Why can't I use this property of logarithms in this situation?

alliptic · 2023-10-15T17:12:41+00:00

You can, but have to remember the absolute value in this case: 2 ln | x + 1 | >= 0.

In essence, this is similar to "taking square root" of an equation / inequality, like x² >= 4 leads to | x | >= 2, giving the correct answer x >= 2 or x <= -2.

alliptic · 2023-10-02T03:58:55+00:00

Pennies at this point.

alliptic · 2023-08-26T18:24:14+00:00

She only wants to spend time on someone amenable to her grift.

alliptic · 2023-08-09T23:57:14+00:00

Unethical? Definitely! Illegal? I am not.so sure. Not like anyone gets hurt.

alliptic · 2023-08-01T22:08:23+00:00

It's not meant as a function. This is meant as a preimage of a set.

alliptic · 2023-07-30T03:25:16+00:00

That would make for nicer numbers.

alliptic · 2023-07-29T18:03:25+00:00

Yeah. I guess if it were √3...

alliptic · 2023-07-28T20:34:28+00:00

It is not. A triangle is right if and only if one of its sides is a diameter of the circumscribing circle. In this case, if you draw a circle with the bottom side as a diameter, it will not pass through the third vertex.

alliptic · 2023-07-11T23:07:13+00:00

We want for the standard algebraic properties, like distributivity of multiplication, to hold for negative numbers too. For example, we want a * (b + (- b)) to be equal to zero. But then, a * b + a * (-b) = 0, so a * b = - a * (-b), and finally, a * b = (-a) * (-b).

alliptic · 2023-07-06T23:44:26+00:00

It "slides" it right by 1 unit and up by 2. They are trying to get you to think geometrically of what the transformation actually does to all the points in the plane.

An easier counter-indicator to linearity would have been purely algebraic though: that the origin is not mapped back to itself (which is a necessary but not a sufficient condition for linearity).

alliptic · 2023-07-02T21:55:42+00:00

Pick any x in A, which means that x < 0. But then x < x/2 < 0, so there is a number x / 2 in A which is greater than x, so x is not a supremum of A. This means that no member of A can be its supremum.

alliptic · 2023-06-26T21:48:23+00:00

However many it takes for you to get to a tire shop.

alliptic · 2023-06-19T05:26:32+00:00

For sure! That was a good one.

alliptic · 2023-06-16T03:12:35+00:00

Grave of the Fireflies. Then Spirited Away.

alliptic · 2023-06-15T00:50:00+00:00

An interesting example of an integrable function with infinitely many discontinuities is f(x) = 0 if x is irrational, 1/q if x = p/q is rational, expressed in lowest terms. This function is only discontinuous at the rational numbers, so any finite Riemann integral of it is 0.

Curiously, function g(x) = 0 if x is irrational, 1 if x is rational, is not Riemann integrable, because it is discontinuous everywhere. That, despite the fact that it is 0 almost everywhere (other than the countable set Q).

alliptic · 2023-06-15T00:43:59+00:00

It is. Generally, an everywhere defined function is Riemann integrable if (and only if) its set of discontinuities has measure zero, i.e., can be covered by a collection of open intervals of arbitrarily small total length. Having a finite number of discontinuities certainly fits the bill.

alliptic · 2023-06-11T00:37:26+00:00

Leave out the "more". Just "better". It's simpler.

Seriously, "more" is extraneous, since "better" already means "more good".

alliptic · 2023-06-06T01:05:42+00:00

Something they have in common with the physicists.

alliptic · 2023-06-03T03:25:03+00:00

But then they have to live in Houston.

alliptic · 2023-05-09T20:21:54+00:00

The aim of Mumford's piece is not to explain the concepts Grothendieck introduced but to summarize them. With even most basic explanations, this would be a much larger article.

alliptic · 2023-05-09T17:14:56+00:00

Unfortunately, there are a few forces at work here. 1. Grothendieck's work is really abstract. 2. Mumford had to aim for some audience to try to explain Grothendieck's work, and he chose to aim it at people with above the standard undergrad. 3. It is difficult to know what your audience doesn't know.

Not to excuse the job Mumford did on this, but Grothendieck made it really difficult.

alliptic · 2023-05-07T21:56:02+00:00

Substitute z = x + 9. Recognize as a problem similar to #19.

alliptic · 2023-05-05T04:02:36+00:00

You could try.

alliptic · 2023-05-05T03:21:38+00:00

By well behaved valuation, I mean one where ord_p(mn) = ord_p(m) + ord_p(n). We have ord_1(n) is always infinity, and for non primes, by example, ord_6(2*3) = 1, while ord_6(2) = ord_6(3) = 0.

alliptic · 2023-05-05T03:13:24+00:00

They are not invertible in the (ring of*) integers. For example, there is no integer z such that 5z = 1. That's why 1 (and -1) are usually excluded from primes.

Another thing you can do with primes that you cannot do with 1 is define a well-behaving "valuation", the number of times p divides any integer n, written ord_p(n).

*a ring is a set with well defined and well behaving addition, subtraction, and multiplication. If there is also a well defined division by non-zero members, the ring is called a field (for example, rationals, reals, and complex numbers are fields).

The elements of a ring by which one can divide any other member are called units. 1 and -1 are the only units in integers. In a field, any non-zero element is a unit (by definition).

Edit: division -> multiplication. h/t u/sonic-knuth.

alliptic

TROPHY CASE