roots of cubics

[R.D. Silverman]

Originally Posted by VBCurtis

Two bad assumptions, even around here.

I think that it is polite to assume that people mean what they say. The alternative
is to assume that they are either or idiots.

[Uncwilly]

Originally Posted by R.D. Silverman

I think that it is polite to assume that people mean what they say. The alternative is to assume

that accidents, typos, and editing errors happen. Good programming deals gracefully with problematic input. If a lecturer misstates something that the audience can error correct in wetware, there is no need to jump up in the audience and call them an idiot. And here we have a chance to ask: "Did you maybe mean the P-axis?", instead of saying "Every moron knows there is no such thing as a O-axis."

[chalsall]

Originally Posted by Uncwilly

And here we have a chance to ask: "Did you maybe mean the P-axis?"

+1

[R.D. Silverman]

Originally Posted by Uncwilly

that accidents, typos, and editing errors happen. Good programming deals gracefully with problematic input. If a lecturer misstates something that the audience can error correct in wetware, there is no need to jump up in the audience and call them an idiot. And here we have a chance to ask: "Did you maybe mean the P-axis?", instead of saying "Every moron knows there is no such thing as a O-axis."

If one sees such an error within a piece of prose that mostly makes sense otherwise,
then assuming the error is a typo or editing error makes sense. This error was made
multiple times.

When one sees such an error within a piece of prose that shows other egregious
and elementary mathematical errors then assuming that the error is what the
writer intended or that the writer is totally confused in general also makes sense.

The OP made the egregious and elementary mistake of not knowing that complex
roots come with their conjugates. This is such a fundamental error that it is also likely
under such circumstances that the O.P. was also confused about the x and y axes.
This is second year secondary school algebra. One of the first things taught about
complex roots is that they come in pairs. Confusion about the math seems
a more likely explanation than repeating the same 'typo' three times.

And noone is calling anyone names here. I said that it was polite to assume that
people mean what they write rather than assuming that they are idiots.

[wildrabbitt]

Hi,


I'm really sorry. I made a dreadful mistake. I should have written x-axis.


I realise I can't expect people to be mindreaders. Thanks for the replies everyone, it's been really helpful.



I've done some algebra and seen that if f(a) = 0 and the derivative is 0 at a then a must be a repeated root but I can't honestly say I understand anything better. I know about the fundamental theory of algebra but I've never seen a proof of it.



Really sorry I caused a row.

[Nick]

Let's call your cubic polynomial f and write f' for its derivative.
We are assuming there is a number c for which f(c)=0 and f'(c)=0
(c is a root of the polynomial and it also has derivative 0 there).
We can divide f by the polynomial \((x-c)^2\) getting a quotient q(x) and a remainder r(x),
.i.e. \(f(x)=q(x)(x-c)^2+r(x)\).
The degree of the remainder is less than the degree of \((x-c)^2\), which is 2,
so \(r(x)=sx+t\) for some constants s and t.
But f(c)=0 and f'(c)=0 so r(c)=0 and r'(c)=0.
As \(r'(x)=s\), it follows that s=0 and t=0 hence \(f(x)=q(x)(x-c)^2\), making c a repeated root of f.

[R.D. Silverman]

Originally Posted by wildrabbitt

Hi,


I'm really sorry. I made a dreadful mistake. I should have written x-axis.


I realise I can't expect people to be mindreaders. Thanks for the replies everyone, it's been really helpful.



I've done some algebra and seen that if f(a) = 0 and the derivative is 0 at a then a must be a repeated root but I can't honestly say I understand anything better. I know about the fundamental theory of algebra but I've never seen a proof of it.



Really sorry I caused a row.

A good habit to get into is to hit 'preview post' and proofread before hitting ;submit reply'.
I've caught many of my own errors that way.

[Dr Sardonicus]

Originally Posted by R.D. Silverman

A good habit to get into is to hit 'preview post' and proofread before hitting ;submit reply'.
I've caught many of my own errors that way.

Now that is a constructive comment. But, as the semicolon before "submit" shows, "many" does not mean "all."


The result you previously mentioned, non-real roots always showing up in complex-conjugate pairs, assumes the polynomial f(z) has real coefficients. Under that assumption, the coefficients are invariant under complex conjugation, and we may write



where the bar indicates complex conjugation.

[R.D. Silverman]

Originally Posted by Dr Sardonicus

Now that is a constructive comment. But, as the semicolon before "submit" shows, "many" does not mean "all."

Yep. Cervical Radiculopathy is a bitch. It makes it hard to type.

[LaurV]

This thread should have been closed. The mistake was that it was posted in the Math subforum, where RDS is entitled to comment

Just kidding, hehe, but there is a part of truth in this, maybe the homework thread would have been more suitable. Not to make an excuse for RDS, his outburst was totally uncalled for.

But well, anyhow, to be on the constructive side, we just watched a beautiful Mathologer video about one week ago (as my one-week-old comment on that video shows, if you sort the comments by time) - the video was posted in August. It exactly addresses how the cubic graphic looks like, and why.

Beautiful one. You (all) should watch it !

[R.D. Silverman]

Originally Posted by LaurV

This thread should have been closed. The mistake was that it was posted in the Math subforum, where RDS is entitled to comment

Just kidding, hehe, but there is a part of truth in this, maybe the homework thread would have been more suitable. Not to make an excuse for RDS, his outburst was totally uncalled for.

You are deluded. I made no outburst whatsoever.