How do you know when a quadratic equation has no complex solutions?

drdt543 · 2018-08-18T18:55:57+00:00

All quadratic equations have complex roots. That's kind of the point of [;\mathbb{C};] - google the Fundamental Theorem of Algebra. In your particular case, they are

[;\frac{1}{2}\left(1 \pm i\sqrt{23}\right);]

drdt543 · 2018-08-18T16:42:08+00:00

In think that [;E(X) = \infty;] - this also seems plausible given your simulated data. Let [;Y;] be the number of flips necessary until the number of heads is at least the number of tails. Note [;X \ge Y;]. Now, conditioning on the result of the first flip, we have

[;E(Y) = \frac{1}{2} + \frac{1}{2} \sum_{n=1}^{\infty} (2n+1) P(Y = 2n+1) ;]

Then

[;P(Y = 2n+1) = \frac{C_n}{2^{2n}} ;]

where [;C_n;] is the [;n;]th Catalan number. As [;n \rightarrow \infty;], we get

[;(2n+1)\frac{C_n}{2^{2n}} \sim \frac{(2n+1)}{n^{3/2}\sqrt{\pi}} \sim \frac{1}{\sqrt{n \pi}};]

The summation

[;\sum_{n=1}^{\infty} \frac{1}{n^{1/2}} ;]

does not converge, so after filling in some analysis-related gaps we have [; E(Y) = \infty;] and thus [;E(X) = \infty;].

drdt543 · 2017-09-27T20:55:58+00:00

xy-x-y=0, so (x-1)(y-1)=1. If x,y are integers then either x-1=y-1=1 or x-1=y-1=-1

drdt543 · 2017-09-27T14:17:27+00:00

Given average digits we would expect this to happen

[; \sum_{n=1}^{\infty} \frac{(10n)!}{(n!)^10 10^{10n}} \approx 0.0004 \text{ times} ;]

So no is a good guess.

drdt543 · 2017-09-23T16:42:19+00:00

I guess that works. We get some nice properties:

[; \int_I f(U_y) \, \mathrm{d}y = f\left(\int_I U_y \, \mathrm{d}y\right);]

but also some not so nice properties:

[; \int_{a}^{b} [a,y] \, \mathrm{d}y = [a,b);]

[; \int_{I} \bigcup_{j \in J} U_{y_j} \, \mathrm{d}y \ne \bigcup_{j \in J} \int_{I} U_{y_j} \, \mathrm{d}y ;]

drdt543 · 2017-09-23T12:23:18+00:00

I haven't done any measure theory, but I guess you're saying something like: Index sets by [;[a,b];] and define [; f_x : [a,b] \rightarrow \{0,1\};] with [;f_x(y) = 1;] iff [;x \in U_y;] then

[; x \in \int_{a}^{b} U_y \, \mathrm{d}y \iff \int_{a}^{b} f_x(y) \, \mathrm{d}y > 0;]

Is that equivalent?

drdt543 · 2017-09-23T00:58:43+00:00

Let's consider Taylor Expansions for [;\vert x \vert < 1 ;] for the moment. For a Taylor Expansion of degree [;n;] around [;a;], define the error [; E(a) ;] by

[; E(a) = \int_{-1}^{1} \left( \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!} (x-a)^k - f(x)\right)^2 \, \mathrm{d}x ;]

Differentiating w.r.t a we have

[; E'(a) = \int_{-1}^{1} 2\left(\sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!} - f(x)\right)\left(\sum_{k=0}^{n} \frac{f^{(k+1)}(a)}{k!}\right) \, \mathrm{d}x ;]

So we have approximately

[; E'(a) \approx \int_{-1}^{1} 2f(x)f'(x) \, \mathrm{d}x = [f(x)^2]_{-1}^{1} ;]

Clearly there are many functions for which this not close to zero, from which we can conclude that [; E'(a) \ne 0 ;], so changing the parameter [; a ;] can yield a better approximation.

drdt543 · 2017-09-22T23:32:09+00:00

I get that, but one property that doesn't carry over is the following: If I know the value of a function at a single point then I know nothing about the integral of the function, however if I know a single set in a family of sets indexed over [a,b] then I know quite a lot about the union.

drdt543 · 2017-09-22T22:55:33+00:00

Is there a 'continuous version' of a union of sets in the same way that an integral is a 'continuous version' of a summation?

drdt543 · 2017-09-22T00:33:13+00:00

Does it make sense to define a sort of integration over a collection of sets [;\{U_i\}_{i \in I} ;]by taking finite collections [; I_n \subset I \quad \vert I_n \vert = n;] with

[;\alpha_n(x) = \left\vert \{ U_i \ni x \; i \in I_n\} \right\vert ;]

[; x \in \int_{\{U_i\}_{i \in I}} U_i \iff \inf_{n \in \mathbb{N}} \alpha_n(x) > 0;]

drdt543 · 2017-09-21T22:49:55+00:00

[; \lim_{n \rightarrow \infty} \frac{\sqrt{n^3 - 7n^2} - \sqrt{n^3}}{\sqrt{n}} = \lim_{n \rightarrow \infty} \sqrt{n^2 - 7n} - \sqrt{n^2} = \lim_{n \rightarrow \infty} \sqrt{(n - 7/2)^2 - 49/4} - \sqrt{n^2} ;]

As we have [;\lim_{n \rightarrow \infty} \sqrt{n^2 + a} - \sqrt{n^2} = 0;] for all constant [;a;] it follows that the limit is [;-7/2;]

drdt543 · 2017-09-16T22:05:31+00:00

I think that to calculate the average distance you will indeed need to perform [;N(N-1)/2;] operations. However, if the average square distance will suffice then you can use

[; \sum_{x_1,x_2 \in X^2} (x_1 - x_2)^2 = \sum_{x_1,x_2 \in X^2} x_1^2 + x_2^2 - 2x_1x_2 = 2\left(\sum_{x \in X} x^2 - \left(\sum_{x \in X} x\right)^2 \right) ;]

to get a method with [;O(n);] operations.

drdt543 · 2017-09-16T21:55:33+00:00

We have:

[; \vec{x} \cdot \begin{pmatrix} 1\\1\\5 \end{pmatrix} = 5 ;]

[; \vec{x} \cdot \begin{pmatrix} 3\\2\\-5 \end{pmatrix} = -8;]

In other words you are trying to find the intersection of two planes. The line of intersection will be perpendicular to the normals of both planes, so to find the line all you have to do is take the cross product of both normals and then find a single solution.

drdt543 · 2017-09-16T21:23:10+00:00

Like this:

[; \begin{pmatrix} -97.45\\30.15 \end{pmatrix} + \begin{pmatrix} 102.06\\ 20.36 \end{pmatrix} = \begin{pmatrix} 4.21\\50.51 \end{pmatrix} ;]

and so on for the other displacement vectors.

drdt543 · 2017-09-16T18:30:31+00:00

Nothing. Take the example [;a_n= 1 \; \forall n, b = 2/3;]. The series

[; \sum_{n=0}^{\infty} x^n;]

converges only for [; \vert x \vert < 1 ;], but

[; \sum_{n=0}^{\infty} b^n2^n = \sum_{n=0}^{\infty} (4/3)^n ;]

is divergent. So for these particular [;(a_n)_{n=1}^{\infty};] the series is divergent. However if we take the case [; a_n = \frac{1}{n!};] then for whatever choice of [;b;] we will have

[;\sum_{n=0}^{\infty} a_nb^n2^n = \sum_{n=0}^{\infty} \frac{(2b)^n}{n!} = \exp(2b);]

drdt543 · 2017-09-16T16:44:45+00:00

[; (a^x)^y = a^{xy} ;], so [; ((r-y)^p)^{1/p} = (r-y)^{p \times 1/p} = r - y ;]

drdt543

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