binomial series

[thereddevils]

1. The problem statement, all variables and given/known data

Expand (1-3x)^{\frac{1}{3}} in ascending power of x , up to the term x^3 . By using an appropriate substitution for x , show that \sqrt[3]3=\frac{33809}{19683}



2. Relevant equations



3. The attempt at a solution

the expansion would be 1-x-x^2-(5/3)x^3 and the range of x whixh make this expansion valid is |-3x|<1/3 .

My question is how do i find this appropriate substitution for x ? Instead the guessing way , is there a proper way ?

[HallsofIvy]

Well, since you have (1- 3x)^{1/3} and want (3)^{1/3} you probably want to make 1- 3x= 3. Solve that for x.

[thereddevils]

Well, since you have (1- 3x)^{1/3} and want (3)^{1/3} you probably want to make 1- 3x= 3. Solve that for x.

i don think so because -1/3<x<1/3 for this expansion to be valid .

[songoku]

Hi thereddevils

Maybe this method can be applied. Since we want to find the cube root and the range for x is -1/3<x<1/3 , we can deduce that (1-3x) should be fraction.

So, the denominator should be a number that is integer and has a cube root, which is also integer , such as 8, 27, etc

Now we can try (1-3x) = 3/8 or (1-3x) = 3/27 , etc..

But cube root of 3 is not 33809 / 19683.
33809 / 19683 is closer to square root of 3. Maybe you can re-check the question :smile: