Confused about taylor approximation

LostInSpace
Messages
21
Reaction score
0
I am a bit confused about taylor approximation. Taylor around [tex]x_0[/tex] yields
[tex] f(x) = f(x_0) + f'(x_0)(x-x_0) + O(x^2)[/tex]

which is the tangent of f in [tex]x_0[/tex], where
[tex] f'(x) = f'(x_0) + f''(x_0)(x-x_0) + O(x^2)[/tex]

which adds up to
[tex] f(x) &=& f(x_0) + (f'(x_0) + f''(x_0)(x-x_0) + O(x^2))(x-x_0)+O(x^2) \\ &=& f(x_0) + f'(x_0)(x-x_0) + f''(x_0)(x-x_0)^2 + O(x^3)[/tex]
But it should be
[tex] f(x) = f(x_0) + f'(x_0)(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2 + O(x^3)[/tex]

Where does the 2! come from? Is this approach completely incorrect?
 
on Phys.org
You have ignored in line 3 the O(x^(2))-term from the expansion of f(x).
Hence line 3 is not accurate to O(x^(3)), it's only accurate to O(x^(2)).
 
Do you remember how to derive, from the limit definition of the derivative, the differential approximation formula:

[tex] f(x+\epsilon ) = f(x) + \epsilon f'(x) + \epsilon \delta(x, \epsilon)[/tex]

Where [itex]\lim_{\epsilon \rightarrow 0} \delta(x, \epsilon) = 0[/itex]?

Try writing the second derivative with limits, and see if any approach suggests itself.
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
6K