Find polynoms, with as least as power possible

[Physicsissuef]

Why 2(m+1) unknowns? And why 2(m+1) = n+m+1?

 

[tiny-tim]

Why 2(m+1) unknowns? And why 2(m+1) = n+m+1?

Because if A and B start x^m, then obvioulsy each has m+1 coefficients.

Since they're all unknown, that makes 2(m+1) unknowns.

And we need 2(m+1) = n+m+1 because we can't (generally) solve a set of simultaneous equations unless there are as many equations as unknowns.

(For an obvious example, if you have two unknowns, one equation is not enough, but two equations would be.)

 

[Physicsissuef]

and where did u find n+m+1, from? Can you give me some simple example, pleasE?

 

[tiny-tim]

and where did u find n+m+1, from? Can you give me some simple example, pleasE?

If the original polynomials start x^n, and A and B start x^m, then when we multiply them, we get polynomials starting x^n+m; and a polynomial starting x^n+m has n+m+1 coefficients.

For example, the original question had quartic polynomials, so n = 4; and our A and B were cubic, so m = 3. And a quartic time a cubic starts x^7, and so has 8 coefficients (which does equal n+m+1), and we need 8 equations to find them (which does equal 2(m+1)).

 

[Physicsissuef]

Wait, wait... We had given case like this, let's say:
$$(x^3+x^+x+1)A(x)+(x^2+2x+1)B(x)=x^3$$
And our mission, is to find does A and B are quadric, cubic or whatever... If we find it, we can substitute for the compatible polynom and solve the equation. How we will do it?

 

[tiny-tim]

Wait, wait... We had given case like this, let's say:
$$(x^3+x^+x+1)A(x)+(x^2+2x+1)B(x)=x^3$$
And our mission, is to find does A and B are quadric, cubic or whatever... If we find it, we can substitute for the compatible polynom and solve the equation. How we will do it?

Quadratic should do it - it's always one less: m = n - 1.

If A is quadratic, then the whole equation is quintic, so it has 6 coefficients, which give us 6 equations, so that should solve 6 unknowns, which is ok because we do have 6 unknowns (3 from a and 3 from B).

However, if A was linear, then the whole equation would be quartic, and so would have 5 coefficients, which give us 5 equations, which will not normally be ok because we only have 4 unknowns (2 from a and 2 from B), and 5 equations in 4 unkowns will normally contradict each other.

 

[Physicsissuef]

Can I always be 100% sure what are A(x) and B(x), looking from the euqation? (I mean are they quadric, cubic...)

 

[tiny-tim]

Can I always be 100% sure what are A(x) and B(x), looking from the euqation? (I mean are they quadric, cubic...)

Hi Physicsissuef!

No, you can't be sure, because the n+m+1 equations might not be independent.

For example, if the equation is $$(x^{100}\,+\,2)A(x)\,+\,(x^{100}\,+\,1)B(x)\,=\,x\,,$$
then n = 100, so you might expect m = 99;
but when you write out all 200 equations, you find that most of them are 0 = 0.

(And the solution is obviously A(x) = x, B(x) = -x.)

So you can't be sure that A and B will be (n-1)-th order.

But you can be sure that A and B will be (n-1)-th order or less.

 

[Physicsissuef]

Ok, and let me write, your way, from the start...

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

$$A(x)=\frac{P+Q}{2} , B(x)=\frac{P-Q}{2}$$

$$(x^4+2x^3+x+1)\frac{P+Q}{2} + (x^4+x^3-2x^2+2x-1)\frac{P-Q}{2}=x^3-2x$$

$$2Px^4+3Px^3+Qx^3-2Px^2+2Qx^2+3Px-Qx+2Q=2(x^3-2x)$$

$$P(2x^4+3x^3-2x^2+3x)\,+\,Q(x^3+2x^2-x+2)\,=\,2(x^3-2x)$$

$$Q=xR$$

and then dividing the whole equation with x

$$P(2x^3+3x^2-2x+3)\,+\,R(x^3+2x^2-x+2)\,=\,2x^2-4\,.$$

then rearranging the whole equation

$$x(x^2\,-\,1)(2P\,+\,R)\,+\,(x^2\,+\,1)(3P\,+\,2R)\,=\,2x^2 \,-\,4$$

substituting

$$xU = 2P + R, V = 3P + 2R$$

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

$$U\,=\,-3,\,V\,=\,3x^2\,-\,4\,.$$

Now, what should, I substitute for?

 

[Physicsissuef]

Ok, I found them:
$$A=3x^3+3x^2-x-2$$

$$B=3x^3+3x^2-14x+4$$

Anyway, how many polynoms are possible the find?

 

[Physicsissuef]

Ok, I have new problem.

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

$$A=\frac{P+Q}{2} , B=\frac{P-Q}{2}$$

$$P(x^4-x^3-4x^2+x-2)+Q(x^4-3x^3-4x^2+11x+4)=2x^4$$

And, I am stuck up in here, I have -2 and +4. Please, help.

 

[tiny-tim]

Ok, I found them:
$$A=3x^3+3x^2-x-2$$

$$B=3x^3+3x^2-14x+4$$

Hurrah!

And you did it all yourself!

(Goodness, that wass a lot of LaTeX! )

But it's not quite correct.

For example, you can see immediately, from the original equation, that if A ends +2, then so must B.

Plus … I checked my solution by multiplying it out, so I do know that mine is correct:

$$A(x)\,=\,(3x^3\,+\,3x^2\,-\,7x\,+\,2)\,;\,B(x)\,=\,(-3x^3\,-6x^2\,+\,x\,+\,2)\,.$$ ​

This is exactly why I think this sort of solution is horrible - it's so easy to make a mistake, and so awkward to work out where the mistake is!

P.S. I have a feeling that the fact that the right-hand side of the original equation, $$x^3\,-\,2x$$, is odd (that is, only odd powers of x) means that there's an easy short-cut that I haven't spotted! ​

Anyway, how many polynoms are possible the find?

There should be only one solution.

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

I chose my original P and Q so as to eliminate the units. P+Q only worked because the original polynomials ended in +1 and -1.

Your new ones end in +1 and -3.

So, to use the same technique, either change the formula for P and Q, or - much easier - put A = 3A', so that:

$$(3x^4-6x^3-12x^2+18x+3)A'(x)+(x^3-5x-3)B(x)=x^4\,,$$​

and proceed as before.

 

[Physicsissuef]

$$A'=\frac{P+Q}{2} , B=\frac{P-Q}{2}$$

$$P(3x^4-5x^3-12x^2+13x)+Q(3x^4-7x^3-12x^2+23x+6)=2x^4$$

$$Q=xR$$

$$P(3x^4-5x^3-12x^2+13x)+R(3x^5-7x^4-12x^3+23x^2+6x)=2x^4$$

and then dividing the whole equation with x

$$P(3x^3-5x^2-12x+13)+R(3x^4-7x^3-12x^2+23x+6)=2x^3$$

rearranging the whole equation:

Hm... how to rearrange the whole equation?

 

[tiny-tim]

… I wasn't concentrating …

Ah … I wasn't concentrating …

It should have been P = A' + B (so A' = P - B), and Q = B (in other words, leave B as it is, and don't make a Q).

That way, P gets a bracket with no units, and B gets what it always had, a bracket with no x^4.

Then proceed as before!

Sorry!

The general principle - I've just worked out - is:

if the first bracket is (ax^4 + … + b), and the second is (cx^4 + … + d),
then you put P = dA - pB, Q = cA - aB (or any multiple of either of them),
and that gives you a p bracket with no units, and a Q bracket with no x^4. ​

In Theofilius' original example, a = c = 1, b = -d = 1, so it was P = A+B, Q = A-B.

In your example, a = 1, c = 0, b = 1, d = -3, so we should put P = 3A-B, Q = -B (or B … it doesn't matter!).

 

[Physicsissuef]

ok, no problem. I should substitute for $$A=\frac{P+B}{3}$$ and $$B=-Q$$,
so $$A=\frac{P-Q}{3}$$ and $$B=-Q$$

right?

 

[tiny-tim]

… don't bother with Q …

Yes … but don't bother with Q, just leave B as it is, and work with P and B.

(And the difference between B and -B doesn't matter: remember, any multiple of dA - pB, or of cA - aB, will do.)

 

[Physicsissuef]

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

$$A(x)=3A'$$

$$(3x^4-6x^3-12x^2+18x+3)A'(x)+(x^3-5x-3)B(x)=x^4$$

$$A'=\frac{P+B}{3}$$

$$P(x^4-2x^3-4x^2+6x+1)+B(x^4-x^3-4x^2+x-2)=x^4$$Hm... Look what I get?

 

[Physicsissuef]

tiny-tim please help!

 

[tiny-tim]

… correction … again …

Ah! in post #74, my:

then you put P = dA - bB, Q = cA - aB (or any multiple of either of them)​

should have been:

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

And the inverse formula for that is:

P = -(bA + dB)/(ad - bc); Q = (aA + cB)/(ad - bc).​

So in your example, a = 1, b = 1, c = 0, d = -3,
and we should have put A = -3Q, B = -(P+Q); or more simply B = A/3 - Q.

Sorry!

 

[Physicsissuef]

Now, should I multiply first with 3 or not?

 

[tiny-tim]

I'm not sure - try it and see, if you want!

 

[Physicsissuef]

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

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

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

$$P=xR$$

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

dividing with x

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

then rearranging the whole equation:

Hmmm...I am stuck up in here. Help!

 

[tiny-tim]

… lucky you … !

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

Hi Physicsissuef!

That looks fine!

Well, you still have no units on the right-hand-side (unlike our original case, which had x^2 - 2 on the right, which is why we had to either change method or put x = y + √2, and keep the same method), so you can actually do it all over again …

This time with a = -1, b = 3, c = -3, d = -13.

Lucky you … !

(or you could just put $$Q\,=\,ex^2\,+\,fx\,+\,g,\,R \,=\,hx^2\,+\,jx\,+\,k$$, and solve the six equations for e f g h j and k)

 

[Physicsissuef]

what is x = y + √ 2? And why I need to do it all over again and how do u know that I will succeed with $$Q\,=\,ex^2\,+\,fx\,+\,g,\,R \,=\,hx^2\,+\,jx\,+\,k$$?

 

[tiny-tim]

… it's the American way …

what is x = y + √ 2? And why I need to do it all over again and how do u know that I will succeed with $$Q\,=\,ex^2\,+\,fx\,+\,g,\,R \,=\,hx^2\,+\,jx\,+\,k$$?

Putting x = y + √ 2 would change x^2 - 2 to y^2 + 2y√2, which has no units. Obviously, it changes everything else also, so it's really long-winded, and a complete waste of time unless we're desperate!

The e f g h j k method will work because of the m = n - 1 we discussed earlier.

Why do you need to do it all over again? You don't - you could use e f g h j k for example. But it will work … and you have to do something … your public is expecting!

 

[Physicsissuef]

and if I like to continue, what should I use next without e f g h j k?

 

[tiny-tim]

Sorry … not following you … the only two methods which come to mind are P Q R U V and e f g h j k.

Without that, what were you thinking of using?

 

[Physicsissuef]

I would like to continue out of here:

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

 

[tiny-tim]

Well, I recommend e f g h j k …

 

[Physicsissuef]

But, can I continue out of here, or not?