Doubt regarding Taylor's theorem

[utkarshakash]

Homework Statement

In Taylor's series, the Lagrange form of remainder is given by
R_n = \dfrac{f^{n+1}(t)(x-a)^{n+1}}{(n+1)!} \\ t \in [a,x]

whereas my book states that it is given by
\dfrac{h^n}{n!} f^n (a+ \theta h) \\ \theta(0,1)

I don't see how these two are interrelated. Can anyone explain ?

 

[Ray Vickson]

Homework Statement

In Taylor's series, the Lagrange form of remainder is given by
R_n = \dfrac{f^{n+1}(t)(x-a)^{n+1}}{(n+1)!} \\ t \in [a,x]

whereas my book states that it is given by
\dfrac{h^n}{n!} f^n (a+ \theta h) \\ \theta(0,1)

I don't see how these two are interrelated. Can anyone explain ?

How is $h$ defined in the expression from your book? Obviously, if $h = x-a$ then your book's expression is the same as the Lagrange version, because if $\theta \in [0,1]$ then $a + \theta h$ is a number in $[a,x]$.

 

[utkarshakash]

How is $h$ defined in the expression from your book? Obviously, if $h = x-a$ then your book's expression is the same as the Lagrange version, because if $\theta \in [0,1]$ then $a + \theta h$ is a number in $[a,x]$.

But the book's version has nth derivative. Shouldn't it be n+1?

 

[Mark44]

It would be helpful to include some of the accompanying text in your book. It's possible that your text is giving the remainder after the terms of degree n - 1, while the more usual form you also show gives the remainder after terms of degree n.

 

[Ray Vickson]

But the book's version has nth derivative. Shouldn't it be n+1?

I agree with Mark44. One form gives the remainder after an (n+1)-term polynomial (where the powers of (x-a) go from 0 to n) while the other (perhaps) gives the remainder after an n-term polynomial (where the powers go from 0 to n-1).