- #1

squenshl

- 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].

How do I show that le^x-(1+x/1!+x^2/2!)l <= e/6 for all x E [0,1].

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- Thread starter squenshl
- Start date

- #1

squenshl

- 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].

How do I show that le^x-(1+x/1!+x^2/2!)l <= e/6 for all x E [0,1].

- #2

Phrak

- 4,265

- 6

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

adriank

- 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

ice109

- 1,714

- 6

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

- 2,119

- 41

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.

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