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    An iteration of the form X n+1 =g(Xn) converges when the gradient of y=g(x) at the point of intersection with the line y=x satisfies the condition |g'(x)|<1, provided a suitable value for X1 is chosen.

    can someone explain this to me with an example
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    (Original post by man111111)
    An iteration of the form X n+1 =g(Xn) converges when the gradient of y=g(x) at the point of intersection with the line y=x satisfies the condition |g'(x)|<1, provided a suitable value for X1 is chosen.

    can someone explain this to me with an example
    The iteration scheme x_{n+1}=g(x_n) produces a staircase/cobweb diagram which you would've covered at A2 level.

    If we wanted to approximate solutions to the equation 0=\log(x+2)-x then we can rearrange this into x=\log(x+2) which graphically can be interpreted as the intersection point(s) between y=x and y=\log(x+2). We proceed by the iteration scheme x_{n+1} = \log(x_n+2). For reference, it looks like this:

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    Now note that g(x)=\log(x+2) with g'(x) = \frac{1}{x+2}. Testing the values of intersection, we get that g'(-1.841) \approx 6.289 and g(1.146) \approx 0.318 so we should expect x_n \rightarrow 1.146 and not towards -1.841 for a suitably chosen x_0.
    This can be clearly observed:
    Starting on x_0 &gt; 1.146 produces a staircase to the root. Starting on -1.841 &lt; x_0 &lt; 1.146 produces a staircase to the expected root as well. But starting x_0 &lt; -1.841 produces a staircase that doesn't stop (more or less). So, we never get convergence to -1.841 as expected.

    The reason and analysis for |g'(x)| &lt; 1 should be covered in your notes.
 
 
 

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