Looking for a proof! (corrected)

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# Looking for a proof! (corrected)

Submitted by Zhangaini on Wed, 08/31/2011 - 19:03

Forums:

Dear PMs

The link: http://www.proofwiki.org/wiki/Category:Boubaker_Polynomials

contains a sub-link about looking for a proof !!:

http://www.proofwiki.org/wiki/Relation_of_Boubaker_Polynomials_to_Dickso...

Can you help on this? (as a math challenge or a simple problem)?

Regards

PS. those links can help:

http://en.wikipedia.org/wiki/Dickson_polynomial

and

http://en.wikiversity.org/wiki/Boubaker_Polynomials

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## Re: Looking for a proof! (corrected)

It is probable that the subscript 'n + 1' on the RHS be 'n + j'.

## Re: Looking for a proof! (corrected)

Sorry, or 'n + j + 1'.

## Re: Looking for a proof! (corrected)

Thank you.

can you correct there?

## Re: Looking for a proof! (corrected)

Hey Zhangaini,

> can you correct there?

You are looking for a proof. How may one correct a mathematical proposition if one pretends to prove an unknown proposition? Logic says us that your above question has no answer.

perucho

## Re: Looking for a proof! (corrected)

Sorry

there is no error.

The result is true, but no proof has been provided,

The request of proof stands.

Best regards!

## Re: Looking for a proof! (corrected)

Now your question makes sense. Let's recall the first relation.

B_{n+1}(x)B_{n+j}(x} - B_{n+j+1}(x)B_n(x) = (3x^2+4)D_{n+1}(x,1/4).

I suppose 'j' ranges over nonnegative integers. Let's take, e.g. the first one j=0. Thus for j=0,

B_{n+1}(x)B_n(x) - B_{n+1}(x)B_n(x) = (3x^2+4)D_{n+1}(x,1/4),

whence

(3x^2+4)D_{n+1}(x,1/4) = 0.

But this equation is absurd as neither (3x^2+4) nor D_{n+1}(x,1/4) vanish identically. Do you want discard j=0? Not problem!

For j=1,

B_{n+1}(x)B_{n+1}(x) - B_{n+2}(x)B_n(x) = (3x^2+4)D_{n+1}(x,1/4). (1)

For j=2,

B_{n+1}(x)B_{n+2}(x) - B_{n+3}(x)B_n(x) = (3x^2+4)D_{n+1}(x,1/4). (2)

Next, if you subtract (1) from (2), the RHS is zero because (3x^2+4)D_{n+1}(x,1/4) is invariant with respect to 'j'. So that, leaving the positive terms on LHS and passing the negative terms on the zero RHS, you will get, in each side product of different Boubaker polynomials each other. However, all Boubaker Polynomials are different, arriving so to a contradiction. Moreover, one can choose any pair of nonnegative integers which makes the contradiction more evident.

On the contrary, the second equation

B_n(x) = D_n(2x,1/4) + 4D_{n-1}(2x,1/4),

at least is consistent. Note that this equation is the reciprocal recursion of the first equation, which is a strong proof that the first equation is incorrect.

perucho

## Re: Looking for a proof! (corrected)

Hi Zhangaini,

There is a mistake in the LHS of the first equation, since assigning different values to the subscript 'j', the RHS doesn't change which is absurd. Please, check it out.

perucho