Limits in sequences (Real analysis questions)

[trickslapper]

Justify the existence of the limit of the sequence and then find the limit

1. x[sub:3u7jrgvc]n+1[/sub:3u7jrgvc]=1/3(2x[sub:3u7jrgvc]n[/sub:3u7jrgvc] + M/x[sup:3u7jrgvc]2[/sup:3u7jrgvc][sub:3u7jrgvc]n[/sub:3u7jrgvc]), x[sub:3u7jrgvc]1[/sub:3u7jrgvc]=M>0, where M is a real number

2. x[sub:3u7jrgvc]1[/sub:3u7jrgvc]= 3 + 1/3, x[sub:3u7jrgvc]2[/sub:3u7jrgvc]= 3 + 1/x[sub:3u7jrgvc]1[/sub:3u7jrgvc],...,X[sub:3u7jrgvc]n+1[/sub:3u7jrgvc] = 3 + 1/x[sub:3u7jrgvc]n[/sub:3u7jrgvc],....

3. x[sub:3u7jrgvc]n+1[/sub:3u7jrgvc]= (x[sub:3u7jrgvc]n[/sub:3u7jrgvc] + A)/4, x[sub:3u7jrgvc]1[/sub:3u7jrgvc]=0, where A is a real number

4. x[sub:3u7jrgvc]1[/sub:3u7jrgvc]=25, x[sub:3u7jrgvc]2[/sub:3u7jrgvc]=arctan(25), x[sub:3u7jrgvc]3[/sub:3u7jrgvc]=arctan(arctan(25)),...


I know to show the limit exists, i have to show that it is increasing and bounded above, and im not sure on how to actually find the limit. Can someone show me how to proceed with these questions?

 

[galactus]

2. x[sub:2k94w58k]1[/sub:2k94w58k]= 3 + 1/3, x[sub:2k94w58k]2[/sub:2k94w58k]= 3 + 1/x[sub:2k94w58k]1[/sub:2k94w58k],...,X[sub:2k94w58k]n+1[/sub:2k94w58k] = 3 + 1/x[sub:2k94w58k]n[/sub:2k94w58k],....

What you have is a continued fraction. Since it is nested, you can replace \(\displaystyle x_{n}\) with L.

Then, we have \(\displaystyle L=3+\frac{1}{L}\)

Multiply by L:

\(\displaystyle L^{2}-3L-1=0\)

solve the quadratic to find the limit.

 

[trickslapper]

I have to prove that it has a limit first, the way we usually do that is to show that it is increasing and that it is bounded above. Any idea how to do this ?

 

[soroban]

Hello, trickslapper!

Justify the existence of the limit of the sequence and then find the limit.

\(\displaystyle (3)\;\;x_{n+1} \;=\; \frac{x_n + A}{4},\;\;x_1\,=\,0,\;\text{ where }A\text{ is a real number.}\)

\(\displaystyle \begin{array}{ccccc}\text{Multiply by 4:}& 4x_{n+1} &=& x_n + 4 & [1] \\ \text{The "next" term:}& 4x_{n+2} &=& x_{n+1} + A & [2] \end{array}\)

\(\displaystyle \text{Subtrct [2] - [1]: }\;4x_{n+2} - 4x_{n+1} \;=\;x_{n+1} - x_n \quad\Rightarrow\quad 4x_{n+2} - 5x_{n+} + x_n \;=\;0\)

\(\displaystyle \text{Let: }\,X^n = x_n: }\;4X^{n+2} - 5X^{n+1} + X^n \;=\;0\)

\(\displaystyle \text{Divide by }X^n\!:\;\;4X^2 - 5X + 1 \;=\;0 \quad\Rightarrow\quad (X - 1)(4X - 1) \:=\:0 \quad\Rightarrow\quad X \:=\:1,\:\tfrac{1}{4}\)

\(\displaystyle \text{The generating function is of the form: }\;f(n) \;=\;C(1^n) + D\left(\tfrac{1}{4}\right)^n \;=\;C + \frac{D}{4^n}\)

\(\displaystyle \text{We have the first two terms: }\:f(1) \,=\,0,\;f(2) \,=\,\tfrac{A}{4}\)

\(\displaystyle \text{Use them to set up a system of equations:}\)


. . \(\displaystyle \begin{array}{cccccccc}f(1) = 0: & C + \frac{D}{4} &=& 0 & [3] \\ \\[-3mm] f(2) = \frac{A}{4}: & C + \frac{D}{16} &=& \frac{A}{4} & [4] \end{array}\)

\(\displaystyle \text{Subtract [3] - [4]: }\;\frac{3}{16}D \:=\:-\frac{A}{4} \quad\Rightarrow\quad D \:=\:-\frac{4}{3}A\)

\(\displaystyle \text{Substitute into [3]: }\;C + \frac{-\frac{4}{3}A}{4} \:=\:0 \quad\Rightarrow\quad C \:=\:\frac{A}{3}\)


\(\displaystyle \text{Therefore: }\;f(n) \;=\;\frac{A}{3} + \left(-\frac{4}{3}A\right)\cdot\frac{1}{4^n} \;=\;\frac{A}{3} - \frac{A}{3\cdot4^{n-1}}\)

. . . . . . . . \(\displaystyle f(n) \;=\;\frac{A}{3}\left(1 - \frac{1}{4^{n-1}}\right)\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


\(\displaystyle \text{The limit of the sequence is: }\;\lim_{n\to\infty}\,\frac{A}{3}\left(1 - \frac{1}{4^{n-1}}\right) \;=\;\frac{A}{3}(1 - 0) \;=\;\frac{A}{3}\)

 

[trickslapper]

Hey i'm sorry, but i'm not seeing how we prove that it is increasing and that it is bounded above in that proof :? .

 

[daon]

Have you heard of contractive sequences? Every sequence that is contractive is Cauchy.

Coming from Soroban's subtraction line:

\(\displaystyle |x_{n+2}-x_{n+1}| = \frac{1}{4}|x_{n+1}-x_n}| < \frac{1}{3}|x_{n+1}-x_n}|\)

Hence it must converge.

 

[trickslapper]

Yea, we haven't gone over that yet. All we have done in class to answer limit questions is first prove that there is a limit by showing that the sequence is increasing and that the sequence is bounded. If both those conditions hold then there is a finite limit and then we find it.