Talk:Square root

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Request for a graph[edit]

Could somebody produce an image of the graph which isn't that humongous and which doesn't use commas as decimal points? AxelBoldt 02:12 Feb 1, 2003 (UTC)

Taylor series[edit]

Can you tell me exactly how to continue the coefficients of the Taylor series? They appear to all have powers of 2 in the denominator.


Hi,

In your Taylor's formula i wrote x^n at the end

62.147.217.21 12:35 May 4, 2003 (UTC)


Why the restriction to normal matrices? I can take any square matrix with a complete set of eigenvectors and just take the square roots of the eigenvalues to get a square root of the matrix. Can't I?

Josh Cherry 01:49, 13 Oct 2003 (UTC)

"Can you tell me exactly how to continue the coefficients of the Taylor series? They appear to all have powers of 2 in the denominator."

I hope my latest edit has addressed this question.

dcljr 23:12, 5 Jul 2004 (UTC)

t762@inforamp.net (J. B. Rainsberger) wrote:

>In article <31C14605.2D3E@laria.u-picardie.fr>,
>   Mustapha Benali  wrote:
>>And what do you think about this :
>>
>>
>> -1 = (-1)^1 = (-1)^(2/2) = [(-1)^2]^(1/2) = 1^(1/2) = 1
>>
>A great deal has been said about "the square root" of one. I had this 
>discussion with a grade five (?) student from Australia on this topic and 
>he understood it perfectly, so I present it here:

>"The" square root oftens refers to the principle square root, which is the 
>positive quantity whose square is the given quantity. There are *two* 
>square roots of any and all real numbers, and they are both complex 
>numbers. (Sometimes strictly real.)

>The expression x^(1/2) refers to the principle square root of a number. 
>This is verified by the relations:

>    x^(1/2) = y
>    1/2 ln x = ln y

>In order to define this properly for x > 0, we require y > 0. If y < 0, 
>then ln y is not strictly a real number, and thus, neither is ln x. If ln x 
>is not strictly real, then x is not positive. As a result, if x >= 0, 
>x^(1/2) >= 0.

>As for why the above equation is not quite correct, it is commonly known 
>that when squaring all members of an equation, extraneous roots are often 
>introduced. As an example:

>    sqrt(x) + 2 = x
>    x + 4 sqrt(x) + 4 = x^2
>    4 sqrt(x) = x^2 - x - 4
>    16 x = x^4 - 2 x^3 - 7 x^2 + 8 x + 16
>    0 = x^4 - 2 x^3 - 7 x^2 - 8 x + 16
>    solving for x, x is in {1, 4, -3/2 +/- sqrt(7)/2 i}

>Now I ask you: given that sqrt(x) is the principle (positive) square root 
>of x, is x = 1 a solution to this equation? No.

>There you have an example of an extraneous root produced by squaring all 
>members of an equation. As a result, if you rearrange to do this:

>-1 = (-1)^1 = (-1)^(2/2) = [(-1)^(1/2)]^2 = i^2 = -1

>then you're cooking.

>Just be careful that you're not adding extraneous roots by squaring both 
>sides of an equation. You need to check all roots thus obtained in the 
>original equation or expression.

Calculating Square Roots - Mentally[edit]

Hi, I find the explanation of calculating square roots using Pell's equation hard to follow. Would it be possible to explain what is happening in each step more clearly? -- MattW, 05 June 2005

Are there two square roots or just one?[edit]

I had recently made a correction on the square root page of wikipedia where I had corrected that -4 is not a square root of 16. It was reverted and I was hoping we could discuss on the topic.

Square root of 16 is only 4. That is why [1] shows answer as just 4 and not -4 too. I think where most people confuse this is the fact that -4 squared is 16 and hence they think square root of 16 is also -4.

See x^2=16 is a polynomial with degree 2. That means the maximum number of roots it can have is 2. Hence x = +4 and -4. On the other hand, x = sqrt(16) is a polynomial in degree 1 which means the maximum number of roots it can have is just 1. Hence only +4 is the square root of 16.

I think a popular argument is stating that there are two square roots and only positive is considered as principal. However, I would like citation on the same from a source. Can you provide that as all other credible sources just consider 4 to be square root of 16.

Additionally, I wanted to edit how principal square root is introduced (if at all such a citation exists) in a way so that people are clear only 4 is square root of 16. Students usually mistake on this.

If we can add a section explaining with the polynomial degree why square root is just 4 and not -4 it will be good.

122.169.19.85 (talk) 02:31, 25 July 2015 (UTC)

References

There is something to be said for the multiple anons who are probably the same person, but not much. The fact that there are two square roots of 16 should precede the discussion of the principal square root. I've reverted to the stable, accurate, text again. — Arthur Rubin (talk) 00:47, 27 July 2015 (UTC)
I agree with the revert. The article clearly states that "the square root" is used to refer to the principal square root, and that "every positive number a has two square roots: a, which is positive, and −a, which is negative." This should be sufficient to prevent students getting confused. Anon, also note that "x=sqrt(16)" is not a polynomial, but an equation. And it is already solved. Going from "x=sqrt(16)" to "x=4" is not solving an equation, but simplifying a constant. - DVdm (talk) 08:36, 27 July 2015 (UTC)