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Homework Help: Analysis: Potentially flawed proof in book describing roots of multiplicity m

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data

    In the book "Friendly introduction to analysis, 2nd Ed." by kosmala there is a definition of the root of a function and subsequent theorem and proof. Either the proof is not directly addressing certain important properties, or is flawed. The definition and theorem are as follows.

    Definition 5.4.6: Consider a function f. Suppose that ## x = r ## is a solution of the equation ## f(x) = 0 ##. Then ## x = r ## is a root of multiplicity m of f, with ## m \in N ##, if and only if m is the smallest value for which f(x) can be written as:

    ## f(x) = (x - r)^m q(x) ##

    with ## x \neq r ##, where ## \lim_{x \to r} q(x) \neq 0 ##. If ## m = 1 ##, then the root is called a simple root. If ## m = 2 ## then the root is called a double root.

    Theorem 5.4.7: Suppose that a function f is m times continously differentiable. The function f has a root of multiplicity m at ## x = r ## if and only if:

    ## 0 = f(r) = f'(r) = ... = f^{(m-1)}(r) ##

    but ## f^{(m)}(r) \neq 0 ##

    Proof: We are only going to prove the theorem in the case of a simple root (i.e. m=1).

    (=>): First, assume that f has a simple root at ## x = r ##. We want to show that ## f(r) = 0 ## and that ## f'(r) \neq 0 ##. Clearly, ## f(r) = 0 ##. Also, by definition, we can write ## f(x) = (x - r)q(x) ##, where ## \lim_{x \to r} q(x) \neq 0 ##. Since f has a continuous first derivative at ## x = r ##, we can write:

    ## f'(r) = \lim_{x \to r} f'(x) = \lim_{x \to r} [q(x) + (x - r)q'(x)] = \lim_{x \to r} q(x) \neq 0 ##

    That is the end of the (=>) proof.

    2. Relevant equations


    3. The attempt at a solution

    The problem I have with the definition and the proof centers around q(x). The definition defines the limit of q(x) to not equal zero, but the definition does not guarantee that the limit exists or is finite. Further, the definition does not address q'(x) at all. The proofs assumption seems to be that it is finite as x-> r, and that may be the case by some property of continuity that I am unaware of, but I would like to understand exactly what properties allow them to assume that q'(x) is either finite, or diverges to infinity slower than x -> r.

    I've spent a number of hours with two different analysis professors looking through half a dozen texts and neither myself nor my professors can reconcile the definition with the proof.

    I've tried writing q'(x) as various forms of limits as x-> r to see if there is some way I can bound it. I've introduced the assumption that q(x) has a finite limit as x -> r and that did not help. I feel like if the definition and proof are 100% correct that there must be some property of continuously differentiable functions that I am overlooking, but I've found nothing in the text to help.

    Given that a continuous function can be the sum of two discontinuous functions I don't see how f'(x) being continuous asserts anything about q(x) or q'(x) individually, and certainly nothing about rates of divergence.

    Anyway, I hope I've presented the problem clearly. If anyone can shed any light I would be much obliged. This isn't a homework problem, just something I'm having trouble reconciling within the text, and the professor also was not able to reconcile this problem.
    Last edited: Nov 8, 2012
  2. jcsd
  3. Nov 8, 2012 #2


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    To make your post more readable, click the Edit button and change all your [; and ;] tags to ## instead. Then click the Go Advanced button at the bottom, preview it, and save it when you have it reading correctly.
  4. Nov 8, 2012 #3


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    @ LCKurtz,

    I was about to PM you, Dick and micromax to look at this post. I'm glad you saw it.

    @ wjamesbonner,

    Hello and welcome to PF !

    As an example of what LCKurtz is telling you:

    Code (Text):

    [; f(x) = (x - r)^m q(x) ;]


    ## f(x) = (x - r)^m q(x) ##
    will give you ## f(x) = (x - r)^m q(x) ##

    instead of [; f(x) = (x - r)^m q(x) ;]
  5. Nov 8, 2012 #4


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    It gives me a headache to try to read it so I'm waiting for it to be fixed.
  6. Nov 8, 2012 #5
    Thanks for the feedback, I changed the latex tags to use ## which seems to have produced the desired effect.

    I actually just finished looking through every analysis book in my schools science library (rudin, funke, hille, haeussler, etc...) over 30 books and not one of them had this particular definition or proof (or not that I could find from the index anyway). I've only ever seen this definition and proof in Kosmalas books on analysis (Advanced calcules, and friendly introduction).

    Let me know if there is something I can do to make the post more readable or the content more clear.
  7. Nov 8, 2012 #6
    First of all, there is no reason why the limit [itex]\lim_{x\rightarrow r} q(x)[/itex] should exist. As example: take [itex]f(x)=x\sin(1/x)[/itex]. But this function is not continuously differentiable.
    I have encountered this definition of multiplicity of roots before, but that was in complex analysis and the functions are all very well-behaved there (= analytic). I think we should utilize Taylor's theorem to solve this.

    If f has a simple root and is continuously differentiable, then Taylor's theorem says that we can write


    for [itex]\lim_{x\rightarrow r}{h(x)}=0[/itex].

    Since we have a simple root, we can write [itex]f(x)=q(x)(x-r)[/itex]. In particular, it follows that f(r)=0. Thus




    taking limits, we see that [itex]\lim_{x\rightarrow r} q(x)=f^\prime(r)[/itex] and thus is nonzero.
  8. Nov 8, 2012 #7
    Thanks for the response, it is a very clear proof via Taylor's theorem. I am still wondering though if it is possible to prove from the definition provided. The problem is that Taylor's theorem is introduced after this particular definition and theorem. I definitely appreciate the proof via Taylor's theorem, but my main concern is with the veracity of the given proof in the book, and whether or not it is correct (and as such whether or not I am correct in my skepticism of it).

    Thanks again for your response!
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