# Convergence of 10^-2^n. Linear, quadratic, cubic, quartic, hectic

• gikiian
In summary: There's no reason for \alpha to be an integer. The interpretation of \lambda = \infty is that |p_n - p|^{\alpha} \to 0 faster than any polynomial. I think it would be more accurate to say that the sequence converges "faster" than quadratic.
gikiian

## Homework Statement

Show that the sequence ${(p_{n})}^{∞}_{n=0}=10^{-2^{n}}$ converges quadratically to 0.

## Homework Equations

$\stackrel{limit}{_{n→∞}}\frac{|p_{n+1}-p|}{|p_{n}-p|^{α}}=λ$

where
• α is order of convergence; α=1 implies linear convergence, α=2 implies quadratic convergence, and so on, provided that 0<λ<1.
• λ is the asymptotic error constant; in scope of the course (Numerical Analysis), 0<λ<1, so I do not have to play with my solution further if I get λ=0 or λ=1

## The Attempt at a Solution

I think the algorithm for the solution should be the following:

Check if the sequence converges for α=1 by substituting α=1 in the equation, and solving for λ.
- If λ turns out to be 0<λ<1, then the order of convergence is 1.
- If λ turns out to be λ=0, λ=1, or 1<λ<∞, then don't play further with the solution.
- If λ turns out to be λ=∞, it means that the sequence diverges for α=1. Hence we must next substitute α=2 in the equation, and solve again for λ.
- If λ turns out to be 0<λ<1, then the order of convergence is 2.
- If λ turns out to be λ=0, λ=1, or 1<λ<∞, then don't play further with the solution.
- If λ turns out to be λ=∞, it means that the sequence diverges for α=2. Hence we must next substitute α=3 in the equation, and then solve again for λ. And so on.

I realize that there is a bug in the above algorithm. I only need someone to make me identify the bug.

Following is the solution according to the procedure stated above:

Substituting for α=1 in the equation:

$⇒\stackrel{limit}{_{n→∞}}\frac{|10^{-2^{n+1}}-0|}{|10^{-2^{n}}-0|^{1}}=λ$

$⇒\stackrel{limit}{_{n→∞}}\frac{10^{-2^{n+1}}}{10^{-2^{n}}}=λ$

$⇒\stackrel{limit}{_{n→∞}}\frac{10^{2^{n}}}{10^{2^{n+1}}}=λ$

$⇒\stackrel{limit}{_{n→∞}}\frac{10^{2^{n}}}{10^{2^{n}.2}}=λ$

$⇒\stackrel{limit}{_{n→∞}}\frac{10^{2^{n}}}{(10^{2^{n}})^2}=λ$

$⇒\stackrel{limit}{_{n→∞}}\frac{1}{10^{2^{n}}}=λ$

$⇒\frac{1}{∞}=λ$

$⇒0=λ$

$⇒λ=0$

... but according to the textbook (Burden Faries), the sequence has to quadratically converge to p=0. In other words, solving for λ with α=2 must give 0<λ<1, but it is clear that this is not going to happen.I just need to know what is wrong with my solution/algorithm.

Last edited:
Try checking how you substitute n+1 for n.

Are you saying that $10^{-2^{n}}$ should equal $10^{(-2)^{n}}$ instead of $10^{-(2^{n})}$?

I'm saying the n+1 term is 10-2(n+1). I don't see that in your steps, or it isn't marked clearly.

gikiian said:

## Homework Statement

Show that the sequence ${(p_{n})}^{∞}_{n=0}=10^{-2^{n}}$ converges quadratically to 0.

## Homework Equations

$\stackrel{limit}{_{n→∞}}\frac{|p_{n+1}-p|}{|p_{n}-p|^{α}}=λ$

where
• α is order of convergence; α=1 implies linear convergence, α=2 implies quadratic convergence, and so on, provided that 0<λ<1.
• λ is the asymptotic error constant; in scope of the course (Numerical Analysis), 0<λ<1, so I do not have to play with my solution further if I get λ=0 or λ=1

## The Attempt at a Solution

I think the algorithm for the solution should be the following:

Check if the sequence converges for α=1 by substituting α=1 in the equation, and solving for λ.
- If λ turns out to be 0<λ<1, then the order of convergence is 1.
- If λ turns out to be λ=0, λ=1, or 1<λ<∞, then don't play further with the solution.
- If λ turns out to be λ=∞, it means that the sequence diverges for α=1. Hence we must next substitute α=2 in the equation, and solve again for λ.
- If λ turns out to be 0<λ<1, then the order of convergence is 2.

This is inconsistent with the wikipedia definition, which says that if $p_n \to p$ and
$$\lim_{n \to \infty} \frac{|p_{n+1} - p|}{|p_n - p|} = \lambda$$
then $p_n \to p$ linearly if $0 < \lambda < 1$. If $\lambda = 0$ then the sequence converges superlinearly, and the order of convergence is $\alpha$ such that
$$\lim_{n \to \infty} \frac{|p_{n+1} - p|}{|p_n - p|^\alpha} > 0.$$

(1) if you find that $\lambda = 0$ with $\alpha = 1$ then you should proceed to $\alpha = 2$, and
(2) if you find that $\lambda > 0$ for some $\alpha > 1$ you stop.

1 person
pasmith said:
This is inconsistent with the wikipedia definition, which says that if $p_n \to p$ and
$$\lim_{n \to \infty} \frac{|p_{n+1} - p|}{|p_n - p|} = \lambda$$
then $p_n \to p$ linearly if $0 < \lambda < 1$. If $\lambda = 0$ then the sequence converges superlinearly, and the order of convergence is $\alpha$ such that
$$\lim_{n \to \infty} \frac{|p_{n+1} - p|}{|p_n - p|^\alpha} > 0.$$

(1) if you find that $\lambda = 0$ with $\alpha = 1$ then you should proceed to $\alpha = 2$, and
(2) if you find that $\lambda > 0$ for some $\alpha > 1$ you stop.

Thanks a million! :!)

pasmith said:
(1) if you find that $\lambda = 0$ with $\alpha = 1$ then you should proceed to $\alpha = 2$, and

Why should I proceed? Wouldn't the sequence be shown to be superlinear already?

Or can it be superlinear and quadratic at the same time?!
(2) if you find that $∞ >$$\lambda > 0$ for some $\alpha > 1$ you stop.

Do you mind this correction?

Last edited:
gikiian said:
Why should I proceed? Wouldn't the sequence be shown to be superlinear already?

Or can it be superlinear and quadratic at the same time?!

Got it! Quadratic convergence is a form of superlinear convergence.

gikiian said:
Do you mind this correction?

I still did not get this. Will we stop even if λ turns out to be ∞?

I am becoming clearer on this. Just correct me if I'm wrong. My new algorithm goes like:

Substitute α=1 into the equation, and solve for λ
- If λ∈(0,1), then the order of convergence is 1
- If λ=1, then the sequence converges sublinearly
- If λ=0, then the order of convergence is greater than 1
- - Substitute α=2 into the equation, and solve for λ
- - If λ>0, then the order of convergence is 2
- - If λ=0, then the order of convergence is greater than 2
- - - Substitute α=3 into the equation, and solve for λ
- - - If λ>0, then the order of convergence is 3
- - - If λ=0, then the order of convergence is greater than 3
- - - - Substitute α=4 ... (repeat until λ turns out to be λ>0)

gikiian said:
Got it! Quadratic convergence is a form of superlinear convergence.

I still did not get this. Will we stop even if λ turns out to be ∞?

There's no reason for $\alpha$ to be an integer.

The interpretation of $\lambda = \infty$ is that $|p_n - p|^{\alpha} \to 0$ faster than $|p_{n+1} - p| \to 0$, and the interpretation of $\lambda = 0$ is the reverse. If $0 < \lambda < \infty$ then $|p_n - p|^\alpha \to 0$ about as fast as $|p_{n+1} - p| \to 0$.

It may turn out that
$$\lim_{n \to \infty} \frac{|p_{n+1} - p|}{|p_n - p|^k} = 0$$
but
$$\lim_{n \to \infty} \frac{|p_{n+1} - p|}{|p_n - p|^{k+ 1}} = \infty$$
in which case $k < \alpha < k + 1$.

gikiian said:
I am becoming clearer on this. Just correct me if I'm wrong. My new algorithm goes like:

Substitute α=1 into the equation, and solve for λ
- If λ∈(0,1), then the order of convergence is 1
- If λ=1, then the sequence converges sublinearly
- If λ=0, then the order of convergence is greater than 1
- - Substitute α=2 into the equation, and solve for λ
- - If λ>0, then the order of convergence is 2
- - If λ=0, then the order of convergence is greater than 2
- - - Substitute α=3 into the equation, and solve for λ
- - - If λ>0, then the order of convergence is 3
- - - If λ=0, then the order of convergence is greater than 3
- - - - Substitute α=4 ... (repeat until λ turns out to be λ>0)

If you are speaking of a general sequence ##\{ p_n \}##, rather than your original sequence, then some of your statements are incorrect. For ##\alpha = 1##, just having ##\lambda = 1## does not guarantee convergence at all; the sequence may diverge (eg. ##p_n = n##). If ##\lambda = 0## the sequence does converge superlinearly, but that does not mean the convergence is quadratic; we could, for example, have order ##\alpha = 1 + p## with ##0 < p < 1##, etc. Many algorithms produce sequences that have quadratic convergence, but non-integer orders do occur as well for some convergent algorithms. For example, the secant method for solving f(x) = 0 has convergence order ##(1+\sqrt{5})/2 \doteq 1.62##; see, eg., http://en.wikipedia.org/wiki/Secant_method or
http://www.ugrad.cs.ubc.ca/~cs302/secant.pdf . Muller's method of solving f(x) = 0 has order of convergence ##\alpha \doteq 1.84##, being the positive root of the equation ##x^3-x^2-x-1=0##; see, eg., http://en.wikipedia.org/wiki/Muller's_method .

Last edited by a moderator:

## 1. What is the significance of the term "convergence" in the expression 10^-2^n?

The term "convergence" in this expression refers to the behavior of the sequence as n (the exponent) increases. In this case, as n approaches infinity, the value of 10^-2^n approaches 0, indicating that the sequence is converging towards 0.

## 2. How does the value of n affect the convergence of 10^-2^n?

The value of n directly impacts the rate of convergence for 10^-2^n. As n increases, the sequence converges at a faster rate towards 0. This is due to the exponential increase in the exponent, resulting in a rapidly decreasing value.

## 3. What are the different types of convergence that can occur in this expression?

The types of convergence that can occur in this expression are linear, quadratic, cubic, quartic, and hectic. These refer to the rate at which the sequence approaches 0 as n increases. Linear convergence means that the sequence decreases at a constant rate, while quadratic, cubic, quartic, and hectic convergence indicate a faster decrease in the sequence as n increases.

## 4. How can the convergence of 10^-2^n be graphically represented?

The convergence of 10^-2^n can be graphically represented by plotting the sequence against the values of n. As n increases, the values of the sequence will decrease towards 0. The type of convergence can also be visualized by observing the slope of the line on the graph. A constant slope indicates linear convergence, while a steeper slope indicates a faster rate of convergence.

## 5. What practical applications does the concept of convergence in this expression have?

The concept of convergence in this expression has various practical applications in fields such as computer science, engineering, and mathematics. It is used in algorithms and equations to determine the rate of convergence for different sequences. It can also be applied in data analysis and modeling to determine patterns and trends in data that follow a converging sequence.

• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
491
• MATLAB, Maple, Mathematica, LaTeX
Replies
2
Views
687
• Calculus and Beyond Homework Help
Replies
4
Views
662
• Calculus and Beyond Homework Help
Replies
5
Views
711
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
626
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
2K