Talk:Formal derivative

The slick alternative definition of the formal derivative, using [ f(Y)-f(X) ] / [Y - X], is due to Bill Dubuque of MIT, if I'm not mistaken. LDH (talk) 04:51, 26 November 2008 (UTC)

Repeated factors or roots

Might it be simpler to say that formal derivatives can be used to detect repeated factors rather then repeated roots? It would save introducing the notions of extensions and closures, and the example would become $x^{6}+1=(x^{2}+1)^{3}{\pmod {3}}$. Also, if the example had factors with different multiplicities, then it could be mentioned that these could be separated out by repeatedly taking the GCD. --catslash (talk) 11:27, 20 July 2009 (UTC)

I agree with Catslash.Hippo.69 (talk) 09:35, 31 December 2012 (UTC)

Noncomutative rings?

How does the polynomials look like for noncomutative rings ... would not $\sum _{i=0}^{n}a_{i}x^{i}b_{i}$ be the standard form? Hippo.69 (talk) 09:35, 31 December 2012 (UTC)

It seems to me, polynomials over noncomutative rings could not be described by easy standard form ... either it is not closed on addition or it is difficult to test that two polynomials are identical. ... Possibility closed on addition would be to maintain list of pairs (a,b) for each power. One could shorten the lists when two pairs with the same a or with the same b are found ... but that could lead to different (list of) lists for the same polynomial.Hippo.69 (talk) 11:25, 2 January 2013 (UTC)

Late reply, but I think the article is all mixed up about this.

The article on the polynomial ring points to Free algebra which clearly describes that polynomials over noncommutative rings have a standard form (under "Contrast with polynomials"). Caleb Stanford (talk) 19:54, 10 April 2023 (UTC)

Better definition for noncomutative rings

I have used for Czech version a'=0 for scalars and x'=1 for atomic terms and for composed terms (a+b)'=a'+b' and (ab)'=a'b+ab' as the definition. One just should prove the derivative gives the same result for terms which are equal according to equations from ring axioms (commutativity for addition, associativity for both addition and multiplication and distributivities). The derivative for polynomial is an easy consequence. Hippo.69 (talk) 22:51, 1 January 2013 (UTC)

I have finally decided to write this to the main article Hippo.69 (talk) 19:39, 1 April 2013 (UTC)

Application to finding repeated factors

To get a root $x=a$ that satisfies $0=f(x)$ of a factorial polynomial as

$f(x)=(x-a)g(x)$

where $g(x)$ is the quotient polynomial, it is irritating if answer has not enough accuracy that the system usually assures. It is most conspicuous when '$a$' above is one of famous numbers(integers, rational numbers, analytic numbers(like ${\sqrt {2}}$)). Actually, to find a root of

$f(x)=(x-a)(x-b)$,

by $0=f(x)$, some sort of repetitive methods (for example Newton-Raphson) is necessary to solve the root. A simple repetitive method is enough if $a\neq b$, however, just small accuracy is expected in multiple root cases. One example is

$0=f(x)=(x-a)^{3}$.

In a procedure of repetitive methods, '$x$' relatively close to the accurate value(say $x=a\pm 0.0001$) gives a result very close to zero ($f(x)=0.000000000001$). So, using value of $f(x)$ itself is not an ideal way to seek the zero point of '$x$'. In a expectation that, although different roots might have the same value(multiple roots), roots of different values shall not be close, using formal derivative is effective to get a punctual value of '$x$'. The second derivative of above polynomial is

$f''(x)=6(x-a)$,

and you can get the root seeking $0=f''(x)$ with a good accuracy. On the contrary, in cases plural roots are of very close values, it is disguisable. For example

$f(x)=(x-a+0.0001)(x-a)(x-a-0.0001)$.

Here, the second derivative will be the same as above

$f''(x)=6(x-a)$,

where small differences $\pm 0.0001$ for '$x$' are neglected.

Thus, using formal derivative is not almighty, however, it assures high probability to get amiable answers in most cases. Tsukitakemochi (talk) 16:03, 18 December 2021 (UTC)

integer mult in a ring

@EuclideanSwag: Hello and thank you for your help editing the article!

Multiplication by an integer in a ring does not need to be written out that way; it's standard defined in any ring as repeated addition. I tried to edit to clarify this. Feel free to post here if you have any questions or suggestions for further improvement. Best, Caleb Stanford (talk) 23:16, 13 April 2023 (UTC)

Edit looks good! I just wanted the multiplication to receive attention as it has can have a big effect depending on the characteristic of the field. Cheers! EuclideanSwag (talk) 02:54, 14 April 2023 (UTC)