Thread: Cube root of 5

Hi Guys,

Need a little hint on this one.

Expand in ascending powers of as far as the term in , simplifying the terms as much as possible. By substituting 0.08 for in your result obtain an approximate value of the cube root of 5, upto four decimal places.



I expanded the problem as a binomial series,



The rest of the question is confusing me. How do I break up 5 to include 0.08?

The only approach I could think of was,



Then using x = -3/8, to put in the above equation.

This gives me 1.7122 which is incorrect, and also does not use the suggested 0.08. Any ideas where I am going wrong?

Thanks for your help.

Can you proceed?

Surely if you're evaluating the cube root of 5, you need to let x = 4...

Originally Posted by alexmahone

Can you proceed?

Thanks for the quick reply!

I am missing something here, Wouldn't that be ? How does that lead to ?

Originally Posted by mathguy80

Thanks for the quick reply!

I am missing something here, Wouldn't that be ? How does that lead to ?

Of course. I'm sorry, that was a typo.

Anyway,

Thanks! That's exactly what I was looking for. Didn't make the connection with the fractional indices. Fits nicely!

One more question if you don't mind. The method I choose is clearly inaccurate but using 0.08 and your suggestion gives a much more accurate answer. How would you approach this problem if 0.08 was not provided and you were asked to find the cube root? I mean the 0.08 is the key but what train of thought would get you to make this assumption?

@Prove It, I am not sure I follow your suggestion.

Originally Posted by mathguy80

Thanks! That's exactly what I was looking for. Didn't make the connection with the fractional indices. Fits nicely!

One more question if you don't mind. The method I choose is clearly inaccurate but using 0.08 and your suggestion gives a much more accurate answer. How would you approach this problem if 0.08 was not provided and you were asked to find the cube root? I mean the 0.08 is the key but what train of thought would get you to make this assumption?

Originally Posted by mathguy80

@Prove It, I am not sure I follow your suggestion.

Surely if you have a series for (1 + x)^(1/3), if you let x = 4, then you have a series for (1 + 4)^(1/3) = 5^(1/3)...

Originally Posted by Prove It

Surely if you have a series for (1 + x)^(1/3), if you let x = 4, then you have a series for (1 + 4)^(1/3) = 5^(1/3)...

But the binomial series (when the index is a fraction) converges only if |x|<1.

Originally Posted by alexmahone

But the binomial series (when the index is a fraction) converges only if |x|<1.

Point taken. I'm sure, however, that a series can be found which is centred somewhere near x = 4 so that you can evaluate it at x = 4.

Nice! This is why I love this site! Thanks @Archie Meade.

Interesting point @Prove it. The |x| < 1 for the convergent binomial series has tripped me up as well. Don't know enough Math to know if there is such a series.. Thanks for all the help today, guys, much appreciated.