Michael Seifert
• Member for 6 years, 5 months
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• New London, CT

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It's helpful to think of each quadratic polynomial as corresponding to a point in "coefficient space", an abstract 3D space whose coordinates are $(a,b,c)$. The question "What fraction of quadratic ...

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Here's how I would go about dividing the expenses: Tally up the expenses. This should be the total of rental fees, fuel, tolls, and an agreed-upon amount for amortization &amp; insurance for the ...

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Your intuition is correct for a slightly different statement: There is no function $f(x)$ on $x \ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)&lt;x^2$ for $x&gt;0$ and that the third derivative of $f(... View answer 2 answers 6 votes 878 views Accepted answer 16 votes I think you're misunderstanding the problem. The problem is not to show that functions$\{\mathbf{u}(\mathbf{x},t), P(\mathbf{x},t)\}$exist which satisfy the Navier-Stokes equations; that much is ... View answer 6 answers 20 votes 4k views 15 votes Consider the two points$(-\frac12,0)$and$(\frac12,0)$. For any real value of$y_0$, we can draw a circular arc between these two points which is centered at$(0,y_0)$and which lies entirely in ... View answer 3 answers 5 votes 481 views Accepted answer 14 votes Summary: There is an uncountably infinite family of$\mathcal{C}^\infty$functions on$(0, +\infty)$satisfying this equation, and these functions are also continuous at$x = 0$if we demand$f(0) = 1$... View answer 3 answers 8 votes 616 views Accepted answer 12 votes It means "the product over all pairs$i, j$where$i &lt; j$." For example, if$p(i)$is defined for$i = 1, 2, 3, then $$\prod_{i&lt;j} \;[\,p(j) - p(i)\;] = [ p(2) - p(1) ] [p(3) - p(1)] [p(3) - ... View answer 2 answers 2 votes 404 views 12 votes You just didn't go far enough. While (p-1)! + p is prime for p = 2, 3, 5, 7, and 11, it is composite for p = 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47. p = 53 is the next prime for ... View answer 2 answers 7 votes 241 views Accepted answer 9 votes It sounds like what you did was to write$$ \int_0^\infty \left( \sin(1/x) - \frac{\sin(\pi/x)}{\pi} \right) \,dx = \int_0^\infty \sin(1/x)\, dx - \int_0^\infty \frac{\sin(\pi/x)}{\pi} \,dx You ... View answer 6 answers 2 votes 60 views Accepted answer 8 votes The quick-and-dirty method: \begin{align*} f(x) = \frac{1 + a x^2}{1 + bx^2} &amp;= (1 + a x^2) \left( 1 - b x^2 + b^2 x^4 - b^3 x^6 + \cdots \right) \\&amp;= 1 - (b - a) x^2 + (b^2 - ab) x^4 - (b^3 - ... View answer 1 answers 6 votes 533 views 7 votes This is not a complete answer (yet), and I encourage other folks to build on it: A set of autonomous coupled linear first-order ODEs can always be written in the form y' = A y $$where y stands ... View answer 2 answers 4 votes 65 views 6 votes The &quot;soccer ball&quot; polyhedron is formally called a truncated icosahedron. According to that Wikipedia article, if the sides are of length a, then the circumscribing sphere has a radius of ... View answer 2 answers 3 votes 175 views 6 votes Counterexample: Let$$ A = \begin{bmatrix} 0 &amp; -1 \\ 0 &amp; 1 \end{bmatrix} \qquad B = \begin{bmatrix} 1 &amp; 0 \\ 0 &amp; 0 \end{bmatrix} $$Both matrices are idempotent. Then AB = 0 but ... View answer 3 answers 6 votes 187 views Accepted answer 6 votes Idea of counterexample: Construct an oscillating function whose "rate of oscillation" decreases as n increases. The idea is to make the differences between successive values smaller due to the ... View answer 3 answers 6 votes 199 views 6 votes You have proven that for any distance \delta, we can find two points at that distance whose colors match. What you haven't proven (and what is requested) is that we can always find two points at ... View answer 1 answers 5 votes 2k views Accepted answer 6 votes Maybe I'm missing something, but I'm not sure why it would be any more complicated than$$ \mathcal{P}(\vec{x};\vec{x_0},r) = \frac{1}{V_n} \Theta(r - ||\vec{x} - \vec{x_0}||) $$for the interior of a ... View answer 3 answers 3 votes 7k views 6 votes No. As an example, consider the vector field \vec{F} = y \hat{i} - x \hat{j} + \hat{k}, whose curl is \vec{\nabla} \times \vec{F} = 2 \hat{k}. This is obviously not perpendular to \vec{F} ... View answer 2 answers 8 votes 2k views 6 votes It does appear to be symmetric, though the proof I came up with requires the introduction of a covariant derivative operator. There may be another proof out there that doesn't require quite so much ... View answer 1 answers 1 votes 43 views Accepted answer 5 votes This is a mathematical technique known as taking the total differential of both sides of the previous equation. On the left-hand side, we have \mathrm{d}(r^2) = 2 r \, \mathrm{d}r, and similarly ... View answer 1 answers 5 votes 107 views Accepted answer 5 votes HINT: Note that$$ \begin{bmatrix}283 \\ 3136 \\ 6776 \\ 2464\end{bmatrix} - 56 \begin{bmatrix}5 \\ 56 \\ 121 \\ 44\end{bmatrix} = \begin{bmatrix} 3 \\ 0 \\ 0 \\ 0 \end{bmatrix} $$which allows you ... View answer 1 answers 1 votes 81 views Accepted answer 5 votes If we express our metric in polar coordinates, it becomes a bit easier to see what to do. Let's suppose that our desired metric is related to the Euclidean metric by a conformal transformation of the ... View answer 3 answers 5 votes 218 views 5 votes You can use the negative square root if you want; it all works out the same way in the end. If we denote the positive square root of x by \sqrt{x}, we can substitute t = -\sqrt{x} instead. We ... View answer 2 answers 0 votes 115 views Accepted answer 5 votes To do this formally, expand x(t) in a "power series at infinity", multiplied by an unknown power of t:$$ x(t) = t^\alpha \sum_{n=0}^\infty \frac{a_n}{t^n}. $$We can assume WLOG that a_0 \neq 0,... View answer 5 answers 28 votes 4k views 5 votes Another observation too long for a comment: It's not actually that good of an approximation if you look at the ratio of f_k(x) = \sin^k(x) to the value of the g_k(x) = \exp ( - k (x- \pi/2)^2) for ... View answer 1 answers 1 votes 109 views 5 votes No. Consider the sequence of functions defined by$$ f_n(x) = \frac{1}{n} \cos (n^2 x). $$on [0, 2\pi]. We have \max |f_n(x)| = \frac{1}{n}, but \max|f'_n(x)| = n. Thus, f_n \to 0 as n \... View answer 15 answers 13 votes 6k views 5 votes Just for fun: we note that since we have a "depressed cubic" of the form t^3 + pt + q = 0 (i.e., no quadratic term), the roots have a nice geometric interpretation:$$ t_k = 2 \sqrt{ -\frac{p}{3}} \... View answer 1 answers 3 votes 2k views Accepted answer 5 votes I'm a physics professor, and more mathematically inclined than your average physicist. Arnol'd is a beautiful textbook, but it is most emphatically not the one to learn Lagrangian mechanics from. (I ... View answer 2 answers 3 votes 2k views Accepted answer 5 votes If the divergence of a vector field is zero (\nabla \cdot \vec{v} = 0\$), then the flux of that vector field through any closed surface is zero. This is a consequence of the divergence theorem: for ...
The general relationship you need is Eqn. 10.22.63 in the Digital Library of Mathematical Functions: $$\int_{0}^{\infty}J_{\mu}\left(ax\right)J_{\mu-1}\left(bx\right)\mathrm{d}x=% \begin{cases}b^{\mu-... View answer 2 answers 4 votes 129 views Accepted answer 4 votes If I'm not mistaken, you've done your substitutions incorrectly. For the straight-line segment of the integral in the upper half-plane (call it \gamma_1), for example, the integral is$$ I_1 = \...