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Math Help - Math Exam, due in about 18 hours

  1. #1
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    Math Exam, due in about 18 hours

    I'm new here, and wanted to ask this about a week ago, but I got the other 6 problems that I was going to ask for on my own, and with friends.

    The alst problem bothering us is:

    For what integers p > 0 is

    f(x) = x^p*sin(1/x)   x =/= 0<br />
0 x=0

    Differentiable? For what p is the derivative continuous?

    My work: All I have for the most part is the general derivative of the top equation.

    x^p*((cos(1/x)*(-1/x^2))+(sin(1/x)*px^(p-1)) (that is, in the last part, p times x raised to the power of (p-1)

    That factors if we pull out the x raised to the power of (p-1) to:

    x^(p-1)*(-cos(1/x)*(1/x)+sin(1/x)*p)

    Notice that if we pull out that x from the left side, we have reduced from 1/x^2 to 1/x.


    Help? I think it is whenever p > 2, that it will be continuous because by that, the function would converge to zero as the functions approaches x=0 from the left or right sides.

    Would the derivative be continuous for p = anything?
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  2. #2
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    Quote Originally Posted by ZevWolf View Post
    I'm new here, and wanted to ask this about a week ago, but I got the other 6 problems that I was going to ask for on my own, and with friends.

    The alst problem bothering us is:

    For what integers p > 0 is

    f(x) = x^p*sin(1/x)   x =/= 0<br />
0 x=0

    Differentiable? For what p is the derivative continuous?

    My work: All I have for the most part is the general derivative of the top equation.

    x^p*((cos(1/x)*(-1/x^2))+(sin(1/x)*px^(p-1)) (that is, in the last part, p times x raised to the power of (p-1)

    That factors if we pull out the x raised to the power of (p-1) to:

    x^(p-1)*(-cos(1/x)*(1/x)+sin(1/x)*p)

    Notice that if we pull out that x from the left side, we have reduced from 1/x^2 to 1/x.


    Help? I think it is whenever p > 2, that it will be continuous because by that, the function would converge to zero as the functions approaches x=0 from the left or right sides.

    Would the derivative be continuous for p = anything?
    I was mystified by your definition of f(x) at first but eventually realized you meant f(x) is defined piecewise as x^p*sin(1/x) when x \neq 0 and 0 when x = 0. (I'm not sure how to make it look nice with a big left brace in LaTeX.)

    I'll edit this post in a bit to address the problem itself, but thought it might help others to know what f(x) is in the meantime.

    EDIT 1: It's been a while since I've done these, so first, a sketch:

    f(x) is differentiable at x=a if both left and right derivatives exist at a and they are equal.

    We'll only need to check the case x = 0 because it's pretty clear that the derivative is well defined when x \neq 0.

    I'll edit again when I get a bit further.

    EDIT 2: I realized just now that it's more convenient for everyone if people make new replies rather than editing existing posts. That way, the poster gets to use the "preview" feature, and everyone else can get email notification. Sorry, I'm new here.

    Anyway, this is what I came up with:

    You found that

    f'(x)=x^p*((cos(1/x)*(-1/x^2))+(sin(1/x)*px^{p-1})
    =x^{p-1}*(-cos(1/x)*(1/x)+sin(1/x)*p)

    when x \neq 0. I agree with this part.

    In what follows, when I assign a name to a limit (like L_1), I mean that this name applies only in the case that the limit exists.

    We should use the definitions for right-derivative

    L_1 = lim_{h->0^+}(f(a+h)-f(a))/h

    and left-derivative,

    L_2 = lim_{h->0^-}(f(a+h)-f(a))/h

    to find out whether f(x) is differentiable at x=a=0. If both limits exist and L_1=L_2 then the derivative exists and f'(0)=L_1=L_2. We should not think only in terms of

    L_3 = lim_{x->0}f'(x)

    because it's possible that f'(0) exists but either (1) L_3 doesn't exist, or (2) L_3 exists but f'(0) \neq L_3, and in both cases (1) and (2), f(x) will be differentiable, but the derivative will not be continuous. It is also generally possible for L_3 to exist but f'(0) to not exist, in which case f(x) will not be differentiable but we might mistakenly think it is. This is a bit tricky to realize, but I believe it's the whole point of the problem.

    So first, the right derivative:

    L_1 = lim_{h->0^+}(f(a+h)-f(a))/h
    =lim_{h->0^+}(f(0+h)-f(0))/h
    =lim_{h->0^+}(h^p*sin(1/h)-0)/h
    =lim_{h->0^+}h^{p-1}*sin(1/h)

    If p = 1, then we have

    L_1=lim_{h->0^+}h^0*sin(1/h)
    =lim_{h->0^+}sin(1/h)

    which does not exist since sin(1/h) oscillates an infinite number of times in any closed interval [0, b], b > 0.

    If p > 1, then we'll be able to use the sandwich theorem (aka squeeze theorem) to determine that both right- and left-derivatives exist and equal 0.

    So the first part of the question is answered: f(x) is differentiable if and only if p > 1.

    For the second part we need to check

    L_3 = lim_{x->0}f'(x)

    I believe it's most useful to express f'(x) like this

    f'(x)=-x^{p-2}*cos(1/x)+p*x^{p-1}*sin(1/x)

    because otherwise we might fall into a certain trap: if lim_{x->0}g(x)=0 and lim_{x->0}h(x) is undefined, it would be wrong to conclude that lim_{x->0}(g(x)h(x))=0, for example consider g(x)=x and h(x)=x^{-4/3}.

    Anyway it's my understanding that

    lim_{x->0}(-x^{p-2}*cos(1/x))

    only exists when p > 2, in which case it will be 0 by the sandwich theorem. So overall f'(x) will be continuous only in the case p > 2.

    I think this is right but again it's been a while, so please check my work and see if you agree.

    By the way in LaTeX the way to get superscripts with more than one character is like this, without spaces: [ math]x^{p-1}[ /math].
    Last edited by undefined; April 6th 2010 at 04:49 PM.
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  3. #3
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    Oh, sorry about that - meant to define it as a piece-wise function, such that:
    if x = 0, f(x) = 0
    x not equal to 0, f(x) is that long equation
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  4. #4
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    We should know that the derivative should be continuous wherever it exists.

    based on my work of factoring out an x^(p-1) from the derivative, whenever x = 0, one of the factors from the multiplication is 0, so the whole derivative should be 0, regardless of what p is.

    From what I understand (and I may very well be wrong), the only sticking point is whether the function itself is continuous for p (and we are pretty sure that it isn't all values of p, because I'd submitted it as a homework a week before, and he marked it wrong).


    Ah, yes, and I also didn't notice the new edit. Thanks for the help with the exponents by the way.
    Last edited by ZevWolf; April 6th 2010 at 08:00 PM.
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    Triple posting is not allowed, but I can't message for a response to my questions. I just noticed the response 5 minutes ago. So if a moderator can, please edit and consolidate my posts.

    Quote Originally Posted by undefined
    EDIT 2: Anyway, this is what I came up with:

    You found that

    f'(x)=x^p*((cos(1/x)*(-1/x^2))+(sin(1/x)*px^{p-1})
    =x^{p-1}*(-cos(1/x)*(1/x)+sin(1/x)*p)

    when x \neq 0. I agree with this part.
    Good to know I did something right :S


    In what follows, when I assign a name to a limit (like L_1), I mean that this name applies only in the case that the limit exists.

    We should use the definitions for right-derivative

    L_1 = lim_{h->0^+}(f(a+h)-f(a))/h

    and left-derivative,

    L_2 = lim_{h->0^-}(f(a+h)-f(a))/h

    to find out whether f(x) is differentiable at x=a=0. If both limits exist and L_1=L_2 then the derivative exists and f'(0)=L_1=L_2. We should not think only in terms of

    L_3 = lim_{x->0}f'(x)

    because it's possible that f'(0) exists but either (1) L_3 doesn't exist, or (2) L_3 exists but f'(0) \neq L_3, and in both cases (1) and (2), f(x) will be differentiable, but the derivative will not be continuous. It is also generally possible for L_3 to exist but f'(0) to not exist, in which case f(x) will not be differentiable but we might mistakenly think it is. This is a bit tricky to realize, but I believe it's the whole point of the problem.

    I follow you on this as well, that the limit may exist on both sides, but not necessarily equal

    So first, the right derivative:

    L_1 = lim_{h->0^+}(f(a+h)-f(a))/h
    =lim_{h->0^+}(f(0+h)-f(0))/h
    =lim_{h->0^+}(h^p*sin(1/h)-0)/h
    =lim_{h->0^+}h^{p-1}*sin(1/h)

    If p = 1, then we have

    L_1=lim_{h->0^+}h^0*sin(1/h)
    =lim_{h->0^+}sin(1/h)

    which does not exist since sin(1/h) oscillates an infinite number of times in any closed interval [0, b], b > 0.

    If p > 1, then we'll be able to use the sandwich theorem (aka squeeze theorem) to determine that both right- and left-derivatives exist and equal 0.

    So the first part of the question is answered: f(x) is differentiable if and only if p > 1.

    That makes sense.



    For the second part we need to check

    L_3 = lim_{x->0}f'(x)

    I believe it's most useful to express f'(x) like this

    f'(x)=-x^{p-2}*cos(1/x)+p*x^{p-1}*sin(1/x)

    Okay, took a minute to realize, but yes - I agree

    because otherwise we might fall into a certain trap: if lim_{x->0}g(x)=0 and lim_{x->0}h(x) is undefined, it would be wrong to conclude that lim_{x->0}(g(x)h(x))=0, for example consider g(x)=x and h(x)=x^{-4/3}.

    Anyway it's my understanding that

    lim_{x->0}(-x^{p-2}*cos(1/x))

    only exists when p > 2, in which case it will be 0 by the sandwich theorem. So overall f'(x) will be continuous only in the case p > 2.

    What about p = 2?

    I think this is right but again it's been a while, so please check my work and see if you agree.

    The work is very complicated, and it will take my mind 10 minutes to understand, but for now, let's agree that the work is correct. I thank you for your help, greatly.
    Last edited by ZevWolf; April 6th 2010 at 08:54 PM. Reason: Part 2: The derivative
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  6. #6
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    Quote Originally Posted by ZevWolf View Post
    Triple posting is not allowed, but I can't message for a response to my questions. I just noticed the response 5 minutes ago. So if a moderator can, please edit and consolidate my posts.
    I don't know the rules on triple posting, but hopefully you will be able to reply to this if you still have questions.

    Quote Originally Posted by ZevWolf
    What about p = 2?
    When p = 2, f'(0) = 0, but lim_{x->0}f'(x) does not exist. I'm attaching a picture I made with Mathematica showing f'(x) when p = 2. As you can see, this oscillates more and more the closer x gets to 0. You may think of f'(x) defined piecewise: f'(x)=x^{p-1}*(-cos(1/x)*(1/x)+sin(1/x)*p) when x \neq 0, and f'(x)=0 when x = 0.

    I hope that over time my answer it will seem less complicated; my basic method was to start with the definitions of differentiable and continuous and then apply them to the problem.

    If you have more questions, I'll be happy to answer them, and if I'm hard to understand then maybe someone better at explaining can help me out

    EDIT: I forgot to attach the file, and now I realize I don't know how to attach files. So you can find the file here.
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  7. #7
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    No, no - that note was to an admin / moderator because I triple-posted....
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