1
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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? $\endgroup$
    – stochasticboy321
    Commented Sep 22, 2015 at 5:29
  • $\begingroup$ 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? $\endgroup$
    – Shammy
    Commented Sep 22, 2015 at 5:30
  • $\begingroup$ 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. $\endgroup$
    – Empiricist
    Commented Sep 22, 2015 at 5:31
  • 1
    $\begingroup$ In the "I know I need to use" formula, simultaneously replace $n+1$ by $n$, and $n$ by $n-1$, and $n-1$ by $n-2$. $\endgroup$
    – André Nicolas
    Commented Sep 22, 2015 at 5:39
  • 1
    $\begingroup$ 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 $x_{n} = x_{n-1} - [ \frac{x_{n-1} - x_{n-2}}{f(x_{n-1}) - f(x_{n-2})} ] * f(x_{n-1})$ $\endgroup$
    – stochasticboy321
    Commented Sep 22, 2015 at 5:41

1 Answer 1

Reset to default
2
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .