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Taking $x_1=3$ as my initial estimate.

My work so far

The derivative of $f$, which is

$$f'(x)=\frac{d}{dx}(x^5-3x^2+1)=5x^4-6x$$

And, applying Newton's method to the table below.

\begin{array}{|c|c|c|c|} \hline x_n& f(x_n) & f'(x_n) & \frac{f(x_n)}{f'(x_n)} & x_n-\frac{f(x_n)}{f'(x_n)} \\ \hline x_1=3.000000 & 217.000000 & 387.000000 & 0.560724 & 2.439276 \\ \hline x_2=2.439276 & 69.508302 & 162.380993 & 0.428057 & 2.011220 \\ \hline x_3=2.011220 & 21.772682 & 69.742981 & 0.312185 & 1.699035 \\ \hline x_4=1.699035 & 6.498158 & 31.471553 & 0.206477 & 1.492558 \\ \hline x_5=1.492558 & 1.724044 & 15.858533 & 0.108714 & 1.383844 \\ \hline x_6=1.383844 & 0.329922 & 10.033520 & 0.032882 & 1.350962 \\ \hline x_7=1.350962 & 0.024737 & 8.549144 & 0.002894 & 1.348068 \\ \hline x_8=1.348068 & 0.000181 & 8.424276 & 0.000021 & 1.348047 \\ \hline x_9=1.348047 & 0.000000 & 8.423353 & 0.000000 & 1.348047 \\ \hline \end{array}

After $x_8$, it becomes apparent there is a convergence to about 1.348. A root for the polynomial. Is my process correct? Also, would there be a better way in computing this?

EDIT - The table has been amended. Thanks Alexey Burdin!

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    $\begingroup$ Your $x_1-\frac{f(x_1)}{f'(x_1)}$ looks off. $\endgroup$
    – Arthur
    Jul 11 '20 at 16:24
  • $\begingroup$ Thanks Arthur. I've amended my original table with Alexey Burdin's corrected table, which fixes my original error. Apologies for this! $\endgroup$
    – Laufen
    Jul 11 '20 at 16:51
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This should really be a comment. Consider this python script. Now I let myself to borrow your table structure and the header. The script produces this:

\begin{array}{|c|c|c|c|} \hline x_n& f(x_n) & f'(x_n) & \frac{f(x_n)}{f'(x_n)} & x_n-\frac{f(x_n)}{f'(x_n)} \\ \hline x_1=3.000000 & 217.000000 & 387.000000 & 0.560724 & 2.439276 \\ \hline x_2=2.439276 & 69.508302 & 162.380993 & 0.428057 & 2.011220 \\ \hline x_3=2.011220 & 21.772682 & 69.742981 & 0.312185 & 1.699035 \\ \hline x_4=1.699035 & 6.498158 & 31.471553 & 0.206477 & 1.492558 \\ \hline x_5=1.492558 & 1.724044 & 15.858533 & 0.108714 & 1.383844 \\ \hline x_6=1.383844 & 0.329922 & 10.033520 & 0.032882 & 1.350962 \\ \hline x_7=1.350962 & 0.024737 & 8.549144 & 0.002894 & 1.348068 \\ \hline x_8=1.348068 & 0.000181 & 8.424276 & 0.000021 & 1.348047 \\ \hline x_9=1.348047 & 0.000000 & 8.423353 & 0.000000 & 1.348047 \\ \hline \end{array}

There's no place for a mistake in the script, you see?)

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  • $\begingroup$ That's brilliant! Many thanks for that. I'm in the beginning stages on learning python, but this applicability helps me see how it can be used for Newton's method. $\endgroup$
    – Laufen
    Jul 11 '20 at 16:42

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