# Thread: [Binomial Series]Find 25^1/3 with precision of 3 decimals.

1. ## [Binomial Series]Find 25^1/3 with precision of 3 decimals.

Question. Find 25^1/3 with precision of 3 decimals.

My book says:

$\sqrt[3]{25}=\sqrt[3]{27}\sqrt[3]{\frac{25}{27}}.$Then we can say: $\sqrt[3]{25}=\sqrt[3]{3(1-\frac{2}{27})}$

I didnt understand the logic. I mean, how did he go from the first affirmation to the second?

2. ## Re: [Binomial Series]Find 25^1/3 with precision of 3 decimals.

Is that actually what is in your book? It is incorrect. What is true is that $\sqrt[3]{27}= 3$, of course. And $\frac{25}{27}= \frac{27}{27}- \frac{2}{27}= 1- \frac{2}{27}$. So $\sqrt[3]{27}\sqrt[3]{\frac{25}{27}}= 3\sqrt[3]{1- \frac{2}{27}}$. That is, the first "3" is not inside the cube root.

3. ## Re: [Binomial Series]Find 25^1/3 with precision of 3 decimals.

Originally Posted by frank1
Question. Find 25^1/3 with precision of 3 decimals.My book says:
$\sqrt[3]{25}=\sqrt[3]{27}\sqrt[3]{\frac{25}{27}}.$Then we can say: $\sqrt[3]{25}=\sqrt[3]{3(1-\frac{2}{27})}$

I absolutely agree with Halls in reply # 2 on this one.

Here is the standard approximation:
$f\left( {x_0 + \Delta x} \right) \simeq f'\left( {x_0 } \right)\Delta x + f\left( {x_0 } \right)$.

In this problem $f(x) = \sqrt[3]{x},\,x_0 = 25,\;\& \,\Delta x = - 2\\ &\simeq \frac{1}{3}\left( {27} \right)^{\frac{{ - 2}}{3}} \left( { - 2} \right) + 3\\ &\simeq \frac{-2}{27}+3\\ &\simeq 3-\frac{2}{27}$.

4. ## Re: [Binomial Series]Find 25^1/3 with precision of 3 decimals.

You're right HallsofIvy,

But why did you go from a cube root to 25/27?

5. ## Re: [Binomial Series]Find 25^1/3 with precision of 3 decimals.

Originally Posted by frank1
You're right HallsofIvy, But why did you go from a cube root to 25/27?

6. ## Re: [Binomial Series]Find 25^1/3 with precision of 3 decimals.

I don't agree with Plato's reply on two counts.

First with the standard approximation stated, how do you know whether or not you will get three decimal accuracy ? You might get lucky (with the size of the correction term), but really, you need at least one more term in the Taylor expansion.

Secondly I don't think that the title is misleading.

We can write $25^{1/3}=(27-2)^{1/3}=3(1-2/27)^{1/3},$ and now use the standard Binomial expansion for $(1-x)^{n}.$

7. ## Re: [Binomial Series]Find 25^1/3 with precision of 3 decimals.

Plato, I'm sorry but I didnt understand what you are doing in Reply 3.

BobP, Thanks a lot! With your answer I understand the example of the book