Thread: Binomial expansion question

Prove that , if x is so small that terms in x^3 and higher powers may be neglected, then .

By substituting a suitable value of x in your result, show that (11)^1/2 is approximately equal to 663/200.



I can solve the first part, but i have difficulty in the second part (show that (11)^1/2 is approximately equal to 663/200). Can anyone help me? Thanks.

If , then what is ?

Originally Posted by Prove It

If , then what is ?

By solving that, i get x= 5/6. Substituting x=5/6 into , i get (11)^1/2=157/72 which is incorrect...

Indeed, I was getting this too. I think there is something wrong with the question. The issue is surely due to the fact that 5/6 is not small enough for x^3 and higher powers to be neglected.

Either that, or I'm missing something.

"Approximately equal" is a loose term. Yes, using x= 5/6 gives while approximately. That's not very close but then, as Unknown008 said, 5/6 is not all that "small"! is larger than 1/2- hardly "negligible".

It must eirler be very late, or very early where you are HallsofIvy

Hint: If you knew the square root of 11/9, you could easily compute the square root of 11.

Originally Posted by awkward

Hint: If you knew the square root of 11/9, you could easily compute the square root of 11.

But how do i get 'square root of 11/9'? And how to compute 'square root of 11' after i get 'square root of 11/9'? Please enlighten me..

Originally Posted by MichaelLight

But how do i get 'square root of 11/9'?

The same way you found the square root of 11 previously.

And how to compute 'square root of 11' after i get 'square root of 11/9'? Please enlighten me..

Simplify .