I want to understand the geometrical interpretation of Convergence & Divergence of Fixed point iteration method.
The figure is here IMAGE: CLICK HERE
In Figure 2.4(a) Left panel: if we draw a tangent at the point of intersection for the curve $g(x)$, the angle, say $\theta_1$, made between tangent and x-axis is less $45^o$, so slope $<tan(45)=1$. Therefore, $g'(P)<1$
In Figure 2.5(a) Right panel: if we draw a tangent at the point of intersection for the curve $g(x)$, the angle, say $\theta_2$, made between tangent and x-axis is greater than $45^o$, so slope $>tan(45)=1$. Therefore, $g'(P)>1$
If I am correct to understand the above two, then please help me to understand the below two.
In Figure 2.4(b) Left panel: if we draw a tangent at the point of intersection for the curve $g(x)$, the angle, say $\theta_3$, made between tangent and x-axis is greater than $90^o$, so slope $\tan(\theta_3)=tan(90+\phi)=-cot \phi$, $0<\phi<90^o$. But this does not imply $ cot \phi<1$ for $0<\phi<90^o$ So $\tan(\theta_3)>-1$ is not true always. Then, how to get $-1<g'(P)<0$?
Also In Figure 2.5(b) Right panel: How to get the $g'(P)<-1$?
Edit: Angles $\theta_3$ and $\theta_4$ are shown.
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Best Answer
In Figure 2.4(b) you have $\theta_3>135°$, so that $-1<\tan\theta_3<0$. In Figure 2.5(b) you have $90°<\theta_4<135°$, so that $\tan\theta_4<-1$.