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\textbf{First name:} Monika \\
\textbf{Last name:} Wunderlich\\
\textbf{Date:} 2.1.04\\
\textbf{Homeworknumber:} 5 \\
\textbf{Homework Title:} 5.11. \\
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\textbf{Problem description:} 
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Suppose you are using the secant method to find a root $x^*$ of a nonlinear equation $f(x) =0$.
Show that if at any iteration it happens to be the case that either $x_k = x^*$ or $x_{k-1} = x^*$ (but
not both), then it will also be true that $x_{k+1} = x^*$.\\
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\textbf{Problem solution:} 
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In the secant method the next iteration is found with:
$$ x_{k+1} = x_k - f(x_k)\frac{x_k - x_{k-1}}{f(x_k) - f(x_{k-1})}.$$
Case $x_k$ has already the value of a root $x^*$: 
$$ x_{k+1} = x^* - f(x^*)\frac{x^* - x_{k-1}}{f(x^*) - f(x_{k-1})} = x^* - 0 \cdot \frac{x^* - x_{k-1}}{0 - f(x_{k-1})} = x^*.$$
Case $x_{k-1}$ has already the value of a root $x^*$: 
$$ x_{k+1} = x_k - f(x_k)\frac{x_k - x^*}{f(x_k) - 0} = x_k - x_k + x^* = x^*. $$
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\textbf{Results:} 
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We see that in both cases the iteration will not loose the root $x^*$ and $x_{k+1}= x^*$.
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