Recent content by NihilTico

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    Spivak & Dimension of Manifold

    Hey all, thanks for the replies and comments and I apologize for getting back a little late. Not really sure why I threw that statement in there, there are plenty of counter-examples. I appreciate the help, thanks!
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    Spivak & Dimension of Manifold

    1. Homework Statement I'm taking a swing at Spivak's Differential Geometry, and a question that Spivak asks his reader to show is that if ##x\in M## for ##M## a manifold and there is a neighborhood (Note that Spivak requires neighborhoods to be sets which contain an open set containing the...
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    Understanding the Coproduct in Grp as a Universal Object

    (Edit: I realize I am relying heavily on Lang's prior proof of the existence of the free group (apparently owed to J. Tits). Please let me know if I should fill it out in detail here if it is not readily accessible to you via the internet or by the text.) I had some time to think about this...
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    Prove that the integral of sup f_n is infinite

    The DCT implies that , ##\lim_{n\to\infty}\int_{0}^{1}f_n d\mu=\int_{0}^{1}gd\mu## for a sequence ##f_n## that converges pointwise to ##g##. All you have is a pointwise supremum, ##g=\sup_n f_n##, which is easily seen to be measurable. In any case, I can't see how the DCT would imply here that...
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    Show that a normal subgroup <S> is equal to <T>

    I think I misinterpreted your question the first time around. Do you mean by 'the normal subgroup generated by ##T##' the normal closure of ##T##? The normal closure of ##T## is the minimal normal subgroup containing ##T##. If so, then let's claim that...
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    Question about summation in power spectral density formula

    As a Fourier series, the first term is ##\frac{A^2/2}{2}=\frac{A^2}{4}## (i.e., divided by ##2##). That, I believe, is what's going on here. EDIT (Again D: ) OK! My bad. What's happening here is this. The sum converges, so we rearrange a finite number of terms (i.e., one term) and notice that...
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    Show that a normal subgroup <S> is equal to <T>

    If by this, you mean that ##\langle gTg^{-1}\rangle=\langle{T}\rangle## for any subset of a group, then I dispute your claim. Just consider a group ##G## where ##G'## is a subgroup of ##G## that is not normal (i.e., ##gG'g^{-1}\ne G##). Indeed, if the set ##\{g_1,\ldots,g_n\}## generate...
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    Understanding the Coproduct in Grp as a Universal Object

    Mea culpa, by 'his Algebra' I mean Lang's.
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    Understanding the Coproduct in Grp as a Universal Object

    Homework Statement Coproducts exist in Grp. This starts on page 71. of his Algebra. Homework Equations [/B] Allow me to present the proof in it's entirety, modified only where it's convenient or necessary for TeXing it. I've underlined areas where I have issues and bold bracketed off my...
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    [itex]\hom_A(-,N)[/itex]Functor Takes Coproducts to Products

    Thanks for the verification. What does this say about hom-functors? Is there a word for this property?
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    How does -2^(5/2) -2^(5/2) = -2^(7/2)?

    Well, yes, I figured that when Rosebud replied with the same notation ;)
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    How does -2^(5/2) -2^(5/2) = -2^(7/2)?

    If the negative sign is outside the power, you can pull it out all together until you're done simplifying. If the negative sign is being raised to a power as well, you must bring it along. So, if you have the negative signs all out front, and not being raised to a power, then yes, because...
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    How does -2^(5/2) -2^(5/2) = -2^(7/2)?

    (-2)^{5/2}+(-2)^{5/2}=2(-2)^{5/2} It might help to think about it like this. We write x=y^{a/b} to mean roughly that x is the number equal to y multiplied together with itself "a/b times". In this sense, how might you think we should write 2(-2)^{5/2}?
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    [itex]\hom_A(-,N)[/itex]Functor Takes Coproducts to Products

    A couple of notes first: 1. \hom_{A}(-,N) is the left-exact functor I'm referring to; Lang gives an exercise in the section preceeding to show this. 2. This might be my own idiosyncrasy but I write TFDC to mean 'The following diagram commutes' 3. Titles are short, so I know that the hom-functor...
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