thomas5267
  • thomas5267
A homework question for a student (not me) preparing for Chinese Gao Kao. Let \(f(x)\) be a function with domain \(\mathbb{R}\). For all \(x_1,x_2\in \mathbb{R}\), there exist a real L such that \(|f(x_1)-f(x_2)|\leq L|x_1-x_2|\). For \(0
Mathematics
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SOLVED
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chestercat
  • chestercat
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thomas5267
  • thomas5267
If I am not mistaken, f(x) is Lipschitz continuous. According to wikipedia, f(x) is also a contraction.
thomas5267
  • thomas5267
The first part is simple. The second part is extraordinarily hard. And this is a homework question for students preparing for Chinese Gao Kao (A Level equivalent)...
Kainui
  • Kainui
Just curious, what is Lipschitz continuous?

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thomas5267
  • thomas5267
A function \(f(x)\) with \(|f(x_1)-f(x_2)|\leq L|x_1-x_2|\) for all \(x_1,x_2\) in the domain of f. L is a fixed positive real number.
thomas5267
  • thomas5267
The second part reminds me of Cesaro summation.
Kainui
  • Kainui
I totally see that Cesaro sum thing. Here's an idea to try to alter the sum a little. \[A_{n+1} = \frac{1}{n+1} \sum_{k=1}^{n+1} a_k = \frac{1}{n+1}\left(a_{n+1}+n* \frac{1}{n} \sum_{k=1}^{n} a_k \right)\] \[A_{n+1} = \frac{1}{n+1}\left(a_{n+1}+nA_n \right) = \frac{a_{n+1}}{n+1} + \frac{n}{n+1} A_n\] \[|A_{n+1}-A_n| = \left|\frac{a_{n+1}}{n+1} + \left(\frac{n}{n+1} -1 \right)A_n \right| = \frac{|a_{n+1}-A_n|}{n+1} \]
thomas5267
  • thomas5267
If I can prove \(a_{k+1}\leq a_k\) then the second part follows.
thomas5267
  • thomas5267
@Kainui Did you not try the first part or couldn't be bothered to type out the answer?
Kainui
  • Kainui
I didn't look at it cause you said it was simple so I assumed you solved it already.
thomas5267
  • thomas5267
Yes I solved it already because it is really simple.
Kainui
  • Kainui
I don't know how to solve any of this though myself, so I have no clue. I don't really care for analysis personally lol
thomas5267
  • thomas5267
For part 1, look at \(|a_k-a_{k+1}|\), then look at the definition of \(a_{k+1}\).
thomas5267
  • thomas5267
Victory is close... Part 1 \[ \begin{align*} &\phantom{{}={}}|a_k-a_{k+1}|\\ &=|f(a_{k-1})-f(a_k)|&\text{Definition of }a_k\\ &\leq L|a_{k-1}-a_k|&\text{Property of }f(x)\\ &\leq L^2|a_{k-2}-a_{k-1}|&\text{Do it again}\\ &\leq L^{k-1}|a_1-a_2| \end{align*} \] \[ \begin{align*} &\phantom{{}={}}\sum_{k=1}^n|a_k-a_{k+1}|\\ &\leq \sum_{k=1}^n L^{k-1}|a_1-a_2|\\ &=|a_1-a_2|\sum_{k=0}^{n-1} L^{k} &\text{Index shift}\\ &<|a_1-a_2|\sum_{k=0}^\infty L^{k}\\ &=\frac{1}{1-L}|a_1-a_2|&0
thomas5267
  • thomas5267
@ganeshie8 Help! Finished part 1 but part 2 just so hard!
ParthKohli
  • ParthKohli
what
thomas5267
  • thomas5267
Is it too confusing or what?
thomas5267
  • thomas5267
The question is in Chinese so I have to translate it into English.

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