Recent content by yyat

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    Undergrad Irrational numbers and repeating patterns

    The rational numbers are exactly those with repeating decimal expansion: Given a decimal number with repeating digits, there is a very easy way to find a representation as a rational number (see also http://en.wikipedia.org/wiki/Repeating_decimal#Fraction_from_repeating_decimal"): For...
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    Partition groups into subcollection

    What is 3z? Is it \mathbb{Z}_3? Think again about cardinality: The rationals are countable, the reals are uncountable. Also consider my hint about the exponential function (Post #4), it relates addition and multiplication.
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    Proof that In^-1=In | Linear Algebra

    Just use the definition. You want to check that the inverse of I_n is I_n itself (this is just another way of saying I_n^{-1}=I_n). What it comes down to is that I_nI_n=I_n.
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    Research on learning abilities of mice

    Check your formula for the conditional probability, there should be a P(B) in the denominator.
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    Differentiation under the integral sign

    The RHS not a function of y, since you already integrated y from d to c. It should just be F(t). F is not the antiderivative of f with respect to y in this case, so the last step is not correct. The final answer is what you had before that (without the y on the LHS).
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    Prove the sequence converges uniformly

    Any function defined on a finite set of points is continuous. Why? You want to prove uniform convergence on [0,1], which is not a finite set. The Lipschitz continuity is crucial here.
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    Proof that In^-1=In | Linear Algebra

    The inverse matrix A^{-1} of A is by definition the matrix such that A^{-1}A=I_n and AA^{-1}=I_n. So is I_n the inverse of I_n?
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    Prove the sequence converges uniformly

    Hint: Pointwise convergence implies uniform convergence on any finite set of points. Since [0,1] is compact, you can choose points x1,...,xk such that the distances between consecutive points is arbitrarily small.
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    Differentiation under the integral sign

    Hi celtics2004! The double integral defines a real-valued function of t, which turns out to be differentiable. So you can compute its derivative d/dt. To compute the derivative, you will need http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign" (as you mentioned) and the...
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    Graduate Antisymmetric but non-symplectic

    A symplectic form is by definition an anti-symmetric, nondegenerate, bilinear form. I am not sure what you mean by "scalar product" in this case, but if it should be nondegenerate then the answer to your question is no.
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    Complex Roots - Not sure I did this right

    Hi Atena! Your answer is correct. You can check this by taking the fourth powers of the solutions you got (using deMoivre).
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    What's difference between (∂y/∂x) and (dy/dx)

    ∂y/∂x and fx are http://en.wikipedia.org/wiki/Partial_derivative" . They are applied to functions of several real variables, for example f(x,y)=x2+y2. dy/dx or y' is the ordinary derivative of a function of a single real variable such as y(x)=1/x.
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    Graduate The generalized rank-nullity theorem

    The third one is really the definition of scalar multiplication, not a relation, and I think the last one should be just (nx,y)-(x,ny) for n an integer. More generally, abelian groups and \mathbb{Q} are both \mathbb{Z}-modules, so one can form the...
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    Undergrad Diagonalizability and Invertibility

    Yes, an nxn-matrix (no need for it to be diagonal) is invertible if and only if 0 is not an eigenvalue. This is easy to see from the definition of an eigenvalue. No, this is wrong. If T is invertible, then dim(V)=dim(W), but the converse is false. The zero-map from V to W is invertible only...
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    Graduate The generalized rank-nullity theorem

    An abelian group is the same thing as a \mathbb{Z}-module. This is why the tensor product, which only makes sense for modules, can be defined. When tensoring with \mathbb{Q} you get a \mathbb{Q}-module, which is of course just a vector space over the field \mathbb{Q}. Moreover, the homorphisms...