Can I do this?

All_Is_Number · 2006-08-01 05:55:55

Here is my proposal:

Example:

If this is not a valid method, why not? I know I have factored out a function of x from an integral, but I have done so under very restrictive conditions so that it will not result in an incorrect answer.

Last edited by All_Is_Number (2006-08-01 10:43:18)

Ricky · 2006-08-01 07:03:13

In general, if you have:

integral of f(x)g(x)

You can not do:

f(x) times integral of g(x).

A simple example:

Let f(x) = x and g(x) = x.

All_Is_Number · 2006-08-01 07:43:37

Ricky wrote:

In general, if you have:
integral of f(x)g(x)
You can not do:
f(x) times integral of g(x).
A simple example:
Let f(x) = x and g(x) = x.

Right. I understand that generally f(x) cannot be factored out of the integral. But what if f(x) has a derivative of zero everywhere it is defined? Can't we then treat it as a constant if it is defined everywhere g(x) is defined?

From my understanding of Calculus, the reason we cannot normally factor out f(x) is because f(x) generally does not have a derivative of zero, and it is this non zero derivative which causes problems.

Ricky · 2006-08-01 08:09:27

You can take a constant out of such integrals because constants don't change as x changes. The same is not true for cos(x)/|cos(x)|.

Ricky · 2006-08-01 08:13:41

Wait a minute. How can you factor out cos(x)/|cos(x)|? That isn't factoring our anything, that's completely changing the function. When tan(x) is negative, there will be times when cos(x)/|cos(x)| is positive. Thus, if such where the case:

tan(x) * cos(x)/|cos(x)| = negative * postive = negative.

But if we are to assume tan(x)*cos(x)/|cos(x)| = |tan(x)|, then it would be that negative = positive.

All_Is_Number · 2006-08-01 08:21:19

Ricky wrote:

You can take a constant out of such integrals because constants don't change as x changes. The same is not true for cos(x)/|cos(x)|.

But, at any defined point, cos(x)/|cos(x)| has a derivative of zero. Why can't it be treated as a constant? It is defined everywhere tan(x) is, so it shouldn't cause problems.

I've used this method to solve differential equations, and ended up with the correct answer, but my answer was correct on all intervals, whereas the book's and instructor's answers were defined only on certain intervals.

I'm not trying to be difficult or argumentative. I'm trying to understand why I can or cannot do something that will always work, given the proper (recognizable) conditions.

Ricky · 2006-08-01 10:05:28

I tried for about a half hour (the amount of time it takes to drive home from work) to find a counter example. I'm glad I stopped there:

Good thinking, All, you were completely right. The proof also holds for points of discontinuity, you just have to adjust some words here and there.

Ricky · 2006-08-01 10:10:25

Regarding your (new) first post:

tan(-3)*cos(-3) / |cos(-3)| ~ -.1425, and thus, the equality can't hold since |tan(x)| must be positive.

All_Is_Number · 2006-08-01 10:52:23

Ricky wrote:

Wait a minute. How can you factor out cos(x)/|cos(x)|? That isn't factoring our anything, that's completely changing the function. When tan(x) is negative, there will be times when cos(x)/|cos(x)| is positive. Thus, if such where the case:
tan(x) * cos(x)/|cos(x)| = negative * postive = negative.
But if we are to assume tan(x)*cos(x)/|cos(x)| = |tan(x)|, then it would be that negative = positive.

You're right. In the third quadrant, tan(x) is positive, yet cos(x) is negative. I chose a bad example. This situation came up several months ago, and I do not have the exact problem handy, so I made one up. Unfortunately, I did not make up a very applicable one.

I have edited my original post to correct that issue, but it still is not an example that shows the principle as perfectly as the one I had months ago.

Last edited by All_Is_Number (2006-08-01 11:55:04)

All_Is_Number · 2006-08-01 10:57:04

Ricky wrote:

Regarding your (new) first post:
tan(-3)*cos(-3) / |cos(-3)| ~ -.1425, and thus, the equality can't hold since |tan(x)| must be positive.

I've edited the first post since then, since I can't, for the life of me, seem to recall the exact problem where I first noticed this.

All_Is_Number · 2006-08-01 11:58:36

Ricky wrote:

I tried for about a half hour (the amount of time it takes to drive home from work) to find a counter example. I'm glad I stopped there:

Good thinking, All, you were completely right. The proof also holds for points of discontinuity, you just have to adjust some words here and there.

Thanks for the proof, Ricky. You were much clearer and more concise than I was.

Ricky · 2006-08-01 13:10:50

Just as a future recommendation, make a new post stating the mistake and correcting it, don't go back and edit it. It makes it much easier for others reading the topic.

All_Is_Number · 2006-08-02 03:16:10

Ricky wrote:

Just as a future recommendation, make a new post stating the mistake and correcting it, don't go back and edit it. It makes it much easier for others reading the topic.

You're right. My mistake.

Math Is Fun Forum

#1 2006-08-01 05:55:55

Can I do this?

#2 2006-08-01 07:03:13

Re: Can I do this?

#3 2006-08-01 07:43:37

Re: Can I do this?

#4 2006-08-01 08:09:27

Re: Can I do this?

#5 2006-08-01 08:13:41

Re: Can I do this?

#6 2006-08-01 08:21:19

Re: Can I do this?

#7 2006-08-01 10:05:28

Re: Can I do this?

#8 2006-08-01 10:10:25

Re: Can I do this?

#9 2006-08-01 10:52:23

Re: Can I do this?

#10 2006-08-01 10:57:04

Re: Can I do this?

#11 2006-08-01 11:58:36

Re: Can I do this?

#12 2006-08-01 13:10:50

Re: Can I do this?

#13 2006-08-02 03:16:10

Re: Can I do this?

Board footer