Solving 31.2^(1/5) ≈ 197/99 with Binomial Expansion

[crays]

Hi, its me again.

$$\left(1 - x\right)^{\frac{1}{5}}$$

show that $$31.2^{\frac{1}{5}} \approx \frac{197}{99}$$


how can i know what value should x be ?

 

[Fightfish]

In order for the binomial expansion to hold, the absolute value of x must be less than 1. In this case, since we are going to approximate it as a fraction, we would want to end up with nice simplifiable roots. This would point us towards 2^5, 32. From there, you can get x after manipulation.

 

[crays]

Still don't get it. so i let 1 - x = 32 ?

or i extract 32 out so 31.2/32 =0.975
and then let it be

[32(0.975)]^1/5
(2)(1-0.025)^1/5

then let x = 0.025?

 

[icystrike]

Hi, its me again.

$$\left(1 - x\right)^{\frac{1}{5}}$$

show that $$31.2^{\frac{1}{5}} \approx \frac{197}{99}$$


how can i know what value should x be ?

Try Maclaurin series (=

 

[Fightfish]

[32(0.975)]^1/5
(2)(1-0.025)^1/5
then let x = 0.025?

Yup, that's the way to do it.
Bleh, I hate Maclaurin's lol, never use it unless necessitated.

 

[Mark44]

There is a forum for calculus problems. This one is for precalculus problems.

 

[crays]

I'm sorry, actually i can't differentiate which is calculus and which is not ._. In my country its just Maths S and Maths T i don't even know what it stands for. I never heard of Maclaurin series O-o