Find polynoms, with as least as power possible

[tiny-tim]

You can use P Q R U V again, if you want to!

Is that what you meant?

 

[Physicsissuef]

a = -1, b = 3, c = -3, d = -13

A = cP + dQ, B = -(aP + bQ)

A=-3P-13Q, B=-(-P+3Q)

A=-3P-13Q, B=P-3Q

btw- How do u know that I should do the whole process again? Because there can't be any rearranging?

 

[tiny-tim]

… Looks good so far! …

Looks good so far!

btw- How do u know that I should do the whole process again? Because there can't be any rearranging?

Well, I can't see any convenient re-arrangement: in Theofilius' example, there was a pattern with x^2, which there isn't here.

 

[Physicsissuef]

$$(x^4-2x^3-4x^2+6x+1)A(x)+(x^3-5x-3)B(x)=x^4$$

$$A=-3P-13Q ; B=P-3Q$$

$$(x^4-2x^3-4x^2+6x+1)(-3P-13Q)+(x^3-5x-3)(P-3Q)=x^4$$

Actually, it didn't worked.
Because in the result I get -6P and -4Q

 

[tiny-tim]

… you've gone back to square one …

$$(x^4-2x^3-4x^2+6x+1)A(x)+(x^3-5x-3)B(x)=x^4$$

erm … you've just gone back to the beginning!

You were at the Q R stage:

$$R(-x^3+5x+3)+Q(-3x^3+5x^2+12x-13)=x^3$$
​

You need to put $$R=-3U-13V ; Q=U-3V$$

 

[Physicsissuef]

$$V(22x^3-15x^2-101x)+U(5x^2-3x-22)=x^3$$

$$U=xF$$

$$V(22x^3-15x^2-101x)+F(5x^3-3x^2-22x)=x^3$$

dividing with x

$$V(22x^2-15x-101)+F(5x^2-3x-22)=x^2$$

how to rearrange now?

 

[tiny-tim]

… shame on you …

$$U=xF$$

F? What about poor old W?

There's W, standing at the end of the line thinking "I'm fat, so no-one will choose me", and then he sees P and Q being chosen, then R, then S and T get skipped over, then U and V, and he thinks "it must be me next" - and suddenly you don't even go on to X or Y, you take one look at W and run all the way back to F!

Shame on you!

$$V(22x^2-15x-101)+F(5x^2-3x-22)=x^2$$

how to rearrange now?

Well, either do it again, with a = 22, b = -101, c = 5, d = -22.

Or put V = lx + m, F = nx + p, and solve the four simultaneous equations!

 

[Physicsissuef]

(ax^2 + … + b), and the second is (cx^2 + … + d)

a=22 , b=-101 , c=5 , d=-22

V = cW + dG, F = -(aW + bG)

V=5W-22G , B=-(22W-101G)

V=5W-22G , B=-22W+101G

$$G(21x^2+27x)+W(-9x-21)=x^2$$

$$W=Ix$$

and dividing with x:

$$G(21x+27)+I(-9x-21)=x$$

$$G=\frac{7}{108} , I=\frac{1}{12}$$

btw- by using V = lx + m, F = nx + p how do u know that both of them will be (lx+m and nx+p), can one of them be V=lx+m and the other just F=p?

 

[tiny-tim]

… you have redeemed yourself! …

$$G=\frac{7}{108} , I=\frac{1}{12}$$

… aah! … you used W ! … ​

Now all you have to do is to undo all the conversions until you get to A and B again!

erm … but isn't it $$G=\frac{7}{66} , I=\frac{9}{66}$$ ?

btw- by using V = lx + m, F = nx + p how do u know that both of them will be (lx+m and nx+p), can one of them be V=lx+m and the other just F=p?

Yes … could be!

That would be the same case - the technique still works, but you find out at the end that n = 0.

 

[Physicsissuef]

Yes, you're right $$G=\frac{7}{66} , I=\frac{9}{66}$$

and I thought like you, but in the case before:
$$x^2(x^2\,-\,1)U\,+\,(x^2\,+\,1)V\,=\,2x^2\,-\,4\,;$$

I tried to substitute $U=ax^2+bx+c$ and $V=dx^2+ex+f$

and it didn't worked.

 

[tiny-tim]

… it worked for me …

and I thought like you, but in the case before:
$$x^2(x^2\,-\,1)U\,+\,(x^2\,+\,1)V\,=\,2x^2\,-\,4\,;$$

I tried to substitute $U=ax^2+bx+c$ and $V=dx^2+ex+f$

and it didn't worked.

hmm … well, it worked for me.

btw, that was an equation with only x^2s and no xs on their own, so you could have left out bx and ex (and I did).

 

[Physicsissuef]

hmm … well, it worked for me.

btw, that was an equation with only x^2s and no xs on their own, so you could have left out bx and ex (and I did).

Yes, when I left them, it worked. Ok, now I will continue with the equation:

$$(x^4-2x^3-4x^2+6x+1)A(x)+(x^3-5x-3)B(x)=x^4$$

$$A=-3Q , B=-(P+Q)$$

$$P(-x^3+5x+3)+Q(-3x^4+5x^3+12x^2-13x)=x^4$$

$$P=xR$$

$$R(-x^4+5x^2+3x)+Q(-3x^4+5x^3+12x^2-13x)=x^4$$

dividing with x

$$R(-x^3+5x+3)+Q(-3x^3+5x^2+12x-13)=x^3$$

$$R=-3U-13V ; Q=U-3V$$

$$V(22x^3-15x^2-101x)+U(5x^2-3x-22)=x^3$$

$$U=xF$$

$$V(22x^3-15x^2-101x)+F(5x^3-3x^2-22x)=x^3$$

dividing with x

$$V(22x^2-15x-101)+F(5x^2-3x-22)=x^2$$

$$V=5W-22G , F=-22W+101G$$

$$G(21x^2+27x)+W(-9x-21)=x^2$$

$$W=Ix$$

and dividing with x:

$$G(21x+27)+I(-9x-21)=x$$

$$G=\frac{7}{66} , I=\frac{9}{66}$$

$$A=9x^2-26x-21$$

$$B=-9x^3+44x^2-39x-7$$

Am I right?

 

[tiny-tim]

… by George! … he's got it! …

By George!, you've got it! ​

Yes, I multiplied it out, and it's all completely correct!

hmm … now you've mastered that method, you really ought to try the e f g h j k method, for comparison …

 

[Physicsissuef]

Thank you very much tiny-tim. Just want to ask you, how did you discover:

A = cP + dQ, B = -(aP + bQ).

or you found it somewhere?

 

[tiny-tim]

… now where's that cornflakes packet … ?

Thank you very much tiny-tim. Just want to ask you, how did you discover:

A = cP + dQ, B = -(aP + bQ).

or you found it somewhere?

You're very welcome, Physicsissuef!

hmm … let's see … I still have the cornflakes packet I worked it out on …

(I tried it in my head first, and got stuck … )

I wrote:

$$(ax^n\,+\,...\,+\,b)(\alpha P\,+\,\beta Q)\,+\,(cx^n\,+\,...\,+\,d)(\gamma P\,+\,\delta Q)$$

$$=\,P((...+\,(\alpha b\,+\,\gamma d))\,+\,Q((\beta a\,+\,\delta c)x^n\,+\,...)\,;$$

so we want $$\alpha b\,+\,\gamma d\,=\,0\,=\,\beta a\,+\,\delta c\,;$$​

so the obvious solution (there are plenty of others) is:

$$\frac{\alpha}{\gamma}\,=\,\frac{-d}{b}\,;\,\frac{\beta}{\delta}\,=\,\frac{-c}{a}\,;$$

so A = dP + cQ, B = -(bP + aQ);​

and then I decided to swap P and Q round, to make it look neater!

 

[Physicsissuef]

I understand everything, just this part:
A = dP + cQ, B = -(bP + aQ);
How did u find it?

So actually our mission is to contract the polynom (the 1st and the last element), right?

 

[tiny-tim]

… your mission, if you choose to accept it, is …

So actually our mission is to contract the polynom (the 1st and the last element), right?

That's right!

I understand everything, just this part:
A = dP + cQ, B = -(bP + aQ);
How did u find it?

Because, since we know:

$$\frac{\alpha}{\gamma}\,=\,\frac{-d}{b}\,;\,\frac{\beta}{\delta}\,=\,\frac{-c}{a}\,;$$​

the obvIous thing is to put:

$$\alpha\,=\,-d\,,\,\beta\,=\,-c\,,\,\gamma\,=\,b\,,\,\delta\,=\,a\,,$$​

though I instead decided to multiply them all by minus-one (that's ok because it leaves the ratios, in the previous line, the same):

$$\alpha\,=\,d\,,\,\beta\,=\,c\,,\,\gamma\,=\,-b\,,\,\delta\,=\,-a\,,$$​

(because that made A positive, which I think looks neater ),

which is the same as A = dP + cQ, B = -(bP + aQ). ​

 

[Physicsissuef]

You're very intelligent. And what is one power less than $x^4-x^2$, is it $x^3-x$?

 

[tiny-tim]

… fourth-order in x, but only second-order in x^2 …

And what is one power less than $x^4-x^2$, is it $x^3-x$?

Ah! You're referring to:

$$x^2(x^2\,-\,1)U\,+\,(x^2\,+\,1)V\,=\,2x^2\,-\,4\,;$$

I tried to substitute $U=ax^2+bx+c$ and $V=dx^2+ex+f$

and it didn't worked.

I agree that $\LARGE x^4\,-\,x^2$ is fourth-order in x, so we should normally use cubics for the e f g etc method - but I treated that equation as an equation in x^2 rather than x (because there's no odd powers of x in it), and it's only second-order in x^2, so m = 1, and I could use the linear-in-x^2 ax + b, cx + d.

 

[Physicsissuef]

Ok, I understand. I have one question more. You write this:
$$=\,P((...+\,(\alpha b\,+\,\gamma d))\,+\,Q((\beta a\,+\,\delta c)x^n\,+\,...)\,;$$
Which is absolute genius. I want to ask you, can we contract the polynom much more, using $x^n^-^1$?

 

[tiny-tim]

… remainders … ?

I want to ask you, can we contract the polynom much more, using $x^n^-^1$?

Hi Physicsissuef

If you mean, can we reduce from x^n to x^n-1, and then reduce again from x^n-1 to x^n-2, and so on, then yes … in fact, it's what you just did with you own example.

But please remember that, although this has been good practice at manipulating polynomials, it's rather slow, and easy to make mistakes in, and the a b c d e f g h j k etc method right from the beginning is probably better in nearly all cases!

And I still suspsect that there's some trick involving remainders that we haven't spotted …

 

[Physicsissuef]

Hi Physicsissuef

If you mean, can we reduce from x^n to x^n-1, and then reduce again from x^n-1 to x^n-2, and so on, then yes … in fact, it's what you just did with you own example.

But please remember that, although this has been good practice at manipulating polynomials, it's rather slow, and easy to make mistakes in, and the a b c d e f g h j k etc method right from the beginning is probably better in nearly all cases!

And I still suspsect that there's some trick involving remainders that we haven't spotted …

-------------------------------------------------------------

I thought like this:

$$(ax^n+bx^n^-^1+cx^n^-^2+dx^n^-^3+ex^n^-^f)(\alpha P+\beta Q)+(gx^n+hx^n^-^1+ix^n^-^2+jx^n^-^3+kx^n^-^w)(\gamma P +\delta Q)$$

To find out some forumula
So we can "jump" through 2-3 steps (the steps that we went all over again).

-------------------------------------------------------------------

Other question.
Can I use just $$\alpha P$$ instead of $$\alpha P+\beta Q$$ and $$\gamma P +\delta Q$$?

----------------------------------------------------------------------

Another question.

What will happen if I have something like this?
$$(x^4-2x^3-4x^2+6x+1)A(x)+(x^3-5x-3)B(x)=x-2$$

Another question.

---------------------------------------------------------------------

What will happen if d=c=1 and a=b=-1
Lets say the polynom is:
$$(-x^3+2x^2+3x-1)A(x) + (x^3-5x^2+7x+1)B(x)=x^2-4x$$

$$A(x)=B(x)$$

?

 

[tiny-tim]

I thought like this:
$$(ax^n+bx^n^-^1+cx^n^-^2+dx^n^-^3+ex^n^-^f)(\alpha P+\beta Q)+(gx^n+hx^n^-^1+ix^n^-^2+jx^n^-^3+kx^n^-^w)(\gamma P +\delta Q)$$
To find out some forumula so we can "jump" through 2-3 steps (the steps that we went all over again).

There obviously is a formula for doing 2 steps together - but it would be much more complicated, and doing the steps separately is easier and has less risk of mistakes!

Can I use just $$\alpha P$$ instead of $$\alpha P+\beta Q$$ and $$\gamma P +\delta Q$$?
What will happen if I have something like this?
$$(x^4-2x^3-4x^2+6x+1)A(x)+(x^3-5x-3)B(x)=x-2$$

Well, that's a = 1, b = 1, c = 0, d = -3, which is ok!

What will happen if d=c=1 and a=b=-1
Lets say the polynom is:
$$(-x^3+2x^2+3x-1)A(x) + (x^3-5x^2+7x+1)B(x)=x^2-4x$$

$$A(x)=B(x)$$ ?

Now, that is a very good question!

You're right - it doesn't work if the ratio a/b is the same as c/d (you gave a = -c, b = -d; but it's just as bad if a = c, b = d, or a = 3c, b = 3d …).

And we certainly can't put A(x) = B(x) (or any multiple of B(x))!

So we'd have to go back to the a b c d e f g etc method (which, I repeat, is probably the best method anyway)

 

[Physicsissuef]

I think in some ways work.

$$(x^4-2x^3-4x^2+6x+1)A(x)+(x^3-5x-3)B(x)=x-2$$

In this case I can not divide by x , because I have x-2 on the right hand. In this case also will not work?

 

[tiny-tim]

Yes, of course: we only used this method because the right-hand side had no units (only positive powers of x).

If the right-hand side has units, as in your present example, then eliminating units from the left-hand side and dividing by x (the first step) would be useless!

(btw, I did a google search on … something … and I now know who StMartin is, and what he's doing! … the secret is safe with me! )​

 

[Physicsissuef]

Yes, of course: we only used this method because the right-hand side had no units (only positive powers of x).

If the right-hand side has units, as in your present example, then eliminating units from the left-hand side and dividing by x (the first step) would be useless!

(btw, I did a google search on … something … and I now know who StMartin is, and what he's doing! … the secret is safe with me! )​

Hmmm... I can't see any secret Why secret?
And I want to ask you something else.

Can I use just $$A(x) = \alpha P$$ and $$B(x) = \beta Q$$ instead of
$$A(x) = \alpha P+\beta Q$$ and $$B(x) = \gamma P +\delta Q$$?

 

[tiny-tim]

Can I use just $$A(x) = \alpha P$$ and $$B(x) = \beta Q$$ instead of
$$A(x) = \alpha P+\beta Q$$ and $$B(x) = \gamma P +\delta Q$$?

Well, we can make any substitution we like!

The resulting new equation will still be valid - but will it be any easier to solve?

I suppose $$A(x) = \alpha P$$ and $$B(x) = \beta Q$$ might sometimes be useful - were you thinking of a particular example?

 

[Physicsissuef]

No, I don't. But I think it is not useful, because, often can happen A(x)=B(x).

And if we want to "jump" 2 or 3 steps or maybe more, do we must know the power of the polynom (ex. $x^3$, or $x^4$) ??
Like this:
$$(ax^4+bx^3+cx^n^2+dx^1+e)(\alpha P+\beta Q)+(gx^4+hx^3+ix^2+jx^1+k)(\gamma P +\delta Q)$$
So we will find the formula out of here, or we can find general formula out of here ?
$$(ax^n+bx^n^-^1+cx^n^-^2+dx^n^-^3+ex^n^-^f)(\alpha P+\beta Q)+(gx^n+hx^n^-^1+ix^n^-^2+jx^n^-^3+kx^n^-^w)(\gamma P +\delta Q)$$

 

[tiny-tim]

If there is a formula, then it will be a general formula, yes, and we needn't know the power (strictly, "degree" or "order") of the polynomial (eg. $x^3\,+\,...$, or $x^4\,+\,...$).

 

[Physicsissuef]

can you tell me please, or start writing that formula?

P.S I tried only A(x)=P and B(x)=Q and it didn't work, because of

$$P(\alpha ax^n + \alpha b) + Q(\beta cx^n + \beta d)$$

so

$$\frac{\alpha}{0}=\frac{0}{\beta}$$ and $$\frac{\beta}{0}=\frac{0}{d}$$