Find the Approximate Value of this Cube Root

[littlemathquark]

Homework Statement

Find approximate value of $\sqrt[3] 63$

Relevant Equations

Find approximate value of $\sqrt[3] 63$

I know a lot of solution of that problem; for example Newton-Ralphson or Taylor expansion.But I saw a solution but I don't understand its theory. The solution like this: $\sqrt[3]{63}=\sqrt[3]{64-1}\approx 4-\dfrac{4}{3.4^3-1}\approx 3.97905759162...$

 

[fresh_42]

What is $3.4^3$? Bernoulli is $(1+x)^r \approx 1+rx$ which yields in this case
$$
\sqrt[3]{1-\dfrac{1}{4^3}}\approx 1- \dfrac{1}{3\cdot 4^3}
$$

 

[littlemathquark]

What is $3.4^3$? Bernoulli is $(1+x)^r \approx 1+rx$ which yields in this case
$$
\sqrt[3]{1-\dfrac{1}{4^3}}\approx 1- \dfrac{1}{3\cdot 4^3}
$$

I don't know. According to your solution $63^{1/3}\approx 4-1/48\approx 3.979166$

 

[fresh_42]

I don't know. According to your solution $63^{1/3}\approx 4-1/48\approx 3.979166$

Forget the factor $4$ on both sides. Bernoulli's inequality says for numbers greater than $-1$
\begin{align*}
0,99476430197409796491613199693719&\approx \dfrac{1}{4}\sqrt[3]{63}=\left(1-\dfrac{1}{64}\right)^{\frac{1}{3}}\leq 1- \dfrac{1}{3}\cdot \dfrac{1}{64}\\&\approx 0,99479166666666666666666666666667
\end{align*}
with an error of about $2,73647 \cdot 10^{-5}.$ If we use $\dfrac{1}{3\cdot 4^3 -1}=\dfrac{1}{191}$ instead, then we get $0,99476439790575916230366492146597$ which varies by $10^{-7}.$ It is slightly better but not significantly. So I don't know where the irrelevant $-1$ in the denominator comes from, but the rest is Bernoulli's inequality.

 

[littlemathquark]

Forget the factor $4$ on both sides. Bernoulli's inequality says for numbers greater than $-1$
\begin{align*}
0,99476430197409796491613199693719&\approx \dfrac{1}{4}\sqrt[3]{63}=\left(1-\dfrac{1}{64}\right)^{\frac{1}{3}}\leq 1- \dfrac{1}{3}\cdot \dfrac{1}{64}\\&\approx 0,99479166666666666666666666666667
\end{align*}
with an error of about $2,73647 \cdot 10^{-5}.$ If we use $\dfrac{1}{3\cdot 4^3 -1}=\dfrac{1}{191}$ instead, then we get $0,99476439790575916230366492146597$ which varies by $10^{-7}.$ It is slightly better but not significantly. So I don't know where the irrelevant $-1$ in the denominator comes from, but the rest is Bernoulli's inequality.

Can be used Halley Method?

 

[fresh_42]

Can be used Halley Method?

No.

It is Maclaurin of $(1-x)^r$ up to the second derivative with an error of $\dfrac{1}{9\cdot 4^6(3\cdot 4^3-1)}+O(4^{-27})=O(4^{-11})=O(10^{-7}).$

 

[Mark44]

I don't know. According to your solution $63^{1/3}\approx 4-1/48\approx 3.979166$

The step that was omitted is this:

$63^{1/3} = (64 - 1)^{1/3} = 64^{1/3}(1 - \frac 1{4^3})^{1/3} = 4(1 - \frac 1{4^3})^{1/3}$
In the final expression above he's working only with the $(1 - \frac 1{4^3})^{1/3}$ part.