Lagrange & Taylor

Jan 2010
72
2
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
Apr 2005
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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.
 
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Jan 2010
72
2
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?