Inspired by Pi Han Goh

Calculus Level 5

$1^2-2^2+3^2-4^2+\cdots$

The series above diverges, of course. Its Abel sum is defined as

$A=\lim_{x\to 1^-}\sum_{n=0}^{\infty}(-1)^{n+1} n^2 x^n$

Find $A$ to three significant figures. Enter 666 if you come to the conclusion that no such Abel sum $A$ exists.

Inspiration .

Twin Problem .

The answer is 0.00.

1 solution

Arturo Presa
May 15, 2016

Let $S(x)=\sum_{n=0}^{\infty}(-1)^{n+1}n^2x^n$ and $f(x)= x-x^2+x^3-x^4+x^5-...=\frac{x}{1+x}.$ It is easy to see that both power series have radius of convergence of 1, and for any $x$ for which $|x|<1,$ the following is true $S(x)=x\frac{d}{dx}(x\frac{d}{dx}f(x))=\frac{x(1-x)}{(1+x)^3}.$ Therefore, $A=\lim_{x\rightarrow 1^-}S(x)= \lim_{x\rightarrow 1^-}\frac{x(1-x)}{(1+x)^3}=0.$

Beautiful problem, Otto!

Arturo Presa - 5 years, 1 month ago

Beautiful solution, Compañero Presa! (+1) Now what about the Bonus? ;)

Otto Bretscher - 5 years, 1 month ago

I already posted my answer next to your question in the comment part of the problem, but I copied and pasted it here.
Hello compañero @Otto Bretscher. About your bonus question -very nice, by the way- I found that the generating function of the sequence $1, 1, -3, ...$ is the function $g(x)=1/(4x^2-x+1)$ , whose singular points are $\frac{1}{8}(1\pm i \sqrt{15}).$ The distance from each one of these points to zero is exactly $\frac{1}{2},$ so the power series $x+x^2-3x^3-...$ might diverge or converge at $x=1/2$ . To determine its convergence, we might need to find an expression for the sequence of coefficients. Actually, that sequence is defined by the linear recurrence $x_n=x_{n-1}-4x_{n-2},$ where $x_1=x_2=1.$ By forming the auxiliary equation of the sequence, we get that its closed form is $x_n=\frac{2^{n+1}}{\sqrt{15}} \sin(n \arctan \sqrt{15})$ . Now the series could be written in the form $\sum_{n=1}^{\infty} \frac{x_n}{2^n}= \sum_{n=1}^{\infty}\frac{2}{\sqrt{15}} \sin (n \arctan{\sqrt{15}}),$ that diverges due to the fact that its general term does not tend to zero. Is this right?

Arturo Presa - 5 years, 1 month ago

@Arturo Presa – Yes, compañero Presa, another delightful solution indeed! Thank you! I ended up posting this as a separate problem , and I proposed a solution using Чебышёв polynomials.

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – Nice...! I did not see the connection with the Chebyshev polynomials. I obtained the expression by solving directly the linear recurrence: $x_n=x_{n-1}-4x_{n-2}$ , whose auxiliary equation is $r^2-r+4=0.$ Since the solution of this equation are $r=\frac{1\pm i \sqrt{15}}{2}= 2e^{\pm i \arctan{\sqrt{15}}}.$ Then the solutions of the recurrence can be expressed in the form: $x_n= c_1 2^n \cos(n(\arctan{\sqrt{15}}))+c_{2} 2^n \sin (n(\arctan{\sqrt{15}})).$ Now using that $x_1=x_2=1,$ we find the values $c_1=0$ and $c_2=\frac{2}{\sqrt{15}}.$

Arturo Presa - 5 years, 1 month ago

@Arturo Presa – Yes, yours is the systematic, general approach, of course.

I have always been fond of the Chebyshev polynomials; they prove so useful in so many contexts. The Chebyshev polynomials of the second kind satisfy the recursion $U_{n+1}(x)=2xU_n(x)-U_{n-1}(x)$ . I noticed that we can apply this identity, not just to the numerators but to the terms of the whole series.

As I told you, I learned most of my math from Soviet text books (mostly from the Steklov Institute in Leningrad), where Russian mathematicians were given the prominent place they are due.

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – Otto, I will always remember the 4 years that spent in Leningrad, learning from the Russian mathematicians there and specially from my adviser, Victor Petrovich Havin, who died recently. He was 82 years old when he died, but he seemed older to me for the number of things that he accomplished in his life.

Arturo Presa - 5 years ago

@Arturo Presa – I never had a chance to study in the Soviet Union, but I have learned a lot from Soviet mathematicians. My thesis was based on work of Nazarova and Roiter from Kiev. In the US I had a chance to teach with some great Russian mathematicians such as Edward Frenkel... we taught Linear Algebra together several times and he contributed a lot of great ideas to my linear algebra text.

Leningrad is one of my favourite cities (together with Havana and Istanbul), and I plan to travel there next year to help them celebrate the anniversary of The Revolution. I will make sure to visit Euler's tomb.

Otto Bretscher - 5 years ago

@Otto Bretscher – This sounds like an excellent plan, Otto. In the future, I plan to go to Leningrad also. I have a lot of good memories. I just bought your book from Amazon. I was in need of a good book of Linear Algebra.

Arturo Presa - 5 years ago

@Arturo Presa – Let's meet in Leningrad on November 7, 2017 to commemorate the anniversary of Вели́кая Октя́брьская социалисти́ческая револю́ция !

Or let's meet in Havana, my favourite city in the world (in your home country, I believe), and let's smoke a Cohiba Esplendido on the grounds of the Hotel Nacional de Cuba... it does not get any better than that. I visit Cuba every winter and I can'd wait to be back. I always stop by at the university and listen to a math lecture or two; they are conducted at an amazingly high level.

Otto Bretscher - 5 years ago

@Otto Bretscher – I was born in Cuba. My mother still lives there in a city that is I in the west part of the island. I really appreciate your invitation. Now it is going to be difficult for me to go anywhere, but I really appreciate. By the way, I am going to send you my email here, so that you can contact me at any moment if you need so: [email protected], or [email protected].

Arturo Presa - 5 years ago

@Arturo Presa – Please don't expect too much from my linear algebra text; the publisher forced me to make some compromises (sadly, we are still living under a capitalist system, you know). My advisor Peter Gabriel (whose advisor was Grothendieck) has written a scholarly linear algebra text (Birkhäuser, in German); my text is essentially a popularized version of his. Please tell me honestly (among Compañeros) what you think of my text...

Otto Bretscher - 5 years ago

@Otto Bretscher – Otto, I have not received the book yet. I just ordered it today. But I will let you know for sure my opinion about it.

Arturo Presa - 5 years ago

Yeah did the same

Aditya Kumar - 5 years ago

Inspired by Pi Han Goh

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