• Support PF! Buy your school textbooks, materials and every day products Here!

Derivatives - differentiabilty, continuity, unbounded

  • Thread starter kathrynag
  • Start date
  • #1
598
0
Let g(x) =x^asin(1/x) if x is not 0
g(x)=0 if x=0

Find a particular value for a such that
a) g is differentiable on R but such that g' is unbounded on [0,1].
b) g is differentiable on R with g' continuous but not differentiable at zero
c) g is differentiable on R but and g' is differentiable on R, but such that g'' is not continuous at zero.


I've only started on a) but I got a little stuck
Well I know that a function is differentiable if g' exists for all points.
I started by calculating derivatives:
g'(x)=ax^(a-1)sin(1/x)-x^(a-2)cos(1/x) if x is not zero.
Now for x=0, we have:
g'(0)=limg(x)-g(0)/x-0=limg(x)/x=limx^(a-1)sin(1/x)
I know xsin(1/x) does not exist, but for all other x, g'(0) does exist. Then g'(0) exists and equals 0 if provided a>2.
So g is differentiable provided a>2.
Now the part that confuses me is the unboundedness. I looked in this particular chapter of my book and there was no mention of boundedness at all.
So from previous chapters, I had that a sequence (xn) is bounded if there exists a number M>0 such that |xn|<=M for all n in N. I just am really confused on the unboundedness part.
 

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,237
962
Let g(x) =xasin(1/x) if x is not 0
g(x)=0 if x=0

Find a particular value for a such that
a) g is differentiable on R but such that g' is unbounded on [0,1].
b) g is differentiable on R with g' continuous but not differentiable at zero
c) g is differentiable on R but and g' is differentiable on R, but such that g'' is not continuous at zero.


I've only started on a) but I got a little stuck
Well I know that a function is differentiable if g' exists for all points.
I started by calculating derivatives:
g'(x)=ax(a-1)sin(1/x)-x(a-2)cos(1/x) if x is not zero.
Now for x=0, we have:
g'(0)=limg(x)-g(0)/x-0=limg(x)/x=limx^(a-1)sin(1/x)
I know xsin(1/x) does not exist, but for all other x, g'(0) does exist. Then g'(0) exists and equals 0 if provided a>2.
So g is differentiable provided a>2.
Now the part that confuses me is the unboundedness. I looked in this particular chapter of my book and there was no mention of boundedness at all.
So from previous chapters, I had that a sequence (xn) is bounded if there exists a number M>0 such that |xn|<=M for all n in N. I just am really confused on the unboundedness part.
Hi Kathryn.

Part (a) is rather troublesome.

g(x) is differentiable on R, (except for x = 0, for some values of a). Definitely, for values of a for which g(x) is unbounded on [0, 1], g'(0) does not exist, since in those cases, g(x) is not continuous at x=0.

Your differentiation, g'(x) = ax(a-1)sin(1/x)-x(a-2)cos(1/x) is correct. Of course, this is undefined for x=0.
Another form you can put g'(x) into is:
g'(x) = x(a-2)(a·x·sin(1/x) - cos(1/x))

What you did to check on g'(0), won't work.
Now for x=0, we have:
g'(0)=limg(x)-g(0)/x-0=limg(x)/x=limx^(a-1)sin(1/x)
To determine if g'(0) exists, consider the limit.

limx→0 g'(x).

If this limit exists, you can argue that g is differentiable at x=0.

Graph g(x) and g'(x) for various values of a - I would suggest integer & a few half integer values from a = -1 to a = 4.

In particular look at WolframAlpha.

g(x), a=0 and g'(x), a=0

g(x), a=2 and g'(x), a=2

One sequence to look at is yn = g'(xn), where xn = 1/(π n), especially for a = 2.
 
  • #3
598
0
Ok, but I'm still confused on what they mean by unbounded.
 
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,237
962
...
Now the part that confuses me is the unboundedness. I looked in this particular chapter of my book and there was no mention of boundedness at all.
So from previous chapters, I had that a sequence (xn) is bounded if there exists a number M>0 such that |xn|<=M for all n in N. I just am really confused on the unboundedness part.
OK, but I'm still confused on what they mean by unbounded.
Unbounded: You can say this in any of several ways.

(xn) is unbounded means: For any M > 0, there exists an n ϵ N, such that |xn| > M .

f(x) is unbounded on [0, 1] means that for any M > 0, there exists an x0 ϵ [0,1] such that |f(x0)| > M .



Did you look at g'(1/(n π))?

g'(1/(n π)) = (1/(n π))(a-2){a·x·sin(1/[1/(n π)]) - cos(1/[1/(n π)])}

= (1/(n π))(a-2){a·(1/(n π))·sin(n π) - cos(n π)}

= - (-1)n(1/(n π))(a-2),     because, sin(n π) = 0, and cos(n π) = (-1)n = ±1 .
If the absolute value of the above sequence diverges for any a, then the sequence and g'(x) are unbounded for that value of a.

If the absolute value of the above sequence converges to any L ≠ 0 for any a, then the sequence and g'(x) are bounded (but g'(x) is not continuous at x=0) for that value of a.

If the absolute value of the above sequence converges to 0 for any a, then the sequence and g'(x) are bounded (and g'(x) is continuous at x=0), for that value of a.
 
  • #5
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,237
962
BTW:

Are you sure that part (a) isn't something like:

g is [STRIKE]differentiable[/STRIKE] continuous on R but such that g' is unbounded on [0,1].

That would make [STRIKE]more[/STRIKE] sense.
 
  • #6
598
0
BTW:

Are you sure that part (a) isn't something like:

g is [STRIKE]differentiable[/STRIKE] continuous on R but such that g' is unbounded on [0,1].

That would make [STRIKE]more[/STRIKE] sense.
No, I'm sure
 
  • #7
598
0
So g' unbounded means we want for M>0, there exists x0 in [0,1] such that |g'(x0)|>M
We had g'(x) = ax^(a-1)sin(1/x)-x^(a-2)cos(1/x)
Now we want |ax^(a-1)sin(1/x)-x^(a-2)cos(1/x)|>M
We know limx^(a-1)sin(1/x)=0, so we are only concerned with -x^(a-2)cos(1/x)
So to be unbounded, we want a>2?
 
  • #8
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,237
962
So g' unbounded means we want for M>0, there exists x0 in [0,1] such that |g'(x0)|>M
We had g'(x) = ax^(a-1)sin(1/x)-x^(a-2)cos(1/x)
Now we want |ax^(a-1)sin(1/x)-x^(a-2)cos(1/x)|>M
We know limx^(a-1)sin(1/x)=0, so we are only concerned with -x^(a-2)cos(1/x)
So to be unbounded, we want a>2?
Sorry Kathryn, I've been away for a day.

The answer to your final question in the quoted post is "No." If β>0, then xβ → 0 as x → 0.

Please try g'(1/(n·π)).
 

Related Threads for: Derivatives - differentiabilty, continuity, unbounded

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
878
  • Last Post
Replies
5
Views
1K
Replies
2
Views
795
  • Last Post
Replies
2
Views
2K
Top