Doubt regarding Taylor's theorem

  • #1
utkarshakash
Gold Member
855
13

Homework Statement


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

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

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

Answers and Replies

  • #2
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728

Homework Statement


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

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

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]##.
 
  • #3
utkarshakash
Gold Member
855
13
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?
 
  • #4
34,544
6,248
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.
 
  • #5
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
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).
 

Related Threads on Doubt regarding Taylor's theorem

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
988
  • Last Post
Replies
0
Views
846
  • Last Post
Replies
0
Views
979
  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
16
Views
547
Replies
4
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
6
Views
2K
Top