thanks.

Results 1 to 4 of 4

- Dec 16th 2006, 10:56 PM #1

- Joined
- Dec 2006
- Posts
- 13

- Dec 17th 2006, 12:33 AM #2

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 5

Prove that between two zeros of there is a zeros of

Let:

First observe that:

,

Now if and are consecutive zeros of then the signs of and would be different (the other possibility that one or both of these is zero does not occur as that would imply that for one ,or both, of the zeros that , but such a point cannot be a zero of ).

But the sign of and being different, implies that is positive at one root and negative at the other, so as is continuous, by the intermediate value theorem must be zero at some point between and . which proves the result.

RonL

- Dec 17th 2006, 01:06 AM #3

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 5

Find where the function:

is an increasing and where it is a decreasing function.

First we restrict attention to , as is complex or undefined otherwise.

Rewrite as:

Now is differentiable on and its derivative is:

.

But the term in the square brackets is always positive, as is the term with the exponential. Therefore is increasing on .

RonL

- Dec 17th 2006, 06:19 AM #4

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 10

First you probably mean,

You can write,

Thus,

Let us look at the term inside,

Now,

And,

By using the L'Hopital rule inside and then applying the limit composite rule.

Thus,

by the extended reals.

Thus,*)

If we subtract 1 from this limit we have -1 as the answer to the limit question.

*)I am note sure about this. Because I employed the limit composite rule for infinite limits (limits that do not exists). Thus, it might not work.