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as the title of the question says I need to approximate roots using the binomial series, I am new to this subject and I have had a hard time understanding how to do it. I have been researching and I have found that the binomial series is: $(1 + x)^n = 1 + \frac{n}{1}x + \frac{n(n-1)}{1*2}x^2 + ...$ At this point is my first doubt, can I use this binomial series to approximate roots of index two, three, four and so on? For example, if I want to find an approximation for $\sqrt{67}$ what I would do is: $(1 + 66)^\frac{1}{2} = 1 + \frac{0.5}{1}66 + \frac{0.5(0.5-1)}{1*2}66^2 + ...$ However, even with the second term of the series (Result = $34$) I am far away from the real value ($8.185352772$). Could you help me and if possible give me some examples of how to approximate roots of index two, three, four, five? In a further example, if I want to find an approximation for $\sqrt[3]{264}$ I would do something like: $(1 + 263)^\frac{1}{3} = 1 + \frac{0.3333333333}{1}263 + \frac{0.3333333333(0.3333333333-1)}{1*2}263^2 + ...$

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    $\begingroup$ The series for $(1+x)^n$ only converges when $|x| < 1$. $\endgroup$
    – Toby Mak
    Aug 16, 2021 at 2:44
  • $\begingroup$ @TobyMak Thanks for the clarification, what could I do in that case? $\endgroup$
    – Sasori1264
    Aug 16, 2021 at 2:48
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    $\begingroup$ You could use that $\sqrt{67} = \sqrt{100 \times .67} = 10 \sqrt{.67} = 10\sqrt{1-.33}$ and now apply the binomial series. $\endgroup$
    – jjagmath
    Aug 16, 2021 at 3:02
  • $\begingroup$ @jjagmath For the expression $\sqrt[3]{3}$ could I use something like $\sqrt[3]{1000*0.003}$? $\endgroup$
    – Sasori1264
    Aug 16, 2021 at 14:51
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    $\begingroup$ Yes. However, be aware that this is not the best way to approximate roots. $\endgroup$
    – jjagmath
    Aug 16, 2021 at 16:07

2 Answers 2

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The series for $(1+x)^n$ only converges when $|x| < 1$.

So you need to find the nearest integer root. For $\sqrt{67}$, since $8^2 = 64$, you can write $\sqrt{67} = 8 \sqrt{67/64} = 8 \sqrt{1 + 3/64}$. Likewise, $\sqrt[3]{264} = 6 \sqrt[3]{264/216} = 6 \sqrt[3] {1+2/9}$.

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  • $\begingroup$ I understand, what would happen in the case of $\sqrt[3]{3}$? $\endgroup$
    – Sasori1264
    Aug 16, 2021 at 14:54
  • $\begingroup$ I have found something interesting, I want to calculate the following fourth root: $\sqrt[4]{1250}$, the nearest integer root is: $5^4$=$625$, So $5*\sqrt[4]{1250/625}$, $\frac{1250}{625}-1=1$ , However, the binomial series converges at $|x|<1$; How should I proceed in this case? $\endgroup$
    – Sasori1264
    Aug 16, 2021 at 22:40
  • $\begingroup$ Then you have $5 \sqrt[4] {2}$ and now take a factor of $1.2^4$ out, since $\sqrt[4] {2} = \sqrt{\sqrt{2}} \approx \sqrt{1.4} \approx 1.2$, as $1+2x \approx (1+x)^2$ when $x$ is small. $\endgroup$
    – Toby Mak
    Aug 17, 2021 at 3:32
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With respect to comments from OP we look at approximation using the binomial series expansion in order to calculate \begin{align*} \color{blue}{\sqrt[3]{3}}&\color{blue}{=1.442\,249\,570\ldots}\tag{1}\\ \color{blue}{\sqrt[4]{1\,250}}&\color{blue}{=5.946\,035\,575}\tag{2}\ldots \end{align*}

Approximation of $\sqrt[3]{3}$:

Expansion of $\sqrt[3]{1+x}$ gives \begin{align*} \left(1+x\right)^{\frac{1}{3}}&=\sum_{n=0}^\infty \binom{\frac{1}{3}}{n}x^n\qquad\qquad\qquad\qquad |x|<1\\ &=1+\frac{1}{3}x-\frac{1}{9}x^2+\frac{5}{81}x^3+\cdots\tag{3} \end{align*}

In order to calculate $\sqrt[3]{3}$ we are looking for an appropriate scaling factor $y$ and make an Ansatz \begin{align*} \sqrt[3]{1+x}=\sqrt[3]{3y^3}=y\sqrt[3]{3}\tag{4} \end{align*} We obtain from (4) \begin{align*} 1+x&=3y^3\\ x&=-1+3y^3 \end{align*} We know the expansion is valid for $|x|<1$, so we are looking for $y^3$ near $\frac{1}{3}$ which makes $x$ close to $0$. Since we want to do the approximation (at least in principle) manually we are looking for a rational value \begin{align*} y&=\frac{a}{b}\\ x&=-1+3y^3=-1+3\frac{a^3}{b^3} \end{align*} with small numerator and denominator. A table lookup of small values of $(n,n^3)$ gives \begin{align*} \begin{array}{c|rrrrrrrrrr} n&1&2&3&4&5&6&7&8&9&10\\ n^3&1&8&27&64&125&216&\color{blue}{343}&512&729&\color{blue}{1\,000} \end{array} \end{align*} We observe \begin{align*} \frac{343}{1\,000}=0.343\approx \frac{1}{3} \end{align*} is close to $\frac{1}{3}$ and we choose \begin{align*} \color{blue}{y=\frac{7}{10}} \end{align*} as scaling factor.

We obtain from (3) and (4) with $y=\frac{7}{10}$ and $x=-1+3y^3=-1+3\frac{7^3}{10^3}=0.029$: \begin{align*} \frac{1}{y}\sqrt[3]{1+x}&\approx\frac{10}{7}\left(1+\frac{1}{3}\cdot0.029 -\frac{1}{9}\cdot 0.029^2+\frac{5}{81}\cdot 0.029^3\right)\\ &\,\,\color{blue}{=1.442\,249}61\ldots \end{align*} which is according to (1) accurate to $6$ decimal places. Note that lineare approximation \begin{align*} \frac{1}{y}\sqrt[3]{1+x}&\approx\frac{10}{7}\left(1+\frac{1}{3}\cdot0.029\right)\\ &\,\,\color{blue}{=1.442}\,38\ldots \end{align*} is already accurate to $3$ decimal places.

Approximation of $\sqrt[4]{1\,250}$:

We can go on in the same way as before. Expansion of $\sqrt[4]{1+x}$ gives \begin{align*} \left(1+x\right)^{\frac{1}{4}}&=\sum_{n=0}^\infty \binom{\frac{1}{4}}{n}x^n\qquad\qquad\qquad\qquad |x|<1\\ &=1+\frac{1}{4}x-\frac{3}{32}x^2+\frac{7}{128}x^3+\cdots\tag{5} \end{align*}

In order to calculate $\sqrt[4]{1\,250}$ we are looking for an appropriate scaling factor $y$ and make an Ansatz \begin{align*} \sqrt[4]{1+x}=\sqrt[4]{4y^4}=y\sqrt[4]{4}\tag{6} \end{align*} We obtain from (6) \begin{align*} 1+x&=1\,250y^4\\ x&=-1+1\,250y^4 \end{align*} We know the expansion is valid for $|x|<1$, so we are looking for $y^4$ near $\frac{1}{1\,250}$ which makes $x$ close to $0$. We are looking for a rational value \begin{align*} y&=\frac{a}{b}\\ x&=-1+1\,250y^4=-1+1\,250\frac{a^4}{b^4} \end{align*}

with small numerator and denominator. A table lookup of small values of $(n,n^4)$ gives \begin{align*} \begin{array}{c|rrrrrrrrrr} n&1&2&3&4&5&6&7&8&9&10\\ n^4&\color{blue}{1}&16&81&256&625&\color{blue}{1\,296}&2\,401&4\,096&6\,561&10\,000 \end{array} \end{align*} We observe \begin{align*} \frac{1}{1\,296}=0.343\approx \frac{1}{1\,250} \end{align*} is close to $\frac{1}{1\,250}$ and we choose \begin{align*} \color{blue}{y=\frac{1}{6}} \end{align*} as scaling factor.

We obtain from (3) and (4) with $y=\frac{1}{6}$ and $x=-1+1\,250y^4=-1+1\,250\frac{1^4}{6^4}=-\frac{23}{648}$: \begin{align*} \frac{1}{y}\sqrt[4]{1+x}&\approx 6\left(1-\frac{1}{4}\cdot \frac{23}{648} -\frac{3}{32}\cdot \left(\frac{23}{648}\right)^2+\frac{7}{128}\cdot \left(\frac{23}{648}\right)^3\right)\\ &\,\,\color{blue}{=5.946\,065}\,28\ldots \end{align*} which is according to (2) accurate to $6$ decimal places. Note that linear approximation \begin{align*} \frac{1}{y}\sqrt[4]{1+x}&\approx 6\left(1-\frac{1}{4}\cdot \frac{23}{648}\right)\\ &\,\,\color{blue}{=5.946}\,75\ldots \end{align*} is already accurate to $3$ decimal places.

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