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I'm really confused on how to implement the Secant Method. If the secant method is used on $f(x) = x^5 + x^3 + 3$ and if $x_{n-2} = 0$ and $x_{n-1} = 1$ what is $x_n$?

I know I need to use $x_{n+1} = x_n - \frac{x_n - x_{n-1}}{f(x_n) - f(x_{n-1})} * f(x_n)$

But I am not really sure how to even start on this problem. I Have no idea what these values are except $x_{n-1}$ I mean plugging in what I know ($x_{n-1} = 1$) I would get: $x_{n+1} = x_n - [ \frac{x_n - 1}{f(x_n) - f(1)} ] * f(x_n)$

And I know that $f(1) = (1)^5 + (1)^3 + 3 = 5$

so that would maybe change this to $x_{n+1} = x_n - [ \frac{x_n - 1}{f(x_n) - 5} ] * f(x_n)$ ? But I am still lost on what to do next.

Shammy
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  • The formula above is for a general $n$. For example, given $x_3$ and $x_4$, you can find $x_5$. Does this clear your issue? – stochasticboy321 Sep 22 '15 at 05:29
  • I understand that, but I don't understand exactly how that helps me. I have all these $x_n$'s how will i solve for them? – Shammy Sep 22 '15 at 05:30
  • Look at the indices more carefully you will find you need the previous two iterates to find the new one. Now substitute $n$ by $n-1$ in the formula of the secant method you will get what you want. – Empiricist Sep 22 '15 at 05:31
  • Are you saying I should let $n$ = $n-1$ ? – Shammy Sep 22 '15 at 05:33
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    In the "I know I need to use" formula, simultaneously replace $n+1$ by $n$, and $n$ by $n-1$, and $n-1$ by $n-2$. – André Nicolas Sep 22 '15 at 05:39
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    I think you're just getting confused by the notation. Suppose I had given you $x_{t-1}$ and $x_{t-2}$, would you be able to tell me $x_t$? The relation above is telling you how to calculate $x_{n+1}$ for any $n$. So yes, you can put the transformation $n \to n-1$ to get $x_{n} = x_{n-1} - [ \frac{x_{n-1} - x_{n-2}}{f(x_{n-1}) - f(x_{n-2})} ] * f(x_{n-1})$ – stochasticboy321 Sep 22 '15 at 05:41
  • Thank you for the help @stochasticboy321 & Andre – Shammy Sep 22 '15 at 06:05

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I believe the answer is $x_n = 1 - \frac{1}{f(1) - f(0)} * f(1) = \frac{-3}{2}$

Shammy
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