# Limit of a recursively defined sequence

• Sep 10th 2009, 02:35 AM
Sam1111
Limit of a recursively defined sequence
Hey

The problem is:

a(n+1) = ( a(n)^2 +16) / ( 2a(n) + 6) converges to the limit:...?
for n>=1 and a(1) = -10

Also note (n+1) and (n) are subscripts

If someone could solve this and explain what they are doing it would be extremely helpful. My textbook has no questions like this so I'm completely lost.

Also another question..:

If f(x) is a function that is 8 times continuously differentiable such that the coefficient of x^5 in its 8th MacLaurin polynomial is 0.2, then f^5(0) =

Help would be much appreciated =)
• Sep 10th 2009, 06:43 AM
HallsofIvy
Quote:

Originally Posted by Sam1111
Hey

The problem is:

a(n+1) = ( a(n)^2 +16) / ( 2a(n) + 6) converges to the limit:...?
for n>=1 and a(1) = -10

Also note (n+1) and (n) are subscripts

If someone could solve this and explain what they are doing it would be extremely helpful. My textbook has no questions like this so I'm completely lost.

I answered this on "Physics Forums" also but:
Assuming the sequence has a limit, call it "a".
Taking the limit on both sides of $a_{n+1}= \frac{a_n^2+ 16}{2a_n+ 6}$ gives $a= \frac{a^2+ 16}{2a+ 6}$ so $a(2a+ 6)= a^2+ 16$ which reduces to the quadratic equation $a^2+ 6a- 16= 0$. That has two solutions. Which of them the sequence converges to depends upon the initial value.

By the way, you could then show that the sequence does have a limit by showing it is increasing and has an upper bound.

Quote:

Also another question..:

If f(x) is a function that is 8 times continuously differentiable such that the coefficient of x^5 in its 8th MacLaurin polynomial is 0.2, then f^5(0) =
What is the formula for the coefficients of a MacLaurin polynomial?

Quote:

Help would be much appreciated =)
• Sep 10th 2009, 06:45 AM
Taluivren
Quote:

Originally Posted by Sam1111
Hey

The problem is:

a(n+1) = ( a(n)^2 +16) / ( 2a(n) + 6) converges to the limit:...?
for n>=1 and a(1) = -10

Also note (n+1) and (n) are subscripts

If someone could solve this and explain what they are doing it would be extremely helpful. My textbook has no questions like this so I'm completely lost.

Also another question..:

If f(x) is a function that is 8 times continuously differentiable such that the coefficient of x^5 in its 8th MacLaurin polynomial is 0.2, then f^5(0) =

Help would be much appreciated =)

Hi Sam,

define new sequence $\{b_n\}= \{a_n+8\}$. This means:

$b_1=a_1+8=-2$

$b_{n+1}=a_{n+1}+8=\frac{a_n^2+16}{2a_n+6}+8=\frac{ a_n^2+16+16a_n+48}{2a_n+6}=\frac{(a_n+8)^2}{2(a_n+ 8)-10}=\frac{b_n^2}{2b_n-10}$.

First we show by induction that $b_n<0$ for every positive integer $n$.
$b_1=-2<0$.
Suppose $b_n<0$, then $b_{n+1}=\frac{b_n^2}{2b_n-10}<0$.

Next, we show that for every positive integer $n$ we have $b_{n+1}-b_n>0$.

$b_{n+1}-b_n = \frac{b_n^2}{2b_n-10}-b_n=\frac{b_n^2-2b_n^2+10b_n}{2b_n-10} = \frac{-b_n(b_n-10)}{2b_n-10}>0$ because $b_n<0$ for every positive integer $n$.

We've proved that $b_n for every positive integer $n$. Thus, $b_n$ is an increasing sequence bounded above, so it must have a limit. Denote this limit $r$.
We have $b_n \rightarrow r$ and also $b_{n+1} \rightarrow r$ which means $\frac{b_n^2}{2b_n-10} \rightarrow r$. But since $b_n \rightarrow r$, we also have $\frac{b_n^2}{2b_n-10} \rightarrow \frac{r^2}{2r-10}$.

This means that $r=\frac{r^2}{2r-10}$, after solving this equation we get $r=0$.

We conclude that $a_n \rightarrow -8$.

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As for your another question, its wording is not clear to me. Could you please start a new thread with this question rewritten using LaTeX?