# Sequences of Functions in terms of x

1. Jul 26, 2011

### Lute

1. The problem statement, all variables and given/known data
Determine the values of x for which the function, for n>=1, is increasing, decreasing, bounded below or bounded above. The function is (x^n)/n

2. Relevant equations

3. The attempt at a solution
I thought about taking the derivative of the function, and setting it to 0. To find which values of x is increasing or decreasing.

Edit: This is first forum post, so I apologize if I am forgetting something.

Last edited: Jul 26, 2011
2. Jul 26, 2011

### gb7nash

Sounds good to me.

3. Jul 26, 2011

### Lute

I get x^(n-1), is that right?

4. Jul 26, 2011

### gb7nash

Yes. Now, you know that n>=1, so n-1>=0. The tricky part is this. We have two different cases here.

If n is an integer, (x^n)/n is just have a regular polynomial with one term (and differentiable everywhere). You took the derivative just fine, so do the usual test to see where it's increasing.

If n is not an integer, what does this mean for (x^n)/n for x < 0? You need to be careful in your answer here.

5. Jul 26, 2011

### Lute

It would depend on the denominator of n(it has to be odd)? Like if n is1/2, where x<0. The top part would be undefined. Is that correct?

6. Jul 26, 2011

### gb7nash

You have the right idea. n>=1, so a valid example is 3/2, not 1/2. If you raise x < 0 to a fraction with an even denominator, you obtain an undefined result. Also, if you raise x < 0 to an irrational number, you also obtain an undefined result. The only way we'll obtain a valid result is if the denominator of the power is odd (note that the power must be a reduced fraction).

7. Jul 26, 2011

### Lute

With that in mind. How you would consider the second case in terms of the restrictions on n? Since, there are many fractions that could give you a valid answer if x<0.

8. Jul 26, 2011

### gb7nash

You have to break up your answer into multiple cases. Whoever made the question up might have meant the positive integers greater than or equal to 1, but the only restriction I see is that n is any real number greater than or equal to 1. If you have doubt about it, you can ask your teacher if this was a typo or (s)he does mean all numbers greater than or equal to 1.

9. Jul 26, 2011

### Lute

Possibly. Assuming that they meant positive integers greater or equal than 1, would you only consider is x>0? Since the resulting number would ether be positive or negative on x<0 depending on the exponent. Or is it one of those things, were I have to ask my teacher about.

10. Jul 26, 2011

### gb7nash

from before:
This is a polynomial with one term, so the derivative is defined everywhere and you can consider all x in R.

But yeah, the letter n is usually associated with positive integers, if you've seen recurrence problems, infinite series, sequences, etc. It's easy to misinterpret this as all integers >= 1. I would ask your teacher to clear up any misconceptions.

11. Jul 26, 2011

### Ray Vickson

The wording sounds to me like it is saying 'x' is a parameter, while 'n' is the variable. If so, we want to know the properties of the function f(n) = a^n / n (using the letter 'a' instead of 'x', to make this clearer).

RGV

12. Jul 27, 2011

### HallsofIvy

Staff Emeritus
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I disagree. The problem said specifically "Determine the values of x for which the function, for n>=1, is increasing, decreasing, bounded below or bounded above." It is clear that this problem is regarding x as the variable while n is fixed.

13. Jul 27, 2011

### Ray Vickson

In that case it would have been better to say "determine the values of n for which ... " because asking for a value of x for which the function is unbounded, etc., sounds artificial (although I know that statements like "f is unbounded at x = 1" are sometimes used informally to indicate some limiting behavior). Of course, if the question wants to know if the function is bounded below, that is OK (in which case the "values of x" are "all of them"). Anyway, I can't see the point of dividing by n if n is not a variable. It was just the wording that seemed wrong to me.

RGV

14. Jul 27, 2011

### gb7nash

The only reason I see that being there is for cancellation when taking the derivative.

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