

sqrt(-i^4) = i

sqrt(i^4)*sqrt(-1) = i

i^2*sqrt(-1) = i

-sqrt(-1) = i

sqrt(-1) = -i

You already said that this is false, because the property:

doesn't hold in the complex domain.

The precise definition of i varies from source to source. However, all definitions leads to i^2 = -1, which is what we're after.

Now, (-i)(-i) = (-1)(-1)(i)(i) = (i)(i) = -1
and (i)(i) = -1. Therefore, both -i and i is "the" square root of -1. (That's why people don't say that i = sqrt(-1), because that implies that i = -i, which is untrue. It is however true that i is one possible value of sqrt(-1).)

So, if you have a sqrt(-1) somewhere, you need to either square it to get it to i^2, or you need to substitute it with ±i instead.

Noah