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You could substitute for $r$ and $\theta$ their expressions in terms of $x$ and $y$. But remember that trigonometric functions of $\theta$ have easier expressions than $\theta$ itself.

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Orthogonal transformations preserve the norm of a vector. You can take the product of that group with the one-dimensional multiplicative group that scales the vectors by nonzero real factors (a copy ...

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u is a given function of x and t, so the triplet (x,t,u) is really (x,t,u(x,t)) which leaves two degrees of freedom, namely, x and t. The capital T means "transpose": exchanging rows and columns of a ...

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For fixed \alpha your function \log^{(\alpha)} is not a bijection onto C. Its range is a horizontal strip of height 2\pi.

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You should define what you mean by "inverse square". This is by no means a standard concept. In you formula it means that the radius of the sphere is divided by the square of time, which indeed ...

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If an arbitrary centered Gaussian measure is given on a product of a Banach space X with itself, then it is not guaranteed to be the tensor product of two centered Gaussian measures on X. As an ...

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Hint: if a norm is derived from an inner product in that way, then the inner product is uniquely determined by the norm and there is an explicit algebraic expression for (f,g) in terms of ||f||, ||g||,...

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The concept that you are thinking of is sequence continuity (which is equivalent to continuity in metric spaces such as R^n but that requires proof). If you check the condition "for every given open ...

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You can eliminate answers A and D by noting that the zero matrix B always satisfies that identity.

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Intuitively I understand the first definition but not the second. In fact the standard notation for a polynomial with integer coefficients and variables taken from the set X suggests first taking ...

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Since $\theta$ is a given real number, the arc is oriented. Only one of two possible circles has the right orientation of the angle from $z_1$ via $z$ to $z_2.$

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There is an abstract notion of products in category theory. A category where every finite collection of objects has a product is called a cartesian category. In those categories (and Set is one of ...

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On an abstract level, the composition of two measurable maps $f:X\to Y$ and $g:Y\to Z$ is measurable; this is very easy to verify from the definition of a measurable map. The only remaining step is ...