The following “polynomial-logarithmic” algebraic identity that one encounters on many occasions turns out to have a rather useful set of applications!

**POLYNOMIAL-LOGARITHMIC IDENTITY:** If is a polynomial of degree with roots , then .

PROOF: This one is left as a simple exercise. (*Hint*: Logarithms!)

A nice application of the above identity is found in one of the exercises from the chapter titled *Analysis* (p120) in Proofs from the Book by Aigner, Ziegler and Hofmann.

**EXERCISE:** Let be a non-constant polynomial with only real zeros. Show that for all .

SOLUTION: If is a zero of , then the right hand side of the above inequality equals zero, and we are done. So, suppose is not a root of . Then, differentiating the above identity w.r.t. , we obtain , and we are done.

It turns out that the above identity can also used to prove the well-known *Gauss-Lucas* theorem.

**GAUSS-LUCAS:** If is a non-constant polynomial, then the zeros of lie in the convex hull of the roots of .

PROOF: See this.

HISTORY: The well-known Russian author V.V. Prasolov in his book *Polynomials* offers a brief and interesting historical background of the theorem, in which he points out that Gauss’ original proof (in 1836) of a variant of the theorem was motivated by physical concepts, and it was only in 1874 that F. Lucas, a French Engineer, formulated and proved the above theorem. (Note that the Gauss-Lucas theorem can also be thought of as some sort of a generalization (at least, in spirit!) of Rolle’s theorem.)

Even though I knew the aforesaid identity before, it was once again brought to my attention through a nice (and elementary) article, titled On an Algebraic Identity by Roberto Bosch Cabrera, available at Mathematical Reflections. In particular, Cabrera offers a simple solution, based on an application of the given identity, to the following problem (posed in the 2006 4*th* issue of *Mathematical Reflections*), the solution to which had either escaped regular problem solvers or required knowledge of some tedious (albeit elementary) technique.

**PROBLEM:** Evaluate the sum . (proposed by *Dorin Andrica and Mihai Piticari*.)

SOLUTION: (Read Cabrera’s article.)

There is yet another problem which has a nice solution based again on our beloved identity!

**PROBLEM:** (Putnam A3/2005) Let be a polynomial of degree , all of whose zeros have absolute value 1 in the complex plane. Put . Show that all zeros of have absolute value 1.

SOLUTION: (Again, read Cabrera’s article.)

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## 12 comments

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November 14, 2008 at 1:15 pm

Todd TrimbleThis reminds me of something that I’ve never seen a calculus text point out about partial fraction decompositions, but which is nice to know: in the case where a polynomial has distinct roots and deg() < deg(), the solution to the partial fraction decomposition

is

which specializes to the logarithmic derivative formula of this post (at least

in the case of distinct roots). For some reason, the calculus texts teach students a slightly tedious song-and-dance for solving for the , when there is this nice formula

just sitting there. Plus, teaching why this formula is true gives a nice simple illustration of the things calculus can do for you; seems a shame to pass up such an opportunity.November 14, 2008 at 1:53 pm

Vishal LamaTodd,

The formula you gave for , wouldn’t that be somewhat tedious to compute owing to the fact that one would need to differentiate ? I mean, suppose I had to decompose into partial fractions, then I would have to differentiate , which using product rule, would yield a sum of three terms. Of course, computing such a sum would be easy since all the terms except one would be zero.

On the other hand, perhaps a “faster” technique (that avoids differentiating altogether) is to note that each can be computed from by removing the appropriate factor from . So, in the above example, evaluated at . And so on.

Perhaps, I am just stating what everyone already knows!

November 14, 2008 at 2:22 pm

Todd TrimbleNormally is not already given in factored form. So no, I don’t necessarily agree it’s all that tedious.

I think the larger point looks beyond the mundane computational aspects and more toward theory. This is a nice neat global result, and it comes in handy here and there. More people should be aware of it!

November 14, 2008 at 2:39 pm

Vishal LamaAh, I see what you are saying. In my response above, I assumed the roots of were already given to us, for otherwise we wouldn’t be able to compute ; hence my emphasis on avoiding the differentiation. However, now that you have emphasized once again the importance of theory over the mundane computational aspects, I can see better why you gave that formula. That result is indeed worth knowing! Thanks!

November 14, 2008 at 3:07 pm

Todd TrimbleJust to illustrate the kind of case where you might like to use this result even for a computation: suppose in complex analysis you wanted to decompose

The formula I gave gives, pretty quickly,

where .

November 14, 2008 at 3:17 pm

Vishal LamaTodd,

Thanks for that example. I finally see it now. Hope you will forgive me for being somewhat obtuse earlier.🙂

In the example you gave, indeed your method speeds up computation significantly! I must remember all this. Great!

November 14, 2008 at 4:26 pm

Todd TrimbleI forgive you.🙂

No, seriously — I sometimes have an unfortunate tendency to state things on a theoretical level, and then forget to give illustrative examples that would help drive the points home. [I’m not the only category theorist who does that.] So any discussion like this may be helpful for all concerned, including myself.🙂

November 14, 2008 at 4:51 pm

Vishal LamaI forgive you.The padawan thanks the Master!🙂

I looked once again at your method and realized (albeit slowly) that in the earlier example I provided, in order to compute there is indeed no need to differentiate all the terms; only needs to be differentiated, for the other terms when evaluated at simply vanish. Which means that your method completely subsumes the one I presented. And, further, in the example you provided, computationally it is more efficient to use your method; my method in fact increases labor! And, above all, the theory behind your method reveals so much more to the reader.

I am glad we had this discussion!

May 3, 2009 at 6:50 pm

Qiaochu YuanTodd, that formula is perhaps one of my favorite algebraic tricks of all time, as it immediately implies three closely related results that are often presented isolated from one another:

– the Lagrange interpolation formula (an observation of Terence Tao),

– the form of the Vandermonde determinant, and

– the form of the discrete Fourier transform!

For the first result, let and let ; then P is the interpolating polynomial. The second follows from the first and Cramer’s rule. The third just generalizes your computation, but it is interesting to think of discrete Fourier transforms as solving an interpolation problem on the unit circle.

May 7, 2009 at 5:27 pm

mzargarI would like to point out that Ahlfors demands a proof for this very useful algebraic identity (that Todd Trimble pointed out in his first comment on this page)at the end of his “Polynomials” section in the second chapter of his (standard/well-known) Complex Analysis book. Immediately following this problem, he asks for a proof of the Lagrange Interpolation formula using this algebraic identity. I also have to agree with Qiaochu Yuan that this is very a beautiful formula!

May 14, 2009 at 10:19 pm

Todd TrimbleGosh, sorry guys, I’ve been shamefully ignoring my (ahem, I mean our, Vishal) own blog for quite a while and didn’t see these nice comments. I’ve long been aware of the connection with Lagrange interpolation. I had begun writing a post which explores other connections, but have put it to one side. Hopefully not forever though.

This is one place in the calculus curriculum which really does look out toward larger vistas. Why, oh why, do calculus text authors elide over it??

January 13, 2010 at 11:47 am

Américo TavaresI posted a variant of the first identity

Let be a polynomial of degree . Assume that the highest coefficient of is equal to . Prove that , where are the roots of

here

http://problemasteoremas.wordpress.com/2010/01/06/problema-do-mes-problem-of-the-month-3/