Finding Taylor Polynomial for f(x) = (1+x)^{2/3}

[3.14159265358979]

1. Let f(x) = (1+x)^{2/3}

(a) find the taylor polynomial T_2(x) of f expanded about a = 0.

i got 1 + (1/3)x - (1/9)x^{2}



For the rest, i have no idea how to do...any help would be greatly appreciated.

(b) For the givven f write the lagrange remainder formula for the error term f(x) - T_2(x).

(c) Show that when x>0 the error f(x)-T_2(x) is at most (5/18)x^{3}. and

(d) Write a fraction that estimates (1.2)^{1/3}, and show that the error in your estimate is at most 1/2025.

Thanks for any help you can provide!

 

[shmoe]

What did you get for f'(x) and f''(x)? Your polynomial is a bit off.

Do you know what the Lagrange remainder is? If not look it up and post again with what about it is troubling you.

 

[3.14159265358979]

sorry, T_2(x) is suppose to be 1 + (2/3)x - (1/9)x^{2}

i believe the lagrange remainder is: f^{n+1}(p) * (x-a)^{n+1}
-----------
(n+1)!

where p is between (a,n).
The problem is, i don't know how to find the last term in the sequence or the error (which i believe is also called the lagrange remainder, right?) thanks for your reply!

 

[3.14159265358979]

sorry, the lagrange remainder is suppose to be

f^{n+1}(p) * (x-a)^{n+1}
-----------
(n+1)!

 

[shmoe]

The remainder term you have gives f(x)-T_n(x), the difference between f and it's taylor polynomial of degree n, also known as the error between the Taylor polynomial and the function it's trying to approximate. In your case n=2 and a=0, so you just put these values into the remainder formula you have and work out the 2+1=3rd derivative of your function.

The p point comes from the mean-value theorem and you don't actually know what it is, just the interval it's on. You said it's on (a,n), did you mean (a,x), for x>a? You can use this to give an upper bound for the error when x>0 by noticing that the third derivative of this function is bounded here.