Le^x-(1+x/1!+x^2/2!)l <= e/6 for all x E [0,1].

  • Thread starter squenshl
  • Start date
  • #1
479
4
I have a problem.
How do I show that le^x-(1+x/1!+x^2/2!)l <= e/6 for all x E [0,1].
 

Answers and Replies

  • #2
4,239
1


Show that the equation is true at the boundries. Show that the term on the left is monotonic in x (if, indeed, it is).
 
  • #3
534
1


What a strange coincidence. I saw this thread only after posting my example in this thread.

If you don't want to see the full solution, here's a hint: Taylor's theorem.
 
  • #4
1,707
5


how to show that a function is monotonic increasing? show that f(x+1) >= f(x) for all x?
 
  • #5
dx
Homework Helper
Gold Member
2,011
18


how to show that a function is monotonic increasing? show that f(x+1) >= f(x) for all x?
No, that won't work. A function is monotonic increasing if f(x) ≥ f(y) whenever x ≥ y.
 

Related Threads on Le^x-(1+x/1!+x^2/2!)l <= e/6 for all x E [0,1].

Replies
5
Views
1K
Replies
5
Views
6K
Replies
1
Views
2K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
7
Views
10K
Replies
4
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
10K
Replies
10
Views
4K
Top