1. ## Lagrange & Taylor

Find the Lagrange form of the remainder Rn(x).

f(x) = sqrt(x+1) ; n=3

So I know the remainder in taylor's theorem can be written as...

Rn(x) = [f^(n+1) (c)] / [(n+1)!] *[x^(n+1)]

not sure to actually do the whole application for these kinds of problems.

2. Originally Posted by WartonMorton
Find the Lagrange form of the remainder Rn(x).

f(x) = sqrt(x+1) ; n=3

So I know the remainder in taylor's theorem can be written as...

Rn(x) = [f^(n+1) (c)] / [(n+1)!] *[x^(n+1)]

not sure to actually do the whole application for these kinds of problems.
You have the formula- just plug in the data for this problem.

The only "hard" part is finding the general derivative.

If $f(x)= \sqrt{x+ 1}= (x+ 1)^{1/2}$, then $f'(x)= (1/2)(x+ 1)^{-1/2}$, $f''(x)= -1/4(x+ 1)^{-3/2}$, $f'''(x)= 3/8(x+ 1)^{-5/2}$, $f^{(4)}= -15/16(x+1)^{-7/2}$...

Do you see the pattern? Obviously, there is an alternating + and -: $(-1)^{n+1}$ since this is + for n odd, - for n even.

The denominator of the fraction is a power of 2: $2^n$.

The numerator is a little harder- it is a product of odd integers from 2n-1 down: 1*3*5*7... We can write that more compactly by filling in the missing evens: $\frac{1*2*3*4*5*6*7...}{2*4*6...}= \frac{n!}{2*4*6...}$. And we can write that denominator as $(2*1)*(2*2)(2*3)...= 2^n(1*2*3...)= 2^k k!$.

The fraction is $\frac{(2n-1)!}{(2^n)(2^{n-2}(n-2)!}$ and the exponent on x+ 1 is $\frac{-(2n-1)}{2}$.

That is, $f^(n)(x)= \frac{(-1)^{n+1}(2n-1)!}{2^{2n-2}(n-2)!}(x+ 1)^{-\frac{2n-1}{2}}$.

Using that, find a maximum on $f^{n+1}$ between 0 and x.

3. Originally Posted by HallsofIvy
You have the formula- just plug in the data for this problem.

The only "hard" part is finding the general derivative.

If $f(x)= \sqrt{x+ 1}= (x+ 1)^{1/2}$, then $f'(x)= (1/2)(x+ 1)^{-1/2}$, $f''(x)= -1/4(x+ 1)^{-3/2}$, $f'''(x)= 3/8(x+ 1)^{-5/2}$, $f^{(4)}= -15/16(x+1)^{-7/2}$...

Do you see the pattern? Obviously, there is an alternating + and -: $(-1)^{n+1}$ since this is + for n odd, - for n even.

The denominator of the fraction is a power of 2: $2^n$.

The numerator is a little harder- it is a product of odd integers from 2n-1 down: 1*3*5*7... We can write that more compactly by filling in the missing evens: $\frac{1*2*3*4*5*6*7...}{2*4*6...}= \frac{n!}{2*4*6...}$. And we can write that denominator as $(2*1)*(2*2)(2*3)...= 2^n(1*2*3...)= 2^k k!$.

The fraction is $\frac{(2n-1)!}{(2^n)(2^{n-2}(n-2)!}$ and the exponent on x+ 1 is $\frac{-(2n-1)}{2}$.

That is, $f^(n)(x)= \frac{(-1)^{n+1}(2n-1)!}{2^{2n-2}(n-2)!}(x+ 1)^{-\frac{2n-1}{2}}$.

Using that, find a maximum on $f^{n+1}$ between 0 and x.
I know I have the forumula, but where am I getting the c from? Do I have to solve some sort of equation to get the max?