# Lagrange & Taylor

#### 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.

#### HallsofIvy

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

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

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

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: $$\displaystyle \frac{1*2*3*4*5*6*7...}{2*4*6...}= \frac{n!}{2*4*6...}$$. And we can write that denominator as $$\displaystyle (2*1)*(2*2)(2*3)...= 2^n(1*2*3...)= 2^k k!$$.

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

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

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

WartonMorton

#### WartonMorton

You have the formula- just plug in the data for this problem.

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

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

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

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

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: $$\displaystyle \frac{1*2*3*4*5*6*7...}{2*4*6...}= \frac{n!}{2*4*6...}$$. And we can write that denominator as $$\displaystyle (2*1)*(2*2)(2*3)...= 2^n(1*2*3...)= 2^k k!$$.

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

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

Using that, find a maximum on $$\displaystyle f^{n+1}$$ 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?