cross-posted from: https://lemmy.world/post/45146291

Many of my clients are in a class that doesn’t teach them, and it’s sad because they are my favorite for chain rule and u-sub.

Last night an old idea came back to me, an idea about a function where all the derivatives start from zero and then grow smoothly. I thought it would be impossible, but then I found some interesting stuff on Wikipedia. So, I learned to use SymPy and wasted a lot of time with it. Here's a report of my (non-)findings.

(UPDATE: I did some numerical differentiation, which showed that h(x) does have negative derivatives. See details in this comment. A disappointment, although perhaps not a surprising one. It doesn't however, necessarily mean the goal is impossible.)

So, if anyone knows whether such a function exists and what it looks like, please tell me.

Sharing this as it's one of my favourite things on the web.

cross-posted from: https://lemmy.bestiver.se/post/852619

Comments

cross-posted from: https://lemmy.bestiver.se/post/847259

Comments

Research claiming that following the taught algorithm is more common for women (and those who agree with 'I want to please the teacher' more broadly), while trying to find shortcuts is more common from men. The latter seems correlated with performance on high stress math tests.

cross-posted from: https://lemmy.bestiver.se/post/774752

Comments

Statistical Tests to determine whether a bit-stream can be considered "random"

If we were to flip a coin 10 times, we would expect to see roughly 5 heads and 5 tails. Let's assign 00 to heads and 11 to tails. Therefore, we might see a sequence like this...

Backpropagation is one of the most important concepts in neural networks, however, it is challenging for learners to understand its concept because it is the most notation heavy part

cross-posted from: https://sh.itjust.works/post/48633930

Normally, we use a place-value system. This uses exponentials and multiplication.

1234
^^^^
||||
|||└ 4 * 10^0 = 4
||└ 3 * 10^1 = 30
|└ 2 * 10^2 = 200
1 * 10^3 = 1000

1000 + 200 + 30 + 4 = 1234

More generally, let d be the value of the digit, and n be the digit's position. So the value of the digit is d * 10^n^ if you're using base 10; or d * B^n^ where B is the base.

1234
^^^^
||||
|||└ d = 4, n = 0
||└ d = 3, n = 1
|└ d = 2, n = 2
d = 1, n = 3

---

What I came up with was a base system that was polynomial, and a system that was purely exponential, no multiplication.

In the polynomial system, each digit is d^n^. We will start n at 1.

polynomial:
1234
^^^^
||||
|||└ 4^1 = 4 in Place-Value Decimal (PVD)
||└ 3^2 = 9 PVD
|└ 2^3 = 8 PVD
1^4 = 1 PVD

1234 poly = 1 + 8 + 9 + 4 PVD = 22 PVD

This runs into some weird stuff, for example:

• Small digits in high positions can have a lower magnitude than large digits in low positions
• 1 in any place will always equal 1
• Numbers with differing digits being equal!

202 poly = 31 poly
PVD: 2^3 + 2^1 = 3^2 + 1^1
8 + 2 = 9 + 1 = 10

---

In the purely exponential system, each digit is n^d^. This is a bit more similar to place value, and it is kind of like a mixed-base system.

1234
^^^^
||||
|||└ 1^4 = 1
||└ 2^3 = 8
|└ 3^2 = 9
4^1 = 4

1234 exp = 4 + 8 + 9 + 1 PVD = 22 PVD

However it still runs into some of the same problems as the polynomial one.

• Small digits in high positions can have a lower magnitude than large digits in low positions (especially if the digit is 1)
• The digit in the ones place will always equal 1
• Numbers with differing digits can still be equal

200 exp = 31 exp
PVD: 3^2 = 2^3 + 1^1
9 = 8 + 1

---

So there you have it. Is it useful? Probably not. Is it interesting? Of course!

Crossposted from https://piefed.social/c/science/p/1406456/first-shape-found-that-cant-pass-through-itself

Nice accessible talk from the Simons Institute presenting one researcher's view on where math is and where it's going.

Summary: there is too much to keep up with and modern proofs are almost too big for individuals to grok. AI with automatic theorem provers is a promising immediate term path forward.

I live in a country that uses metric

I'm a bit nearsighted meaning distant objects appear blurry and I want to know how to work out mathematically how far I can see clearly without my glasses from my glass prescription to see If its worth getting prescription lenses for VR because I don't want to waste money if I don't need them

cross-posted from: https://lemmy.sdf.org/post/37414239

I've read the old papers proving that fact, but honestly it seems like some of the terminology and notation has changed since the 70's, and I roundly can't make heads or tails of it. The other sources I can find are in textbooks that I don't own.

Ideally, what I'm hoping for is a segment of pseudocode or some modern language that generates an n-character string from some kind of seed, which then cannot be recognised in linear time.

It's of interest to me just because, coming from other areas of math where inverting a bijective function is routine, it's highly unintuitive that you provably can't sometimes in complexity theory.