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2022-09-21

How do you find the first three iterate of the function $f\left(x\right)=4x-3$ for the given initial value ${x}_{0}=2$?

Malcolm Flores

Beginner2022-09-22Added 8 answers

The Newton's method is $x}_{n+1}={x}_{n}-\frac{f\left({x}_{n}\right)}{f\prime \left({x}_{n}\right)$

Here $f\left(x\right)=4x-3$ and $f\prime \left(x\right)=4$

And ${x}_{0}=2$

$\therefore {x}_{1}={x}_{0}-\frac{f\left({x}_{0}\right)}{f\prime \left({x}_{0}\right)}=2-\frac{5}{4}=\frac{3}{4}$

$x}_{2}={x}_{1}-\frac{f\left({x}_{1}\right)}{f\prime \left({x}_{1}\right)}=\frac{3}{4}-0=\frac{3}{4$

${x}_{3}={x}_{2}-\frac{f\left({x}_{2}\right)}{f\prime \left({x}_{2}\right)}=\frac{3}{4}-0$

We obtain the same valuebecause f(x) is a linear function and the intercept with the x-axis is $(\frac{3}{4},0)$

Here $f\left(x\right)=4x-3$ and $f\prime \left(x\right)=4$

And ${x}_{0}=2$

$\therefore {x}_{1}={x}_{0}-\frac{f\left({x}_{0}\right)}{f\prime \left({x}_{0}\right)}=2-\frac{5}{4}=\frac{3}{4}$

$x}_{2}={x}_{1}-\frac{f\left({x}_{1}\right)}{f\prime \left({x}_{1}\right)}=\frac{3}{4}-0=\frac{3}{4$

${x}_{3}={x}_{2}-\frac{f\left({x}_{2}\right)}{f\prime \left({x}_{2}\right)}=\frac{3}{4}-0$

We obtain the same valuebecause f(x) is a linear function and the intercept with the x-axis is $(\frac{3}{4},0)$

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