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Homework Help: Spivak's Calculus: Online study group

  1. Sep 8, 2011 #1
    Roll up, roll up, one and all! Herricane and I have this plan for working through Michael Spivak's Calculus together. I've dipped into it here and there in the past, but I thought it would do me good to go through it all in order. We've just started on the first problem set. The plan is to each have a go at the problems on our own, then discuss any tricky ones, or anything of especial interest. Anyone else who wants to join in, either working through the exercises, or just to comment, is welcome. The more the merrier! (Finitely more, at least. Although... maybe infinitely more would be infinitely merrier?) Demands on time mean I might be rather intermittent at it, but hopefully we'll find a pace that works.

    I'm on Problem 5 in the first chapter, up to part (viii). So far so good, except that I didn't find the generalisation to n odd in Problem 1 (vi). I decided to press on and come back to that.

    How're you doing Herricane?
  2. jcsd
  3. Sep 8, 2011 #2
    I would be willing to join in and offer some help, if you want that :smile:
  4. Sep 8, 2011 #3


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    for prob 1(vi) spivak gives a hint.
  5. Sep 8, 2011 #4
    I am on problem 4.

    I thought about 1(vi) but all I could come up with was plugging in -y for y in equation 1(iv)
  6. Sep 13, 2011 #5
    Aha, at last, I see what he must mean by that hint for 1.1.iv (Chapter 1, problem 1, part iv)!

    (EDIT: Ack, having typed all this out, I realise Herricane was way ahead of me here. This is what you meant in #4, isn't it? Anyway, here's how I thought it through.)

    I originally just solved 1.1.iv the obvious way, by distributing the terms on the right-hand side and seeing how things cancelled out to give the expression on the left, just as I did for 1.1.iv. Not hard, but the hint suggests there's something neater, some kind of trick...

    Looking at it again, it suddenly occurred to me to do this:

    In 1.1.iv, we established that, for all [itex]x[/itex] and [itex]y[/itex],


    So let [itex]v=y[/itex]. Then, substituting [itex]v[/itex] for [itex]y[/itex] gives

    [tex]x^3-v^3=(x-v)(x^2+xv+v^2). \enspace\enspace\enspace (1)[/tex]

    Now let [itex]y = -v[/itex], so that [itex]-v^3=(-v)^3=y^3[/itex], and [itex]v^2=(-v)^2=y^2[/itex]. And substitute this [itex]y[/itex] into (1), to get

    [tex]x^3+y^3=(x+y)(x^2-xy+y^2). \enspace\enspace\enspace (2)[/tex]

    Since (1) was true for all numbers [itex]x[/itex] and [itex]v[/itex], and since every [itex]y[/itex] is the negative of some other number, and every number [itex]v[/itex] has a negative, [itex]-v=y[/itex], we can be sure that (2) also holds for all numbers. The generalisation to all odd powers comes from applying the same principle to 1.1.v.

    And in answer to micromass, wa-hey, yes please! Nice to know we have the big guns on our side. :smile:
    Last edited: Sep 13, 2011
  7. Sep 16, 2011 #6
    So that is the trick?

    SWEET :D
  8. Sep 16, 2011 #7
    I'm interested in this but time is a problem. How are you having discussion groups, using pf?

    I worked through the first two chapters a few semesters ago, only missing a few problems. I'm using the fourth addition.
  9. Sep 17, 2011 #8
    This is the very first post. We'll see how it goes.

    I am also using the fourth edition.
  10. Oct 5, 2011 #9
    Hey, has anyone done Question 24 from Chapter 5? If so, can you help me out on getting started? I'm trying to prove it using the delta-epsilon definition of limits but I'm not really sure what epsilon to choose.
  11. Oct 20, 2011 #10
    I haven't got this far, but looking ahead, here's what I'm thinking. Consider these sets:

    [tex]A=\left \{ A_n : n \in \mathbb{Z}_+ \right \},[/tex]


    and, for each [itex]a \in \left [ 0,1 \right ][/itex],

    [tex]C_a=\left \{ \left | x-a \right |: x \in X \right \}.[/tex]

    Suppose there's an [itex]x_0 \neq a \in X[/itex] such that [itex]\left | x_0-a \right | = \min(C_a)[/itex]. If, say, [itex]A_m[/itex], then [itex]f(x_0)=1/m[/itex]. Let [itex]\delta[/itex] be any positive real number such that [itex]\left | \delta - a \right | < \left | x_0 - a \right |[/itex]. Then for every [itex]y[/itex] closer to [itex]a[/itex] than [itex]\delta[/itex], [itex]f(y)=0[/itex].

    Alternatively, suppose there's no such [itex]x_0[/itex]. Since each [itex]A_n[/itex] is finite, the only way for this to be possible is if there's an infinite sequence of elements of [itex]A[/itex], and the union of this sequence contains real numbers arbitrarily close to [itex]a[/itex]. But the only way for this to be possible is if [itex]n[/itex] increases without bound for terms in the sequence, in which case, for [itex]x[/itex] in some [itex]A_n[/itex] in the sequence, [itex]f(x) = 1/n[/itex] will get ever closer to 0, so for any epsilon, we can pick a delta closer than all the numbers whose value is too big.

    Does that make sense?
  12. Feb 21, 2012 #11
    Saladsamurai wrote me a little while ago:

    Hi, Saladsamurai! Sorry it's taken me so long to reply. I'm afraid I've been very busy of late, lots of demands on my time. I got part way through the problem set at the end of Ch. 1 before distractions intervened. Not sure how far exactly as I lost my notebook. Hope to get back to it one of these days when (if?) life gets less hectic!

    How's everyone else doing: Herricane, QuarkCharmer?
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