I found that looking at the original motivation of logarithms has been more elucidating than the way the topic is presented in grade-school. Thinking through the functional form that can solve the multiplication problem that Napier was facing (how to simplify multiplying large astronomical observations), f(ab) = f(a) + f(b), and why that leads to a unique family of functions, resonates a lot better with me for why lo…
The Lost Art of Logarithms
51–60 of 204 posts
Re: The Lost Art of Logarithms
#52Re: The Lost Art of Logarithms
#53Earlier quoted context omitted.
How long do I have to study math to understand this?
I love that you asked this, but I think it's not quite the right question. I've been wishing for years that someone would maintain a "dependency graph" for mathematical concepts. I think Khan Academy tried to do something like this at one point but took it down a long time ago. As long as we're far enough from the bleeding-edge of research topics, I feel like maths is the one field where this might be possible to do…
One source of information could be course syllabi, which describe a progression of topics and their prerequisites.
Rather than be the authority on what concepts must precede others, I envisioned making a system that could represent different educational approaches: not every educator agrees that Calculus I ought to be a prerequisite for studying Physics I. Not all bread recipes use yeast.
I had a hard time finding a good domain for this effort. “Tree of knowledge” dot TLD was taken.
Re: The Lost Art of Logarithms
#54[deleted]
Re: The Lost Art of Logarithms
#55Earlier quoted context omitted.
shouldn't it be var actualSize = 1 ?
Nope. I don’t know why the Java folks decided not to use the fairly standard verb “isolate” for this method, but that’s what it is[1]: > public static int highestOneBit(int i) > Returns an int value with at most a single one-bit, in the position of the highest-order ("leftmost") one-bit in the specified int value. Returns zero if the specified value has no one-bits in its two's complement binary representation, that…
Re: The Lost Art of Logarithms
#56Re: The Lost Art of Logarithms
#57I found that looking at the original motivation of logarithms has been more elucidating than the way the topic is presented in grade-school. Thinking through the functional form that can solve the multiplication problem that Napier was facing (how to simplify multiplying large astronomical observations), f(ab) = f(a) + f(b), and why that leads to a unique family of functions, resonates a lot better with me for why lo…
Re: The Lost Art of Logarithms
#58I feel frustrated that we cannot conceive of numbers like 10^80 (atoms in the universe) or 10^4000 (number configurations for a system with 4000 variables having 10 states each). Maybe there are superbrains out there in the universe that can do so.
I'm not sure you can even conceive a number like 1,000, if you're talking about holding an intuitive visual understanding in your mind at once.
Like, I can easily see 100 in my mind's eye as a 10x10 grid of circles. Even if I don't see each one clearly, I have a good sense of the 10 on each edge and the way it fills in. But ask me to imagine 10 of those side-by-side to make 1,000, and I don't think I can. Once I imagine the 10 groups, each one is just a square simplification, rather than any individual pieces within.
But I'm totally familiar with 1,000 as a concept I can multiply and divide with, and I can do math with 10^80 as well. And I can do so fairly "intuitively" as well -- it's just all the numbers up to 80 digits long. Even 4,000 digits fits on a single page of a book.
Re: The Lost Art of Logarithms
#59I feel frustrated that we cannot conceive of numbers like 10^80 (atoms in the universe) or 10^4000 (number configurations for a system with 4000 variables having 10 states each). Maybe there are superbrains out there in the universe that can do so.
Re: The Lost Art of Logarithms
#60I found that looking at the original motivation of logarithms has been more elucidating than the way the topic is presented in grade-school. Thinking through the functional form that can solve the multiplication problem that Napier was facing (how to simplify multiplying large astronomical observations), f(ab) = f(a) + f(b), and why that leads to a unique family of functions, resonates a lot better with me for why lo…
This follows directly from the fact that exp(x+y)=exp(x)exp(y).