Sunday, August 27, 2006

Products of linear factors

Using the Fundamental Theorem of Algebra, we know that it is possible to express any equation of degree n with one variable as a product of linear factors.

In other words:

xn + a1xn-1 + ... + an = (x - r1)(x - r2)*....*(x - rn) where ri represent the n roots for the equation.

In today's, blog, I show that we are not limited to this form.

Lemma 1:

if (ri ≠ 0), then:

(x - ri) = 0 if and only if (1 - x/ri) = 0


(1) Assume that:

(x - ri) = 0

(2) Then dividing both sides by (-ri) gives us:

(1 - x/ri) = 0

(3) Assume that:

(1 - x/ri) = 0

(4) Multiply both sides by (-ri) so that:

(x - ri) = 0


Corollary 1.1:

if (ri2 ≠ 0), then:

(x2 - ri2) = 0 if and only if (1 - x2/ri2) = 0


This follows directly from Lemma 1 above if set x' = x2 and r' = ri2 then we have:

(x2 - ri2) = (1 - x2/ri2) if and only if (x' - r') = (1 - x'/r')



Anonymous said...

THis is a bad site to use for this.

Larry Freeman said...

I'm sorry to hear the negative comment.

If you could elaborate, I'll be very glad to hear your reasons for disliking the site.

Please also consider that the site is free and is based on my best effort. It is not a commercial site.

If you are looking for other math sites, you might consider:


xyz? said...

I like the way you presented your proof. Don't stop. Keep doing what you like to do. You cannot please everybody. I like to see more proofs. More power to you.

Mathemaperson said...

This may be too late but, I have been looking all over for a decent explanation of Euler's solution to the basel problem and this is great! You link your lemmas and are clear. I think the post above saying this is a bad site may have been because they were used to seeing everything in Latex and find it hard to read. I am grateful for your investment in the site. Thank you.