| In this problem, you will prove the rate of convergence for the secant method. x_{k+1}=x_{k}-\frac{x_{k}-x_{k-1}}{f\left(x_{k}\right)-f\left(x_{k-1}\right)} f\left(x_{k}\right) can be rewritten in the form: x_{k+1}=\frac{x_{k} f\left(x_{k-1}\right)-x_{k-1} f\left(x_{k}\right)}{f\left(x_{k-1}\right)-f\left(x_{k}\right)} \psi\left(x_{k}, x_{k-1}\right)=\frac{x_{k+1}-\xi}{\left(x_{k}-\xi\right)\left(x_{k-1}-\xi\right)} where xk+1

is as in (1). Compute (for fixed value of xk-1) \varphi\left(x_{k-1}\right)=\lim _{x_{k} \rightarrow \xi} \psi\left(x_{k}, x_{k-1}\right) (c) Now compute \lim _{x_{E-1} \rightarrow \xi} \varphi\left(x_{k-1}\right) and therefore show thatlim \lim _{x_{k}, x_{k-1} \rightarrow \xi} \psi\left(x_{k}, x_{k-1}\right)=\frac{f^{\prime \prime}(\xi)}{2 f^{\prime}(\xi)} (d) Next, assume that the secant method has convergence order q, that is to say that \lim _{k \rightarrow \infty} \frac{\left|x_{k+1}-\xi\right|}{\left|x_{k}-\xi\right|^{q}}=A<\infty Using the above results, show that q – 1– 1/q = 0, and therefore that q = (1+ v5)/2. (e) Finally, show that this implies that \lim _{k \rightarrow \infty} \frac{\left|x_{k+1}-\xi\right|}{\left|x_{k}-\xi\right|^{q}}=\left(\frac{f^{\prime \prime}(\xi)}{2 f^{\prime}(\xi)}\right)^{q /(1+q)}