1. ## Lagrange Remainder

My trouble is: How should I use it to prove/show what is requested to be so?

With my kind regards,

M.V.S/Kaemper

2. ## Re: Lagrange Remainder

Let's see:
To see the error using 2nd order approximation (since you have P2), we compute f''' (x)/3! * (x-a)^3.
In your case, a=0 as it's centered at a=0, and x=1/2 since we are estimating the error of f(1/2) from P2(1/2).
So we have f'''(1/2)/6 *(1/2)^3 = f'''(1/2)/48.
Since f'''(1/2) < 1, f'''(1/2)/48 < 1/48 as required.

3. ## Re: Lagrange Remainder

How did you calculate the remainder to be 1/6? The problem states the 3rd derivative is between 0 and 1 for all real x. So, you should get a range of possible values for the remainder, shouldn't you?

It should be $\displaystyle 0 \le E_3\left(\frac{1}{2}\right) \le \dfrac{1}{3!}\left(\frac{1}{2}-0\right)^3 = \frac{1}{48}$

Then $\displaystyle f\left(\dfrac{1}{2}\right)-P_2\left(\dfrac{1}{2}\right) \le E_3\left(\dfrac{1}{2}\right)$

4. ## Re: Lagrange Remainder

He probably used x=1 by accident, which I did initially too.

5. ## Re: Lagrange Remainder

I give sincere thanks to the two of you.