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convergence

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
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 xn+1=g(xn)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)−x0=\log(x+2)-x then we can rearrange this into x=log⁡(x+2)x=\log(x+2) which graphically can be interpreted as the intersection point(s) between y=xy=x and y=log⁡(x+2)y=\log(x+2). We proceed by the iteration scheme xn+1=log⁡(xn+2)x_{n+1} = \log(x_n+2). For reference, it looks like this:

Spoiler



Now note that g(x)=log⁡(x+2)g(x)=\log(x+2) with g′(x)=1x+2g'(x) = \frac{1}{x+2}. Testing the values of intersection, we get that g′(−1.841)≈6.289g'(-1.841) \approx 6.289 and g(1.146)≈0.318g(1.146) \approx 0.318 so we should expect xn→1.146x_n \rightarrow 1.146 and not towards −1.841-1.841 for a suitably chosen x0x_0.
This can be clearly observed:
Starting on x0>1.146x_0 > 1.146 produces a staircase to the root. Starting on −1.841<x0<1.146-1.841 < x_0 < 1.146 produces a staircase to the expected root as well. But starting x0<−1.841x_0 < -1.841 produces a staircase that doesn't stop (more or less). So, we never get convergence to −1.841-1.841 as expected.

The reason and analysis for ∣g′(x)∣<1|g'(x)| < 1 should be covered in your notes.

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