Binomial series - Finding square root of number problem

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SUMMARY

The discussion focuses on expanding the expression (1+x)^(1/3) using the binomial series to approximate the cube root of 5. The correct substitution for x is 4, leading to the expression 5 = 4 + 1, which allows for the calculation of the cube root. The expansion yields the series 1 + (1/3)x - (1/9)x^2 + (5/81)x^3, and substituting x = 4 provides a more accurate approximation. Misunderstandings regarding the terminology, specifically confusing "square root" with "cube root," are also addressed.

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Homework Statement



Expand ##(1+x)^(1/3)## in ascending powers of x as far as the term ##x^3##, simplifying the terms as much as possible. By substituting 0.08 for x in your result, obtain an approximate value of the cube root of 5, giving your answer to four places of decimals.


Homework Equations



Binomial series.

The Attempt at a Solution


Expansion as per binomial series :
##1+(1/3)x - (1/9)x^2 + (5/81)x^3##.

No idea how to find cube root of 5 by using x = 0.08.
Help please!
Sorry if I am asking too many questions - I don't have a tutor for this, I'm on my own

Thanks! :)
 
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Are you sure you have read the problem correctly? Putting x= 0.08 into that formula will give you (0.08+ 1)^{1/3}= 1.08^{1/3}, not \sqrt[3]{5}. 5= 4+ 1 so you should set x= 4 to use that formula to find \sqrt[3]{5}.
 
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HallsofIvy said:
Are you sure you have read the problem correctly? Putting x= 0.08 into that formula will give you (0.08+ 1)^{1/3}= 1.08^{1/3}, not \sqrt[3]{5}. 5= 4+ 1 so you should set x= 4 to use that formula to find \sqrt[3]{5}.

That would approximate ##\sqrt[3]{5}## by a few terms of a divergent series, which does not look like a good idea. However, using ##2^3 = 8## we can write
5 = 8 - 3 = 8 \left(1 - \frac{3}{8}\right) \Longrightarrow 5^{1/3} = 2 (1 - 0.375)^{1/3}

BTW: please try not to confuse everybody by using a bad thread title. Don't say "square root" when you mean "cube root".
 
Last edited:

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