Inspired by Sambhrant Sachan

Calculus Level 5

$\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{4}{8}-\dfrac{9}{16}+\dfrac{11}{32}+\dfrac{56}{64}+\dfrac{1}{128}+\cdots+\dfrac{g(n)}{2^n}+\cdots$

The numerators $g(n)$ are given by a linear recursion $g(n)=ag(n-1)+bg(n-2)$ for $n>2$ , with $g(1)=g(2)=1$ .

Find the value of the series above, to three significant figures. Enter 0.666 if you come to the conclusion that the series fails to converge.

Bonus : What if the series is

$\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{3}{8}-\dfrac{7}{16}+\dfrac{5}{32}+\dfrac{33}{64}+\dfrac{13}{128}+\cdots+\dfrac{g(n)}{2^n}+\cdots$

ceteris paribus?

Inspiration .

The answer is 0.666.

2 solutions

Arturo Presa
May 8, 2016

We can find $a$ and $b$ by using the facts that $g(3)=ag(2)+bg(1)=a+b=-4$ and $g(4)=ag(3)+g(2)b=-4a+b=-9.$ By solving the system formed by these two equations, we obtain that $a=1$ and $b=-5$ . Then the linear recurrence can be written as $g(n)=g(n-1)-5g(n-2)\:\:\:\:\:\:\:\:\:(*)$ We can obtain the generating function $f(x)=\sum_{k=1}^\infty g(n)x^n$ of this linear recurrence by multiplying both sides of $(*)$ by $x^n,$ and we add the terms making $n$ take all the values from $n=3$ to $\infty.$ Then we obtain the following equation for $f(x)$ $f(x)-x^2-x=x(f(x)-x)-5x^2f(x).$ Solving for $f(x),$ we obtain $f(x)= \frac{x}{1-x+5x^2},$ whose Maclaurin series is $\sum_{k=1}^\infty g(n)x^n.$ This function $f(x)$ is an analytic function except at two singular points, which are $\frac{1+i\sqrt{19}}{10}$ and $\frac{1-i\sqrt{19}}{10}$ . It easy to find the distance from any of this points to zero to be $1/\sqrt 5\approx 0.447...$ Hence, the radius of convergence of the series is $\frac{1}{\sqrt 5}$ that is less than $\frac{1}{2},$ and, therefore, $\sum_{k=1}^\infty g(n)(\frac{1}{2})^n$ is divergent. So the answer is $\boxed{0.666}.$

Yes, that is exactly what I had in mind, very well explained. (+1) Thank you, compañero!

Otto Bretscher - 5 years, 1 month ago

Thank you, compañero! Very nice problem!

Arturo Presa - 5 years, 1 month ago

Compañero @Arturo Presa , I posted a bonus for your entertainment ;)

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – Uh... i found f(x) in the wake of our last conversation and blindly plugged in 1/2 which is apparently wrong as you need to consider... umm... 'radius of convergence'? I have no idea of this concept and since i can' t ask you to explain it here... could you please refer some website where i can learn this from scratch? PS i dont have any 'standard' books on calculus either....

Milind Blaze - 5 years, 1 month ago

@Milind Blaze – To give you some idea what's going on, here are two simple examples:

$\frac{1}{1-x}=1+x+x^2+x^3+....$

will diverge when $|x|>1$ since 1 is a root of the denominator. Likewise

$\frac{1}{1+x^2}=1-x^2+x^4- x^6+....$

will diverge when $|x|>1$ since $i$ is a root of the denominator, with $|i|=1$ .

More generally, if $f(x)=\frac{p(x)}{q(x)}$ is a rational function and $x_0$ is a complex root of the denominator, then the power series of $f(x)$ will diverge for all complex $x$ with $|x|>|x_0|$ .

For any power series $\sum a_nx^n$ there exists a unique non-negative real number $R$ (or $R=\infty$ ) such that the series converges (absolutely) when $|x|<R$ and it diverges when $|x|>R$ ; this $R$ is called the radius of convergence of the power series.

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – Thanks a lot!!!

PS:Ummm... by the way what is the ratio test?

Milind Blaze - 5 years, 1 month ago

@Milind Blaze – The ratio test is discussed here.

The part that we need to do this problem is quite simple: If a power series $\sum a_nx^n$ converges for some complex $x=c$ , then it will also converge (absolutely) for all $x$ with $|x|<|c|$ . You can show this by comparison with a geometric series. Try to figure it out for yourself! (I will be glad to offer a hint.)

Since our series diverges at $x=\frac{1+i\sqrt{19}}{10}$ , with $|x|=\frac{1}{\sqrt{5}}<\frac{1}{2}$ , it will also diverge at $x=\frac{1}{2}$

Otto Bretscher - 5 years, 1 month ago

@Otto Bretscher – 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

What is the point of writing "... $f(x)$ is an analytic function..." ? Because I don't see the use of that line. I could say that it is a meromorphic and/or holomorphic function (assuming that it is true) and won't make a difference, no?

It's easy to find the distance from any of this point to zero to be $1/\sqrt5 \approx 0.447\ldots$

How do you show that this is true? I apologize if I'm asking any obvious question.

Pi Han Goh - 5 years, 1 month ago

You are right, I can say meromorphic function or analytic function except at the two singular points (in this case, poles) $\frac{1+i\sqrt{19}}{10}$ and $\frac{1-i\sqrt{19}}{10}$ , which are the two complex numbers that make the denominator zero. The thing is, that the result explained by @Otto Bretscher above is also true for any function that is analytic on a certain open set of the plane. So it can be represented by its Taylor series centered at any point $c$ of the open set and the radius of convergence of the series is the distance from $c$ to the nearest singular point of the function. You can read about this result in wikipedia about Radius of Convergence. Then find the topic : "Radius of convergence in complex analysis." To find the distance from a complex number to zero, you know that you just have to find the absolute value of the number. In this case: $|\frac{1+i\sqrt{19}}{10}|= \sqrt{(\frac{1}{10})^2+(\frac{\sqrt{19}}{10})^2}=1/\sqrt{5}.$

Arturo Presa - 5 years, 1 month ago

This is brand new information. Thank you for your reply =D

Pi Han Goh - 5 years, 1 month ago

@Pi Han Goh – You are welcome!

Arturo Presa - 5 years, 1 month ago

Mark Hennings
Oct 21, 2018

While we are at it, the bonus case can be solved to give $g(n) \; = \; \frac{i}{\sqrt{15}}\left(\frac{1 - i\sqrt{15}}{2}\right)^n - \frac{i}{\sqrt{15}}\left(\frac{1+i\sqrt{15}}{2}\right)^n$ and hence, in this case, $g(n) \; = \; \frac{2^n i}{\sqrt{15}}\big[e^{-in\psi} - e^{in\psi}\big] \; = \; \frac{2^{n+1}}{\sqrt{15}}\sin n\psi$ where $\psi = \tan^{-1}\sqrt{15}$ . Thus, in this case $\frac{g(n)}{2^n} \; = \; \frac{2}{\sqrt{15}}\sin n\psi$ Since $0 < \psi < \tfrac12\pi$ , we cannot have $\frac{g(n)}{2^n} \to 0$ as $n \to \infty$ . Thus the series fails to converge in this case, as well.

Inspired by Sambhrant Sachan

The answer is 0.666.

2 solutions

0 pending reports