Show that F'(x) exists for all x [tex]\in[a,b] [/tex]

  • #1

Homework Statement



Consider
F(x) = x2 sin(1/x2) if 0<x[tex]\leq1[/tex]
and = 0 if x[tex]\leq0[/tex]

Show that F'(x) exists for all x [tex]\in[a,b] [/tex] but that F':[0,1] [tex]\rightarrow1[/tex] is not integrable.

Homework Equations


So we have to show we do not have F(1)-F(0) = [tex]\int[/tex] F'(x)dx
(integral going from 0 to 1)



The Attempt at a Solution


I'm having trouble proving this statement.
Where should I start?
To show that F'(x) exists, should I just take the derivative or do I have to go under some long theorems of analysis to PROVE?
Thanks in advance.:shy:
 

Answers and Replies

  • #2
Most likely you'll have to actually *prove* that this derivative exists - it clearly exists for all x other than 0 because [tex]x^{2}sin(\frac{1}{x^{2}})[/tex] is the composition of differentiable functions. The only tricky part is whether F'(0) exists. To show that it does indeed exist you'll need to show that the following limit exists:

[tex]\lim_{h\to 0}\frac{f(h) - f(0)}{h}[/tex]

But this just equals

[tex]\lim_{h\to 0}\frac{h^{2}sin(\frac{1}{h^{2}})}{h} = lim_{h\to 0}hsin(\frac{1}{h^{2}}) = 0[/tex]

Then by application of the product rule and the chain rule,

[tex]F'(x) = 2xsin(\frac{1}{x^{2}}) - \frac{2cos(\frac{1}{x^{2}})}{x^{2}}[/tex]

if x is anything other than 0, and 0 if x=0. So all you need to show is that the above function is unbounded and it will consequently not be integrable.
 
  • #3
right,
But I think you made a little mistake with differentiation,
I think it will be:

F'(x)= 2x sin(1/x2) - 2cos(1/x2)/ x

Which still, limits exists.

But How do I prove that F'(x) is unbounded?
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,847
966
Take [itex]x= 1/\sqrt{2n\pi}[/itex]. Then [itex]1/x^2= 2n\pi[/itex] so [itex]sin(1/x^2)= 0[/itex] and [itex]cos(1/x^2)= 1[/itex]. [itex]F'(x)= F'(1/\sqrt{2n\pi})= -2\sqrt{2n\pi}[/itex]. x will go to 0 as n goes to infinity. What happens to F'(x)?
 
  • #5
F'(x) will be Undefined as x approaches zero [tex]\rightarrow[/tex] unbounded [tex]\rightarrow [/tex] non integrable ??????
 
  • #6
HallsofIvy just defined F'(x) above, so it *will* be defined; however, as n-->infinity, F'(x) also goes to infinity and this implies that the derivative is unbounded.

Also, thanks for pointing out my mistake; I was in an Econ class :P
 

Related Threads on Show that F'(x) exists for all x [tex]\in[a,b] [/tex]

Replies
4
Views
6K
Replies
2
Views
2K
Replies
5
Views
13K
Replies
2
Views
1K
  • Last Post
Replies
2
Views
1K
Replies
4
Views
8K
Top