Where the **** is Physics Clown when you need him...???

I'm not sure what your overall point to me is. Are you agreeing, disagreeing, etc.?

Just in case: Wolfram Alpha is not a average calculator. It's the same engine used in Wolfram Mathematica.

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I'm not sure what your overall point to me is. Are you agreeing, disagreeing, etc.?

Just in case: Wolfram Alpha is not a average calculator. It's the same engine used in Wolfram Mathematica.

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I've done programming contract work for wolfram. It's still using the sqrt as a function which means it will only produce 1 result.

Go here and you'll see it's only capable of giving 1 answer.

https://reference.wolfram.com/language/ref/Sqrt.html

(-4)^(1/2) * (-9)^(1/2) produces the same result.

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Sqrtrs

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Fun for a time or two, but I couldn't see dating one long term. Way too much clean up.

For the love of God, would somebody please throw this into the finest mathematical solving program ever???


Maple

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(-4)^(1/2) * (-9)^(1/2) produces the same result.

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It's likely using the same function (programming function, not mathematical function).

I also don't see why you are disputing my answer on the first page? What is mathematically wrong with what I said? I can also tell you I have a BS in Math and taught high school calculus for 5 years if that makes any difference. (I left the profession to take a software job but I taught for a good number of years)

Oh look at me, I math good

If everybody could please stop blue starring the incorrect answer on the first page and give them to me, that would be great.

Also I have a degree in bachelors

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If everybody could please stop blue starring the incorrect answer on the first page and give them to me, that would be great.

Also I have a degree in bachelors

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Say what?!?!

Not that there's anything wrong with that

You didn't even go back and star my correct answer!

I did, but then I removed it. You didn't say that I had to leave it there

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You didn't even go back and star my correct answer!

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I starred it.

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(-4)^(1/2) * (-9)^(1/2) produces the same result.

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It's likely using the same function (programming function, not mathematical function).

I also don't see why you are disputing my answer on the first page? What is mathematically wrong with what I said? I can also tell you I have a BS in Math and taught high school calculus for 5 years if that makes any difference. (I left the profession to take a software job but I taught for a good number of years)

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It's not using the same function because if you plug in (-4)^x it returns the sinusoidal-ish real and imaginary plots for the answer. The plots are continuous and for any given y, there is only one (complex) solution.

I'm not doubting your math chops, but I think you are remembering things wrong. The equation for x^y where x=a+b*i and y=c+d*i is:

x^y = p^c * e^(d*t) * [ cos( c*t + d*ln(p) ) + i*sin( c*t + d*ln(p) ) ]
where:
p = sqrt(a^2 + b^2)
t = arctan(b/a)

And as you can see for any particular complex x and y (or reals), there is always a single value for x^y.


So where you went wrong is in saying that because:
2i * 2i = -4
-2i * -2i = -4
(which is true) implies:
sqrt(-4) = +- 2i
(that part is untrue).

by plugging -4^(1/2) into the above equation, you get one single answer of 2i, not two answers. Another way of thinking about it is:

sqrt(-4)=x
sqrt(4*-1)=x
sqrt(4)*sqrt(-1)=x
2*sqrt(-1)=x
2*i=x ( the definition of i is sqrt(-1) not -i)

Gracias!

There, happy? Ass

NERD FIGHT!

Thx! It's kind of a weird knowing who gave you the blue stars.

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(-4)^(1/2) * (-9)^(1/2) produces the same result.

---

It's likely using the same function (programming function, not mathematical function).

I also don't see why you are disputing my answer on the first page? What is mathematically wrong with what I said? I can also tell you I have a BS in Math and taught high school calculus for 5 years if that makes any difference. (I left the profession to take a software job but I taught for a good number of years)

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It's not using the same function because if you plug in (-4)^x it returns the sinusoidal-ish(It's quasi-exponential) real and imaginary plots for the answer. The plots are continuous and for any given y, there is only one (complex) solution.

I'm not doubting your math chops, but I think you are remembering things wrong. The equation for x^y where x=a+b*i and y=c+d*i is:

x^y = p^c * e^(d*t) * [ cos( c*t + d*ln(p) ) + i*sin( c*t + d*ln(p) ) ]
where:
p = sqrt(a^2 + b^2)
t = arctan(b/a)
(edit: I will also note that arctan itself gives an infinite amount of answers)

And as you can see for any particular complex x and y (or reals), there is always a single value for x^y.
(I don't doubt your expansion but to say a sqrt value only gives one answer is factually wrong)

So where you went wrong is in saying that because:
2i * 2i = -4
-2i * -2i = -4
(which is true) implies:
sqrt(-4) = +- 2i
(that part is untrue).

by plugging -4^(1/2) into the above equation, you get one single answer of 2i, not two answers. Another way of thinking about it is:

sqrt(-4)=x
sqrt(4*-1)=x
sqrt(4)*sqrt(-1)=x (the sqrt(4) is +/- 2, not just 2 which is where your proof is incomplete)
2*sqrt(-1)=x
2*i=x ( the definition of i is sqrt(-1) not -i)

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Bobo,

Go check out the Cameron avatar thread on entertainment board. Atmag will argue a wrong side until you just quit.

Literally half of optimization in calculus can't be done if you just decide square roots don't give 2 answers.

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(-4)^(1/2) * (-9)^(1/2) produces the same result.

---

It's likely using the same function (programming function, not mathematical function).

I also don't see why you are disputing my answer on the first page? What is mathematically wrong with what I said? I can also tell you I have a BS in Math and taught high school calculus for 5 years if that makes any difference. (I left the profession to take a software job but I taught for a good number of years)

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It's not using the same function because if you plug in (-4)^x it returns the sinusoidal-ish(It's quasi-exponential) real and imaginary plots for the answer. The plots are continuous and for any given y, there is only one (complex) solution.

I'm not doubting your math chops, but I think you are remembering things wrong. The equation for x^y where x=a+b*i and y=c+d*i is:

x^y = p^c * e^(d*t) * [ cos( c*t + d*ln(p) ) + i*sin( c*t + d*ln(p) ) ]
where:
p = sqrt(a^2 + b^2)
t = arctan(b/a)

And as you can see for any particular complex x and y (or reals), there is always a single value for x^y.
(boboguitar: I don't doubt your expansion but to say a sqrt value only gives one answer is factually wrong)


So where you went wrong is in saying that because:
2i * 2i = -4
-2i * -2i = -4
(which is true) implies:
sqrt(-4) = +- 2i
(that part is untrue).

by plugging -4^(1/2) into the above equation, you get one single answer of 2i, not two answers. Another way of thinking about it is:

sqrt(-4)=x
sqrt(4*-1)=x
sqrt(4)*sqrt(-1)=x (the sqrt(4) is +/- 2, not just 2 which is where your proof is incomplete)
2*sqrt(-1)=x
2*i=x ( the definition of i is sqrt(-1) not -i)

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The sqrt(4) is only 2 NOT -2. You are confusing that for the solution of the equation x^2=4. That is NOT the same thing.

If you do not deny my expansion, then you must concede that there is a single answer. a^2+b^2 is always a non-negative number, so even if you were right in that the square root of a negative number has more than one answer, then do you deny that the square root of a positive number has more than one answer?)

No. Just no.

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(-4)^(1/2) * (-9)^(1/2) produces the same result.

---

It's likely using the same function (programming function, not mathematical function).

I also don't see why you are disputing my answer on the first page? What is mathematically wrong with what I said? I can also tell you I have a BS in Math and taught high school calculus for 5 years if that makes any difference. (I left the profession to take a software job but I taught for a good number of years)

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It's not using the same function because if you plug in (-4)^x it returns the sinusoidal-ish(It's quasi-exponential) real and imaginary plots for the answer. The plots are continuous and for any given y, there is only one (complex) solution.

I'm not doubting your math chops, but I think you are remembering things wrong. The equation for x^y where x=a+b*i and y=c+d*i is:

x^y = p^c * e^(d*t) * [ cos( c*t + d*ln(p) ) + i*sin( c*t + d*ln(p) ) ]
where:
p = sqrt(a^2 + b^2)
t = arctan(b/a)

And as you can see for any particular complex x and y (or reals), there is always a single value for x^y.
(boboguitar: I don't doubt your expansion but to say a sqrt value only gives one answer is factually wrong)


So where you went wrong is in saying that because:
2i * 2i = -4
-2i * -2i = -4
(which is true) implies:
sqrt(-4) = +- 2i
(that part is untrue).

by plugging -4^(1/2) into the above equation, you get one single answer of 2i, not two answers. Another way of thinking about it is:

sqrt(-4)=x
sqrt(4*-1)=x
sqrt(4)*sqrt(-1)=x (the sqrt(4) is +/- 2, not just 2 which is where your proof is incomplete)
2*sqrt(-1)=x
2*i=x ( the definition of i is sqrt(-1) not -i)

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The sqrt(4) is only 2 NOT -2. You are confusing that for the solution of the equation x^2=4. That is NOT the same thing.

If you do not deny my expansion, then you must concede that there is a single answer. a^2+b^2 is always a non-negative number, so even if you were right in that the square root of a negative number has more than one answer, then do you deny that the square root of a positive number has more than one answer?)

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Let me give you wolfram alpha's definition of a square root but feel free to find anyones definition.

http://mathworld.wolfram.com/SquareRoot.html

I'll give you a hint, the square root operation gives 2 answers no matter whose definition you chose.

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so even if you were right in that the square root of a negative number has more than one answer, then do you deny that the square root of a positive number has more than one answer?

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Where is this logical conclusion coming from?

Give some thought to why a square root is called a square root, and you'll see why there is a positive and negative.

Don't listen to them, aTmAg! Keep fighting the good fight (which is my entertainment)

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Give some thought to why a square root is called a square root, and you'll see why there is a positive and negative.

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I agree with you, but this is kind of circular reasoning. Give some thought to the definition of i and why it is an imaginary number.

nvm

Sqrt(4) = x
4= x^2

All I did is square both sides. You've admitted x^2=4 has 2 solutions. So what changes?

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Sqrt(4) = x
4= x^2

All I did is square both sides. You've admitted x^2=4 has 2 solutions. So what changes?

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From your link:

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In common usage, unless otherwise specified, "the" square root is generally taken to mean the principal square root.

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(which is the positive one)