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I keep getting the wrong answer for problems involving approximating roots, mostly due to some kind of algebraic oversight I am making. Essentially, given a problem where I need to approximate a value for $4^{3/4}$ using one step of the secant method where $x_0 = 3, x_1 = 2$, I try:

$x - 4^{3/4} = 0$

$f(x) = x- 4^{3/4}$

Thus, use secant method, so

$x_2 = 2 - \frac{2-3}{(2-4^{3/4})-(3-4^{3/4})} \times (2-4^{3/4})$

This is wrong however, because $f(x)$ is actually:

$f(x) = x^4-4^3$

And you get a slightly different answer.

I can't seem to track down the error in my assumption though; maybe I'm just tired, but I'd greatly appreciate any help with this.

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Using the secant method with $f(x) = x-4^{3/4}$ does not work, for you cannot compute any value of $f$ without knowing $4^{3/4}$ a priori. You have to find a function which known coefficients. Note also that you "approximation" has $$ x_2 = 2 - \frac{2-3}{(2-4^{3/4}) - (3 - 3^{3/4})} \cdot (2 - 4^{3/4}) = 2 - (2 - 4^{3/4}) = 4^{3/4} $$ and hence is not of any help.

Using the suggested function $g(x) = x^4 - 4^3$ (note that - in constrast to the above - we can write $g$ without formal powers, $g(x) = x^4 - 64$), we have $$ x_2 = 2 - \frac{2-3}{(16 - 64) - (81 - 64)} \cdot (16 - 64) = 2 - \frac{-1}{-65} \cdot (-48) = \frac{178}{65} $$

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  • $\begingroup$ So, it's simple a case of modifying the function such that you avoid the $4^{3/4}$ term? It makes sense, I didn't think of the a priori part. $\endgroup$
    – A4Treok
    Commented Oct 14, 2015 at 10:06
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    $\begingroup$ Right, the idea of the secant method is to approximate something we cannot compute in terms of something we can ... $\endgroup$
    – martini
    Commented Oct 14, 2015 at 10:07
  • $\begingroup$ Alright, thank you so much for this. I had a fundamental misunderstanding of how I should be applying the secant method to these problems. Thanks again! $\endgroup$
    – A4Treok
    Commented Oct 14, 2015 at 10:09

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