Page 1 of 2 12 LastLast
Results 1 to 15 of 21

Math Help - Fixed-point iteration 'show that' exercise

  1. #1
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1

    Fixed-point iteration 'show that' exercise

    Hi,

    problem:

    The iteration defined by x_{k+1} = \frac{1}{2}(x^2_k+c), where 0<c<1,
    has two fixed points \xi_1, \xi_2 where 0<\xi_1<1<\xi_2. Show that

    x_{k+1} - \xi_1 = \frac{1}{2}(x_k+\xi_1)(x_k-\xi_1), k=0,1,2,\cdots,

    and deduce that \lim_{k\rightarrow \infty}x_k=\xi_1 \;\textrm{if}\; 0\leq x_0<\xi_2. How does the iteration behave for other values of x_0?

    attempt at solution:

    x_{k+1}-\xi_1 = \frac{1}{2}(x^2_k+c) - \xi_1<br />
    and since,

    \frac{1}{2}(\xi^2_1+c)=\xi_1 \Rightarrow c = 2\xi_1-\xi^2_1

    we have that,

    x_{k+1}-\xi_1 = \frac{1}{2}(x^2_k-\xi^2_1) = \frac{1}{2}(x_k+\xi_1)(x_k-\xi_1)

    I am a bit confused regarding the limit-question. If 0\leq x_0 < \xi_2 and 1<\xi_2 it means that  x_0 could be any number larger than 1 but smaller than \xi_2. I figure that I could then choose some x_0 such that x_1 > x_0, in other words we would be moving away from \xi_1.

    Now if 0 \leq x_0 < \xi_1 instead of 0\leq x_0 <\xi_2, I could see how \lim_{x\rightarrow \infty}x_k= \xi_1.

    As I rarely have the insight to find errors in problems, I'm sure there's something I'm not getting here, and therefore I ask for you help.

    Thanks.
    Last edited by Mollier; September 15th 2010 at 06:10 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Mollier View Post
    Hi,

    problem:

    The iteration defined by x_{k+1} = \frac{1}{2}(x^2_k+c), where 0<c<1,
    has two fixed points \xi_1, \xi_2 where 0<\xi_1<1<\xi_2. Show that

    x_{k+1} - \xi_1 = \frac{1}{2}(x_k+\xi_1)(x_k-\xi_1), k=0,1,2,\cdots,

    and deduce that \lim_{k\rightarrow \infty}x_k=\xi_1 \;\textrm{if}\; 0\leq x_0<\xi_2. How does the iteration behave for other values of x_0?

    attempt at solution:

    x_{k+1}-\xi_1 = \frac{1}{2}(x^2_k+c) - \xi_1<br />
    and since,

    g(\xi_1)=\frac{1}{2}(\xi^2_1+c)=\xi_1 \Rightarrow c = 2\xi_1-\xi^2_1

    we have that,

    x_{k+1}-\xi_1 = \frac{1}{2}(x^2_k-\xi^2_1) = \frac{1}{2}(x_k+\xi_1)(x_k-\xi_1)

    I am a bit confused regarding the limit-question. If 0\leq x_0 < \xi_2 and 1<\xi_2 it means that  x_0 could be any number larger than 1 but smaller than \xi_2. I figure that I could then choose some x_0 such that x_1 > x_0, in other words we would be moving away from \xi_1.

    Now if 0 \leq x_0 < \xi_1 instead of 0\leq x_0 <\xi_2, I could see how \lim_{x\rightarrow \infty} = \xi_1.

    As I rarely have the insight to find errors in problems, I'm sure there's something I'm not getting here, and therefore I ask for you help.

    Thanks.
    A fixed point is a solution of:

    x = \frac{1}{2}(x^2+c)

    for obvious reasons

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1
    Quote Originally Posted by CaptainBlack View Post
    A fixed point is a solution of:

    x = \frac{1}{2}(x^2+c)

    for obvious reasons

    CB
    Hi,

    if I solve x=  \frac{1}{2}(x^2+c) for x and call the solutions \xi_1 and \xi_2 I get,

    \xi_1 = 1 - \frac{1}{2}\sqrt{4-4c}

    and

    \xi_2 = 1 + \frac{1}{2}\sqrt{4-4c}

    Since 0<c<1, then \xi_2<2 and so I need to show that \lim_{k\rightarrow\infty}x_k=\xi_1 when 0\leq x_0<2.

    If my function, call it g(x) = \frac{1}{2}(x^2+c) was a contraction in (0,2), I think (not sure) I could find a way to show that it converges to \xi_1 for x_0\in (0,2). The problem as I see it is that  g'(x)=x and so it cannot be a contraction on (0,2).

    Perhaps I could use my previous result, that is
    <br />
x_{k+1}-\xi_1=\frac{1}{2}(x_k+\xi_1)(x_k-\xi_1),

    and rewrite it as,

    |x_k-\xi_1|=\frac{2|x_{k+1}-\xi_1|}{x_k+\xi_1}.

    I would like |x_k-\xi_1| to approach zero as k goes to infinity, but I don't know how to show this...

    I could use a hand here, thanks.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    if:

     \xi_1=1-\sqrt{1-c}

    \xi_2=1+\sqrt{1-c}

    x_1<\xi_2\ (which implies that x_k<\xi_2 for all k=1, 2, .. )

    and:

    x_{k+1} - \xi_1 = \frac{1}{2}(x_k+\xi_1)(x_k-\xi_1)

    Then:

    \varepsilon_{k+1}=x_{k+1}-\xi_1=\frac{1}{2}(x_k+\xi_1)\varepsilon_k

    etc (to complete this you need to show that there exists a k >0 such that |x_k+\xi_1|<k<2)

    CB
    Last edited by CaptainBlack; September 16th 2010 at 02:29 AM. Reason: Note this is not how I would do this from choice
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1
    Quote Originally Posted by CaptainBlack View Post
    if:

     \xi_1=1-\sqrt{1-c}

    \xi_2=1+\sqrt{1-c}

    x_1<\xi_2\ (which implies that x_k<\xi_2 for all k=1, 2, .. )

    CB
    So if x_1<\xi_1 and x_2=\frac{1}{2}x_1+\frac{1}{2}c then I am sure that
    x_2<\xi_2?
    I know for sure that \frac{1}{2}x_1<\xi_2 and I also know that \frac{1}{2}c<\frac{1}{2}, but I am not able to see that x_2 must be smaller than \xi_2.
    All these inequalities are confusing me..

    Thank you for being so patient with me CB!
    Time to donate some cash to MHF
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Mollier View Post
    So if x_1<\xi_1 and x_2=\frac{1}{2}x_1+\frac{1}{2}c then I am sure that
    x_2<\xi_2?
    I know for sure that \frac{1}{2}x_1<\xi_2 and I also know that \frac{1}{2}c<\frac{1}{2}, but I am not able to see that x_2 must be smaller than \xi_2.
    All these inequalities are confusing me..

    Thank you for being so patient with me CB!
    Time to donate some cash to MHF
    x_2=\frac{1}{2}(x_1^2+c)<\frac{1}{2}(\xi_2^2+c)=\f  rac{1}{2}(1+2\sqrt{1-c}+(1-c)+c)=1+\sqrt{1-c}=\xi_2
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1
    Hi,

    2\left(\frac{\epsilon_{k+1}}{\epsilon_k}\right)=x_  k+\xi_1

    2\left(\frac{\epsilon_{k+1}}{\epsilon_k}\right)<\x  i_2+\xi_1=2

    so,

    \frac{\epsilon_{k+1}}{\epsilon_k}<1

    Is this what you had in mind CB?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Not quite since that is insufficient to prove that \varepsilon_n \to 0 as n \to \infty

    CB
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1
    Hi,

    here's another try!

    x_{k+1}-\xi_1 = \frac{1}{2}(x_k+\xi_1)(x_k-\xi_1)

    Since \xi_2+\xi_1 = 2, and we have that x_0<\xi_2 and in fact that x_k<\xi_2 for k=1,2,\cdots, then

    x_k+\xi_1 < 2.

    We end up with,

    \epsilon_{k+1} = \frac{1}{2}(x_k+\xi_1)\epsilon_k,

    and so \epsilon_{k+1} is decreasing and approaching zero as k\rightarrow\infty.

    Perhaps I will approach the correct answer as \textrm{(number of attempts)}\rightarrow\infty.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1
    Here's perhaps a better way of writing this..

    x_{k+1}-\xi_1 < \frac{1}{2}(\xi_2+\xi_1)(x_k-\xi_1)

    since \xi_2+\xi_1 = 2 we have that,

    x_{k+1}-\xi_1 < x_k-\xi_1

    Since x_{k+1}=\frac{1}{2}(x^2_k+c) and
    c = 2\xi_1-\xi^2_1, we have that

    \frac{1}{2}(x_k+\xi_1)(x_k-\xi_1) < (x_k-\xi_1) and so

    x_k+\xi_1 < 2. Good times.

    If I now look at what happens if x_0>\xi_2 in a similar fashion, I get that,
    x_k+\xi_1>2 and so the sequence will not converge to \xi_2.

    I think this is because the derivative of \frac{1}{2}(x^2+c) is x and when x_0>\xi_2 it means that x_0>1 and so the derivative is larger than 1. If our function has a derivative larger than 1 in the interval we are working on, fixed-point iteration will not converge.

    Is this better?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2

    Lightbulb

    Quote Originally Posted by Mollier View Post
    Hi,

    problem:

    The iteration defined by x_{k+1} = \frac{1}{2}(x^2_k+c), where 0<c<1,
    has two fixed points \xi_1, \xi_2 where 0<\xi_1<1<\xi_2. Show that
    ...
    As I rarely have the insight to find errors in problems, I'm sure there's something I'm not getting here, and therefore I ask for you help.

    Thanks.
    May be you are looking for this.
    I shall denote the sequence by \{x_{k}\}_{k\in\mathbb{N}_{0}} for c\in\mathbb{R}_{0}^{+} and define
    x_{k+1}:=\frac{1}{2}\big(x_{k}^{2}+c\big) for k\in\mathbb{N},\rule{2cm}{0cm}(1)
    where x_{0}\in\mathbb{R}_{0}^{+}.
    It is clear that for c,x_{0}\in\mathbb{R}_{0}^{+}, \{x_{k}\}_{k\in\mathbb{N}_{0}}\subset\mathbb{R}_{0  }.
    Now, for k\in\mathbb{N}, compute
    x_{k+1}-x_{k}=\frac{1}{2}\big(x_{k}^{2}-x_{k-1}^{2}\big)
    ................_ =\frac{1}{2}\big(x_{k}+x_{k-1}\big)\big(x_{k}-x_{k-1}\big).
    Let \varphi(\lambda):=\lambda(\lambda-2)+c for \lambda\in\mathbb{R}.
    Repeating the procedure above with the abbreviation that the empty product
    is the unity, we get for all k\in\mathbb{N}_{0} that
    x_{k+1}-x_{k}=\dfrac{1}{2^{k-1}}\Bigg[\displaystyle\prod_{i=2}^{k}\big(x_{i}+x_{i-1}\big)\Bigg]\big(x_{2}-x_{1}\big)
    ................_ =\dfrac{1}{2^{k}}\Bigg[\displaystyle\prod_{i=1}^{k}\big(x_{i}+x_{i-1}\big)\Bigg]\big(x_{1}-x_{0}\big)
    ................_ =\dfrac{1}{2^{k+1}}\Bigg[\displaystyle\prod_{i=1}^{k}\big(x_{i}+x_{i-1}\big)\Bigg]\varphi(x_{0}),\rule{2cm}{0cm}(2)
    which implies that the sequence is either non-decreasing or non-increasing.
    Therefore, \xi:=\lim_{k\to\infty}x_{k} exists (finite or infinite).
    Taking \lim as k\to\infty on both sides of (1), we get
    \xi=\frac{1}{2}(\xi^{2}+c),\rule{2cm}{0cm}(3)
    which gives the equlibrium points by solving the quadratic equation
    \xi_{1}=1-\sqrt{1-c}<1 and \xi_{2}=1+\sqrt{1-c}>1 for c\in(0,1).
    If \xi satisfies (3), then c=2\xi-\xi^{2}, and using (1), we have
    x_{k+1}-\xi=\frac{1}{2}\big(x_{k}^{2}+2\xi-\xi^{2}\big)-\xi=\frac{1}{2}(x_{k}+\xi)(x_{k}-\xi) for all k\in\mathbb{N}_{0}.
    As the term \varphi(x_{0}) in (2) is quadratic, we can solve \varphi(\lambda)=0 for \lambda and get that
    \lambda_{1}=1-\sqrt{1-c}=\xi_{1}<1 and \lambda_{2}=1+\sqrt{1-c}=\xi_{2}>1 for c\in(0,1).
    • If x_{0}\in[\xi_{1},\xi_{2}), then \varphi(x_{0})\leq0, and the sequence is non-increasing.
      Then the sequence can not converge a value greater than x_{0}.
      Hence, if x_{0}\in[\xi_{1},\xi_{2}), then \lim_{k\to\infty}x_{k}=\xi_{1}.

    • Now, let x_{0}\in[0,\xi_{1}), then \varphi(x_{0})>0, and the sequence is strictly increasing.
      By induction, we have x_{k}<\xi_{1} for all k\in\mathbb{N}_{0} implying that \lim_{k\to\infty}x_{k}=\xi_{1}.

    This shows that the equilibrium point \xi_{1} is stable.
    • In the case, x_{0}=\xi_{2}, we have x_{k}=\xi_{2} for all k\in\mathbb{N}_{0}, i.e., \lim_{k\to\infty}x_{k}=\xi_{2}.

    • Finally, if x_{0}\in(\xi_{2},\infty), then \varphi(x_{0})>0, and the sequence is strictly increasing.
      By induction, we have x_{k}>\xi_{2} for all k\in\mathbb{N}_{0} implying that \lim_{k\to\infty}x_{k}=\infty.

    Therefore, the equilibrium point \xi_{2} is unstable.
    Last edited by bkarpuz; September 20th 2010 at 01:15 PM.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1
    Hi bkarpuz,

    thank you for what looks like a wonderful answer. The only problem is that I do not understand it. First of all, some of the notation is giving me problems.
    You first denote the sequence by \{x_k\}_{k\in\mathbb{N}_0}.
    I think k\in\mathbb{N}_0\; \textrm{means that}\; \{k\in\mathbb{Z}:k\geq 0\}.
    You then write c\in\mathbb{R}^+_0. I have not seen this notation before, with the + sign and everything, but from the way c is defined in the problem, I'm assuming that it means 0<c<1 ?

    Again, thanks for taking the time to write out all that good stuff. I just hope it isn't wasted on me. Will try to understand it!
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2
    Quote Originally Posted by Mollier View Post
    Hi bkarpuz,

    thank you for what looks like a wonderful answer. The only problem is that I do not understand it. First of all, some of the notation is giving me problems.
    You first denote the sequence by \{x_k\}_{k\in\mathbb{N}_0}.
    I think k\in\mathbb{N}_0\; \textrm{means that}\; \{k\in\mathbb{Z}:k\geq 0\}.
    You then write c\in\mathbb{R}^+_0. I have not seen this notation before, with the + sign and everything, but from the way c is defined in the problem, I'm assuming that it means 0<c<1 ?

    Again, thanks for taking the time to write out all that good stuff. I just hope it isn't wasted on me. Will try to understand it!
    Hi Mollier,

    \mathbb{N}_{0}:=\mathbb{N}\cup\{0\}
    and
    \mathbb{R}_{0}^{+}:=\mathbb{R}^{+}\cup\{0\}
    I am glad you like the solution.
    Actually, what I have shown you above is the standard procedure to be followed for such kind of problems.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member Mollier's Avatar
    Joined
    Nov 2009
    From
    Norway
    Posts
    234
    Awards
    1
    Quote Originally Posted by bkarpuz View Post

    Now, for k\in\mathbb{N}, compute
    x_{k+1}-x_{k}=\frac{1}{2}\big(x_{k}^{2}-x_{k-1}^{2}\big)
    ................_ =\frac{1}{2}\big(x_{k}+x_{k-1}\big)\big(x_{k}-x_{k-1}\big).
    Let \varphi(\lambda):=\lambda(\lambda-2)+c for \lambda\in\mathbb{R}.
    Repeating the procedure above with the abbreviation that the empty product
    is the unity, we get for all k\in\mathbb{N}_{0} that
    x_{k+1}-x_{k}=\dfrac{1}{2^{k-1}}\Bigg[\displaystyle\prod_{i=2}^{k}\big(x_{i}+x_{i-1}\big)\Bigg]\big(x_{2}-x_{1}\big)
    ................_ =\dfrac{1}{2^{k}}\Bigg[\displaystyle\prod_{i=1}^{k}\big(x_{i}+x_{i-1}\big)\Bigg]\big(x_{1}-x_{0}\big)
    ................_ =\dfrac{1}{2^{k+1}}\Bigg[\displaystyle\prod_{i=1}^{k}\big(x_{i}+x_{i-1}\big)\Bigg]\varphi(x_{0}),\rule{2cm}{0cm}(2)
    which implies that the sequence is either non-decreasing or non-increasing.
    Therefore, \xi:=\lim_{k\to\infty}x_{k} exists (finite or infinite).
    Hi again bkarpuz,

    thanks for the quick intro to set-theory notation

    I can see how you get,

    x_{k+1}-x_k = \frac{1}{2^k}\Bigg[(x_1+x_0)(x_2+x_1)\cdots (x_k+x_{k-1})\Bigg](x_1-x_0),

    which you then write as,

    x_{k+1}-x_k=\dfrac{1}{2^{k}}\Bigg[\displaystyle\prod_{i=1}^{k}\big(x_{i}+x_{i-1}\big)\Bigg]\big(x_{1}-x_{0}\big).

    If I am reading this right, the sequence is decreasing if,

    (x_{k+1}-x_k)<(x_k-x_{k-1}),

    and if I'm doing this right then,

    x_{k}-x_{k-2} = \dfrac{1}{2^{k}}\Bigg[\displaystyle\prod_{i=1}^{k-1}\big(x_{i}+x_{i-1}\big)\Bigg]\big(x_{1}-x_{0}\big).

    Then x_k-x_{k-1} has one more term than x_{k+1}-x_k and is therefore bigger.

    You rewrite x_{k+1}-x_k in several different ways and then say that the sequence is non-decreasing or non-increasing. I do not get that though..

    Thanks a lot for helping me out!
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Senior Member bkarpuz's Avatar
    Joined
    Sep 2008
    From
    R
    Posts
    481
    Thanks
    2

    Lightbulb

    Quote Originally Posted by Mollier View Post
    Hi again bkarpuz,

    thanks for the quick intro to set-theory notation

    I can see how you get,

    x_{k+1}-x_k = \frac{1}{2^k}\Bigg[(x_1+x_0)(x_2+x_1)\cdots (x_k+x_{k-1})\Bigg](x_1-x_0),

    which you then write as,

    x_{k+1}-x_k=\dfrac{1}{2^{k}}\Bigg[\displaystyle\prod_{i=1}^{k}\big(x_{i}+x_{i-1}\big)\Bigg]\big(x_{1}-x_{0}\big).
    Up to now everything is okay.
    But do not forget that we do not yet know how large is the product, i.e., \geq 1 or \leq 1 .

    Quote Originally Posted by Mollier View Post
    If I am reading this right, the sequence is decreasing if,

    (x_{k+1}-x_k)<(x_k-x_{k-1}),
    A sequence is decreasing if each term is greater than preceding one (or succeeding terms are greater than the present one), i.e., x_{k}<x_{k-1} (or x_{k+1}<x_{k}) for all k.
    And conversely, increasing if x_{k+1}>x_{k}.
    Actually, the forward difference operator \Delta x_{k}:=x_{k+1}-x_{k} notation illustrates this better (analogue to differential operator).
    With this notation
    1. A sequence is increasing if \Delta x_{k}>0 for all k\in\mathbb{N}. For instance, x_{n}:=2^{n}.
    2. A sequence is decreasing if \Delta x_{k}<0 for all k\in\mathbb{N}. For instance, x_{n}:=1/n.
    3. A sequence is nondecreasing if \Delta x_{k}\geq0 for all k\in\mathbb{N}. For instance, x_{n}:=\lfloor n/2\rfloor, where \lfloor\cdot\rfloor is the least integer function.
    4. A sequence is nonincreasing if \Delta x_{k}\leq0 for all k\in\mathbb{N}.
    In each case, the sequence is monotonic, and it has a limit. (compare this with derivative).

    Quote Originally Posted by Mollier View Post
    and if I'm doing this right then,

    x_{k}-x_{k-2} = \dfrac{1}{2^{k}}\Bigg[\displaystyle\prod_{i=1}^{k-1}\big(x_{i}+x_{i-1}\big)\Bigg]\big(x_{1}-x_{0}\big).

    Then x_k-x_{k-1} has one more term than x_{k+1}-x_k and is therefore bigger.
    Also I can say that
    5. If \Delta x_{n+1}>\Delta x_{n} ( x_{n+2}-x_{n+1}>x_{n+1}-x_{n}), which is equivalent to \Delta^{2}x_{n}>0 ( \Delta^{2}:=\Delta\Delta), then the sequence is convex. For instance, x_{n}=n^{2}.
    If an increasing sequence is convex, then it tends to infinity.
    6. If \Delta x_{n+1}<\Delta x_{n}, which is equivalent to \Delta^{2}x_{n}<0), then the sequence is concave.

    Including more terms in the product on the right hand side does not mean that the sequence is decreasing or increasing.
    For instance, x_{k}=\prod_{i=1}^{k}\frac{1}{2}, which gives x_{k+1}-x_{k}=\prod_{i=1}^{k+1}\frac{1}{2}-\prod_{i=1}^{k}\frac{1}{2}=-\frac{1}{2}<0.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Fixed point iteration help
    Posted in the Advanced Math Topics Forum
    Replies: 8
    Last Post: November 8th 2011, 12:19 PM
  2. Fixed Point Iteration
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 10th 2010, 11:14 AM
  3. Fixed point iteration
    Posted in the Differential Geometry Forum
    Replies: 13
    Last Post: October 21st 2010, 04:36 AM
  4. Fixed point iteration
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: February 2nd 2010, 08:30 AM
  5. Fixed Point Iteration
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 12th 2009, 01:40 PM

Search Tags


/mathhelpforum @mathhelpforum