Recent content by TwilightTulip

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    Count to 20 with Fingers: Ideas & Tips

    I'd be weary of doing this in class. Especially with 100 and 101
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    Prove Finite Orthogonal Set is Linearly Independent

    You are assuming the x_i's are orthogonal (i.e. <x_i,x_j>=0 if i =/= j), so the only term in the sum which is not necessarily 0 is a_j<x_j,x_j>
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    General question regarding continuous functions and spaces

    Let X be some topological space. Let A be a subspace of X. I am thinking about the following: If f is a cts function from X to X, and g a cts function from X to A, when is the piece-wise function h(x) = f(x) if x is not in A, g(x) if x is in A continuous? My intuition tells me they must agree...
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    Dimension of Vector Space R^X: Finite/Infinite Cases

    Okay, regarding the above, I get an interesting (wrong?) result. So A is an n-dim subspace of R^X, and A~R^Y for |Y|=n. If θ is an isomorphism between R^Y and A, then if R^Y has basis {f1,f2,...fn} we can map this to a basis {a1,...an} of A to get that θ(f1)=a1 is still a function from Y to F...
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    Dimension of Vector Space R^X: Finite/Infinite Cases

    Hmm. So say i had any finite dimensional space A contained in R^X (where X can actually be infinite), say dimension n. Then I know that it is isomorphic to R^n, which is then again isomorphic to functions from some set Y to R, where |Y|=n. So if I'm thinking about this right, then A ~ R^Y. (I...
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    Dimension of Vector Space R^X: Finite/Infinite Cases

    Only when X is infinite though, correct?
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    Dimension of Vector Space R^X: Finite/Infinite Cases

    I see, that's a shame. Thanks for the reply though.
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    Dimension of Vector Space R^X: Finite/Infinite Cases

    The finite case is fine, as a vector space it is easy to show that R^X is isomorphic to R^n. What about when X is infinite? I believe it is true in general that dim(R^X) = #(X), which I hope holds in the infinite case too. I know that the set given by B={b_x; x in X} defined as b_x(y) =...
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    Something interesting concerning parity

    Rewrite n as 10a+b, where 0<=b<10. Then n is even means 2 divides 10a+b, i.e. 2 divides b (let 10a+b=2k and solve for b). Hence b=0,2,4,6,8
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    Groups of prime order are cyclic. (without Lagrange?)

    I guess not then :biggrin: I was thinking that since \phi(1)->g \neq e works, there must be a way to construct such an ismorphism that makes it easy to "play" with the properties of a group to get the isom to work out. I suppose simply showing that \phi(0) = e=g^0,g^1,...,g^{p-1} are all...
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    Groups of prime order are cyclic. (without Lagrange?)

    I know full well the proof using Lagrange's thm. But is there a direct way to do this without using the fact that the order of an element divides the order of the group? I was thinking there might be a way to set up an isomorphism directly between G and Z/pZ. Clearly all non-zero elements...
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    LaPlace's Equation in Sphereical Coordinates

    Homework Statement I want to cook a 4" meatball. The meatball is being stored in the fridge at 35 degrees F. The meatball will go into a convection oven at 350 degrees F (surface is maintained at precisely 350 for the duration of cooking). I want to cook the meatball to 130 degrees F (in the...
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    Divergence of Complex Harmonic

    I'm still stumped :-(
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    Divergence of Complex Harmonic

    Ok... so my "intuition" to split them at zero symmetrically is where I went wrong. I am using Complex Analysis: In the Spirit of Litman Bers. This question must be related to 7.3 "the cotangent function." No tests of any kind are talked about in detail in this chapter. Chapter 7 is about metrics...
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