Find polynoms, with as least as power possible

[Physicsissuef]

Are you talking about the polynom that Theofilius had given, or this is yours polynom?

 

[tiny-tim]

Are you talking about the polynom that Theofilius had given, or this is yours polynom?

Theofilius' of course - if you put my A(x) and B(x) into his original question, and multiply it out, you'll find that it does work (actually, I assumed you'd already done that!)

 

[Physicsissuef]

Can you please start from beginning and write the problem without using latex. Why you used 1/A-B and 1/A+B, and so on... Please!

 

[tiny-tim]

Can you please start from beginning and write the problem without using latex. Why you used 1/A-B and 1/A+B, and so on... Please!

Oh, sorry, when I put P = 1/2(A+B), Q = 1/2(A-B), I meant P = (A+B)/2, Q = (A-B)/2.

No, Physicsissuef, you do it.

You won't understand anything just by seeing my proof.

Start by writing P = (A+B)/2, Q = (A-B)/2 into the original equation.

You should get an equation which makes it clear that Q must be divisible by x. So Put Q = xR.

… and so on …

If you have any particular difficulty, you can get back to me on it.

However, it's not a very sensible proof anyway, and I really don't recommend you trying to understand it …

 

[Physicsissuef]

Where I substitute P and Q exactly and why you get A+B/2 and A-B/2, why not some other numbers?

 

[tiny-tim]

Where I substitute P and Q exactly and why you get A+B/2 and A-B/2, why not some other numbers?

In the original:

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

replace A by (P+Q)/2, and B by (P-Q)/2.

I chose them because it results in P being multiplied by a polynomial ending in x (in other words, it has no units), just as $$x^3 - 2x$$ ends in x …

 

[Physicsissuef]

Ok, what when I will substitute :D?

 

[tiny-tim]

Ok, what when I will substitute :D?

$$(x^4+2x^3+x+1)(P+Q)/2 + (x^4+x^3-2x^2+2x-1)(P-Q)/2\,=\,x^3-2x\,,$$​

and then you rearrange so that all the Ps are in one bracket and all the Qs in another.

 

[Physicsissuef]

I came up with this
$$2Px^4+3Px^3+Qx^3-2Px^2+2Qx^2+3Px-Qx+2Q=2(x^3-2x)$$
What to do?

 

[tiny-tim]

I came up with this
$$2Px^4+3Px^3+Qx^3-2Px^2+2Qx^2+3Px-Qx+2Q=2(x^3-2x)$$
What to do?

Hi Physicsissuef!

Yes, your'e completely right … BUT! …

… tears hair out …​

you haven't tidied it up into separate brackets!

The whole point is to write it like this:

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

Now you can see that the P bracket and the right-hand side are both divisible by x, while the Q bracket isn't - which means that Q itself must be divisible by x!

(This is why we chose A = P+Q/2 etc other choices wouldn't have given a P-bracket with no units!)

So you write Q = xR, and then divide the whole equation by x, giving:

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

So we've reduced a quartic equation to a cubic equation! - now, that's what I call progress!

Can you see where to go from there?

(Hint: look for a pattern involving x^2.)

 

[Physicsissuef]

I understand everything, but I can't understand this part
"(This is why we chose A = P+Q/2 etc other choices wouldn't have given a P-bracket with no units!)"
What do u mean no units?

 

[tiny-tim]

What do u mean no units?

Hi! I just meant it doesn't end in +1 or -2 etc, it only has x, x^2, etc.

(and so it divides by x.)

 

[Physicsissuef]

what if I took A=P+Q ?

 

[tiny-tim]

what if I took A=P+Q ?

And B=P-Q, of course.

Yes, that also works fine. Any multiple of P+Q and P-Q would do.

I chose 1/2 because I was looking forward to when I would have to reconvert P and Q back into A and B.

Experience tells me always to use 1/2 because, in the end, I'm less likely to make a mistake (which is amazingly easy to do in problems like this).

That's because when:

A=(P+Q)/2 AND B=(P-Q)/2,​

then:

P=(A+B)/2 AND Q=(A-B)/2.​

(Check that for yourself, until you're convinced!)

You see … it's the same factors in both directions, so I never get confused as to which way I'm going!

Good question!

 

[Physicsissuef]

Ok let's go on now. What should I do next after this?
$$P(2x^3+3x^2-2x+3)\,+\,R(x^3+2x^2-x+2)\,=\,2x^2-4\,.$$

 

[tiny-tim]

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

Hi Physicsissuef!

What I would normally do here is to try to simplify it again, probably by putting y = x + √2, so that the right-hand side has no units again (only powers of y) - and then doing the same A+B/2-type trick all over again.

BUT, just as I was about to (and not looking forward to all those √2s), I noticed a pattern involving x^2.

So I used that instead.

Can you see what it is … ?

 

[Physicsissuef]

Hmmm... This seems to much complicated. Can you write down what do you mean, please?

 

[tiny-tim]

I rearranged it to:

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

and then I rearranged the left-hand side again, to … ?

 

[Physicsissuef]

$$(x^2-1)(P2x+Rx)+(x^2+1)(3P+2R)=2x^2-4$$

like this?

 

[tiny-tim]

That's right!

Except I wrote it:

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

which made me notice that the factor of 2P + R contained only odd powers of x, while the factor of 3P + 2R and the right-hand side contained only even powers of x.

So I got rid of the odd x (on the far left) by changing 2P + R and 3P + 2R to … ?

 

[Physicsissuef]

A and B? And how will I find A and B? :D

 

[tiny-tim]

A and B? And how will I find A and B? :D

First, you must use different letters, or you'll get really confused.

I decided to use U and V.

Remember how, in post #40, we reduced a quartic equation to a cubic equation by getting rid of x?

And we did it by writing Q = xR?

Can you see a similar trick here? What should we change 2P + R and 3P + 2R to?

 

[Physicsissuef]

U and V, and what's next? :D

 

[tiny-tim]

U and V, and what's next? :D

No! That doesn't do anything about the odd x, does it?

Remember our trick of writing Q = xR.

We'll do the same here … but which way round … ?

Try again!

 

[Physicsissuef]

Hmm... I don't know... I tried.. Please help!

 

[tiny-tim]

Half-way point …

Hmm... I don't know... I tried.. Please help!

ok, put xU = 2P + R, V = 3P + 2R.

Then

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

so

$$U\,=\,-3,\,V\,=\,3x^2\,-\,4\,.$$​

This is the half-way point: we've actually solved it now, in the sense that we know exactly what the polynomials U and V are.

All we have to do now is to work backwards, and "unconvert" U and V to P and R, then to Q and R, then to A and B.

Do you want to do that?

 

[Physicsissuef]

How did u find that U=-3 and V=3x^2-4?

 

[tiny-tim]

How did u find that U=-3 and V=3x^2-4?

Oh come on … that's easy!

From the equation, with only x^2 and units on the right, the obvious thing to try was U constant, and V constant(x^2) - 4.

Try it!

 

[Physicsissuef]

Ok, what's next? :D

 

[tiny-tim]

Ok, what's next? :D

Nothing … that's it …

As I said before, all you have to do now is to work backwards, and "unconvert" U and V to P and R, then to Q and R, then to A and B.

And I'm not going to do that for you … that's very straightforward … you just use the definitions of V U R Q and P.

Well, what do you think? Pretty horrible, wasn't it?

I haven't actually tried it, but I suspect simply putting A = ax^3 + bx^2 + cx + d, and B = ex^3 + fx^2 + gx + h, and then multiplying out and getting 8 equations for 8 variables, is both easier and quicker.

In hindsight, we should have know to start with two cubic formulas for A and B - if the original polynomials start x^n, and A and B start x^m, then there are n + m + 1 equations (one for x^n+m, one for x^n+m-1 … one for units), but only 2(m+1) unknowns. So we needed 2(m+1) = n+m+1, or m = n - 1.