Finding rational approximations using quadratic Taylor Polyn

[needhelpwithcalculus]

When using Taylor Polynomials, we are given f(x), n, and a.

I'm asked to solve a rational approximation of 26^1/3 (cubed root of 26). I was told to set a = 27, and n = 2. I understand we set a = 27 because it's the closet thing to 26 to take the cubed root of, but why does n = 2?

thanks in advance!

 

[tkhunny]

Re: Finding rational approximations using quadratic Taylor P

What powers of 'x' are in the general #2 term in a Taylor Polynomial?

If you want quadratic, pick n = 2. If you want quartic, pick n = 4.

 

[DrMike]

Re: Finding rational approximations using quadratic Taylor P

Ok, it's not enough to say "(26)^(1/3) is about (27)^(1/3), so I'm done". Hint : this is a Taylor series question.

So, say instead : (26)^(1/3) = (27-1)^(1/3) = 27^(1/3) . (1 - 1/27)^(1/3) = 3 . (1-1/27)^1/3.

Then, find the taylor series for (1-x)^(1/3) (how many terms? Hint : you are given a mysterious number 'n'. What might that be for??) and substitute 1/27 for x.

Let me know how you go.